Explanation:
Okay, so I need to find the largest possible area of a rectangular field that's enclosed in an athletic track. The track has a total length of 400 meters. The Asymptote code shows a rectangle with two semicircular ends, which makes me think it's like a running track. The problem says the track consists of two edges of the field and two semicircles. Wait, maybe the track is the entire perimeter? Let me visualize this.

The Asymptote code draws a rectangle with length 3 units and width 2 units, then adds semicircles on the left and right ends. So, the track is the rectangle plus the two semicircles. But the problem says the track consists of two edges of the field and two semicircles. Maybe the track is just the two straight sides of the rectangle and the two semicircles? So, the total length of the track is the sum of the two straight sides and the circumference of the two semicircles. Wait, the two semicircles would make a full circle. So, the total length of the track would be the perimeter of the rectangle (but only the two straight sides) plus the circumference of a circle with diameter equal to the width of the rectangle. Let me confirm.

Let me denote the length of the rectangle as L and the width as W. The track has two straight sides, each of length L, and two semicircular ends with diameter W. So, the total length of the track is 2L + 2*(π*(W/2)) = 2L + πW. Since the problem says the length of the track is 400 meters, that equation would be 2L + πW = 400.

But the question is about the area of the field, which is the rectangle. The area A = L * W. So, we need to maximize A = L * W given that 2L + πW = 400.

This seems like an optimization problem with a constraint. I can use calculus here. Let me express one variable in terms of the other using the constraint equation. Let me solve for L in terms of W.

From 2L + πW = 400, subtract πW: 2L = 400 - πW. Then divide by 2: L = (400 - πW)/2.

So, substitute this into the area formula: A = L * W = [(400 - πW)/2] * W = (400W - πW²)/2.

So, A(W) = (400W - πW²)/2. To find the maximum area, take derivative of A with respect to W, set it to zero.

First, simplify A(W): A(W) = 200W - (π/2)W².

Then, dA/dW = 200 - πW.

Set derivative equal to zero: 200 - πW = 0 => πW = 200 => W = 200/π.

Then, substitute back to find L: L = (400 - π*(200/π))/2 = (400 - 200)/2 = 200/2 = 100.

So, the maximum area is A = L * W = 100 * (200/π) = 20000/π.

Wait, but let me check if this is correct. Let me verify the steps again.

First, the total length of the track is 2L + πW = 400. Then, area is L*W. Express L in terms of W: L = (400 - πW)/2. Then area is [(400 - πW)/2] * W. Which is 200W - (π/2)W². Taking derivative: 200 - πW. Setting to zero gives W = 200/π. Then L = (400 - π*(200/π))/2 = (400 - 200)/2 = 100. Then area is 100*(200/π) = 20000/π.

But wait, 20000 divided by π is approximately 6366.2 square meters. But is this the maximum? Let me check if I did the derivative correctly.

Yes, derivative of 200W - (π/2)W² is 200 - πW. Correct. So critical point at W = 200/π.

To confirm it's a maximum, check second derivative. The second derivative of A with respect to W is -π, which is negative, so it's a maximum.

Therefore, the maximum area is 20000/π.

But let me think again. The problem says the track consists of two edges of the field and two semicircles. Wait, maybe I misinterpreted the track's length. Maybe the track is the entire perimeter of the field? But the Asymptote code shows that the track is the rectangle with semicircular ends. So, the total length of the track would be the perimeter of the rectangle plus the circumference of the circle? Wait, no. Wait, the track is the two straight sides and the two semicircles. So, the total length of the track is 2L + 2*(π*(W/2)) = 2L + πW. Which is what I had before.

Alternatively, if the track was the entire perimeter of the rectangle, that would be 2L + 2W, but since there are semicircular ends, maybe the track is designed such that when you run around it, you have the two straight sides and the two semicircular ends. So, the total length of the track is 2L + πW.

But let me check with standard athletic tracks. In reality, a standard running track has two straight sides and two semicircular ends. The length of one lap is usually 400 meters. The formula for the length is 2L + 2*(πr), where r is the radius of the semicircular ends. But since the diameter is equal to the width of the rectangle, which is 2r. So, if the width is W, then radius is W/2. Therefore, the length of the track is 2L + 2*(π*(W/2)) = 2L + πW. Which matches my previous equation.

Therefore, my initial setup is correct. Therefore, the maximum area is 20000/π.

But let me check if there's another way to think about this. Maybe using the concept that for a given perimeter, the maximum area is a circle. But here, the shape is a rectangle with semicircular ends. However, the constraint is not on the perimeter of the rectangle, but on the length of the track, which is different.

Alternatively, if we think of the track as a closed loop, the total length is 400 meters. But the field is the rectangle, so maybe the area is the rectangle.

Alternatively, maybe the problem is similar to optimizing the area of a rectangle with semicircular ends, given the total length of the track. Which is exactly what I did.

Alternatively, maybe I can use the method of Lagrange multipliers. Let me try that.

We need to maximize A = L*W, subject to the constraint 2L + πW = 400.

Set up the Lagrangian: L = L*W - λ(2L + πW - 400).

Taking partial derivatives:

∂L/∂L = W - 2λ = 0 => W = 2λ

∂L/∂W = L - πλ = 0 => L = πλ

∂L/∂λ = -(2L + πW - 400) = 0 => 2L + πW = 400

From the first equation, W = 2λ. From the second equation, L = πλ. Therefore, substituting into the third equation:

2*(πλ) + π*(2λ) = 400 => 2πλ + 2πλ = 400 => 4πλ = 400 => λ = 400/(4π) = 100/π

Therefore, W = 2λ = 200/π, and L = πλ = π*(100/π) = 100.

So, same result. Therefore, area is 100*(200/π) = 20000/π.

Therefore, the maximum area is 20000 divided by π.

But let me check if the problem says "the largest possible area of the field". The field is the rectangle, so the area is L*W. So, yes, that's what I calculated.

Alternatively, if the field was the entire track, but the problem says "a rectangular field is enclosed in an athletic track", so the field is the rectangle, and the track is around it. So, the area of the field is the rectangle.

Therefore, the answer is 20000/π square meters.

But let me check if there's a different interpretation. For example, maybe the track is the entire perimeter of the field, but the field is a rectangle with semicircular ends? Wait, no. The Asymptote code shows a rectangle with semicircular ends, so the field is the rectangle, and the track is the path around it.

Alternatively, maybe the track is the rectangle itself? But the Asymptote code shows that the track is drawn with the rectangle and the semicircles.

Alternatively, maybe the problem is similar to a rectangle with semicircular ends, and the total length of the track is 400 meters. So, the track is the perimeter of the rectangle plus the circumference of the circle? Wait, no. If you have a rectangle with semicircular ends, the total length of the track (the perimeter) would be 2L + 2*(π*(W/2)) = 2L + πW. Which is what I had.

Alternatively, if the track was a full circle, but no, the Asymptote code shows a rectangle.

Alternatively, maybe the problem is that the track is the rectangle, and the two semicircles are part of the track? Wait, the problem says "the track consists of two edges of the field, and two semicircles". So, the track is made up of two edges of the field (so the two straight sides of the rectangle) and two semicircles. Therefore, the total length of the track is 2L + πW = 400.

Therefore, my original interpretation is correct.

Therefore, the maximum area is 20000/π.

But let me compute that value numerically to check if it's reasonable.

π is approximately 3.1416, so 20000 / 3.1416 ≈ 6366.2 square meters.

Is that a reasonable maximum area? Let me think. If the track was a circle with circumference 400 meters, the radius would be 400/(2π) ≈ 63.66 meters, and area would be πr² ≈ π*(63.66)^2 ≈ 12732.4 square meters. But here, the field is a rectangle with semicircular ends. Wait, but the area of the field is only the rectangle, not the entire track.

Wait, if the track is 400 meters, and the field is the rectangle, then the area of the field is L*W. If we had a circular track, the area would be πr², but here, the field is only the rectangle. So, 6366 is less than 12732, which makes sense because the field is only the rectangle.

Alternatively, if we had a square field with semicircular ends, but that's not necessarily the case.

Alternatively, maybe there's a different way to maximize the area. Let me think.

Suppose we consider the track as a closed loop, which is a rectangle with semicircular ends. The total length is 400 meters. The area of the field is the area of the rectangle.

Alternatively, if we think of the entire track as a shape, but the problem says the field is enclosed in the track. So, the field is the rectangle, and the track is around it.

Alternatively, maybe the problem is similar to optimizing the area of a rectangle with a fixed perimeter, but here the perimeter is not fixed, but the length of the track is fixed.

But in our case, the track's length is fixed, and we need to maximize the area of the rectangle.

Alternatively, maybe I can think of the problem as follows: the track is a closed loop with length 400 meters, which is made up of two straight sides of length L and two semicircular sides with diameter W. Therefore, the total length is 2L + πW = 400. The area of the field is the area of the rectangle, which is L*W.

Therefore, the problem reduces to maximizing L*W with 2L + πW = 400. Which is exactly what I did.

Therefore, the answer is 20000/π.

But let me check if there's a different approach. For example, using the AM-GM inequality.

We have 2L + πW = 400. Let me write this as 2L + πW = 400.

We need to maximize L*W.

Let me denote variables. Let me set variables such that I can apply AM-GM.

But AM-GM is for positive numbers. Let me think.

Let me consider 2L and πW as two terms.

But the product is L*W.

Alternatively, express the product in terms of 2L and πW.

Let me set x = 2L and y = πW. Then, x + y = 400.

We need to maximize (x/2)*(y/π) = (x*y)/(2π).

Therefore, maximizing x*y given that x + y = 400.

The maximum of x*y occurs when x = y = 200.

Therefore, maximum x*y = 200*200 = 40000.

Therefore, maximum area is 40000/(2π) = 20000/π.

Same result!

Therefore, this confirms the answer.

Therefore, the largest possible area of the field is 20000 divided by π square meters.

So, the answer is \boxed{\dfrac{20000}{\pi}}.

**Final Answer**
The largest possible area of the field is \boxed{\dfrac{20000}{\pi}} square meters.

To determine the **largest possible area of the rectangular field** enclosed in an athletic track, we begin by analyzing the structure of the track and the given constraint.

---

### **Understanding the Track Structure**

The track consists of:

- **Two straight sides** of the rectangle, each of length $ L $,
- **Two semicircular ends**, each with diameter $ W $ (the width of the rectangle).

The **total length of the track** is the sum of the two straight sides and the circumference of the full circle formed by the two semicircles:

$$
\text{Total length} = 2L + \pi W = 400 \text{ meters}
$$

---

### **Objective: Maximize the Area of the Field**

The **area of the rectangular field** is:

$$
A = L \cdot W
$$

We are to **maximize** this area under the constraint:

$$
2L + \pi W = 400
$$

---

### **Step 1: Express One Variable in Terms of the Other**

From the constraint:

$$
2L = 400 - \pi W \Rightarrow L = \frac{400 - \pi W}{2}
$$

Substitute into the area formula:

$$
A = L \cdot W = \left( \frac{400 - \pi W}{2} \right) \cdot W = \frac{400W - \pi W^2}{2}
$$

---

### **Step 2: Maximize the Area Function**

Let’s define the area as a function of $ W $:

$$
A(W) = \frac{400W - \pi W^2}{2}
$$

Take the derivative with respect to $ W $:

$$
\frac{dA}{dW} = \frac{400 - 2\pi W}{2} = 200 - \pi W
$$

Set the derivative equal to zero to find the critical point:

$$
200 - \pi W = 0 \Rightarrow W = \frac{200}{\pi}
$$

---

### **Step 3: Find the Corresponding $ L $ and Compute the Area**

Substitute $ W = \frac{200}{\pi} $ into the expression for $ L $:

$$
L = \frac{400 - \pi \cdot \frac{200}{\pi}}{2} = \frac{400 - 200}{2} = 100
$$

Now compute the area:

$$
A = L \cdot W = 100 \cdot \frac{200}{\pi} = \frac{20000}{\pi}
$$

---

### **Step 4: Confirm Maximum Using Second Derivative**

$$
\frac{d^2A}{dW^2} = -\pi < 0
$$

This confirms that the critical point is a **maximum**.

---

### **Final Answer**

$$
\boxed{\dfrac{20000}{\pi}}
$$

This is the **largest possible area** of the rectangular field, in square meters.

[Assistant]
Explanation:
Okay, so I need to find the largest possible area of a rectangular field that's enclosed in an athletic track. The track has a total length of 400 meters. The Asymptote code shows a rectangle with two semicircular ends, which makes me think it's like a running track. The problem says the track consists of two edges of the field and two semicircles. Wait, maybe the track is the entire perimeter? Let me visualize this.

The Asymptote code draws a rectangle with length 3 units and width 2 units, then adds semicircles on the left and right ends. So, the track is the rectangle plus the two semicircles. But the problem says the track consists of two edges of the field and two semicircles. Maybe the track is just the two straight sides of the rectangle and the two semicircles? So, the total length of the track is the sum of the two straight sides and the circumference of the two semicircles. Wait, the two semicircles would make a full circle. So, the total length of the track would be the perimeter of the rectangle (but only the two straight sides) plus the circumference of a circle with diameter equal to the width of the rectangle. Let me confirm.

Let me denote the length of the rectangle as L and the width as W. The track has two straight sides, each of length L, and two semicircular ends with diameter W. So, the total length of the track is 2L + 2*(π*(W/2)) = 2L + πW. Since the problem says the length of the track is 400 meters, that equation would be 2L + πW = 400.

But the question is about the area of the field, which is the rectangle. The area A = L * W. So, we need to maximize A = L * W given that 2L + πW = 400.

This seems like an optimization problem with a constraint. I can use calculus here. Let me express one variable in terms of the other using the constraint equation. Let me solve for L in terms of W.

From 2L + πW = 400, subtract πW: 2L = 400 - πW. Then divide by 2: L = (400 - πW)/2.

So, substitute this into the area formula: A = L * W = [(400 - πW)/2] * W = (400W - πW²)/2.

So, A(W) = (400W - πW²)/2. To find the maximum area, take derivative of A with respect to W, set it to zero.

First, simplify A(W): A(W) = 200W - (π/2)W².

Then, dA/dW = 200 - πW.

Set derivative equal to zero: 200 - πW = 0 => πW = 200 => W = 200/π.

Then, substitute back to find L: L = (400 - π*(200/π))/2 = (400 - 200)/2 = 200/2 = 100.

So, the maximum area is A = L * W = 100 * (200/π) = 20000/π.

Wait, but let me check if this is correct. Let me verify the steps again.

First, the total length of the track is 2L + πW = 400. Then, area is L*W. Express L in terms of W: L = (400 - πW)/2. Then area is [(400 - πW)/2] * W. Which is 200W - (π/2)W². Taking derivative: 200 - πW. Setting to zero gives W = 200/π. Then L = (400 - π*(200/π))/2 = (400 - 200)/2 = 100. Then area is 100*(200/π) = 20000/π.

But wait, 20000 divided by π is approximately 6366.2 square meters. But is this the maximum? Let me check if I did the derivative correctly.

Yes, derivative of 200W - (π/2)W² is 200 - πW. Correct. So critical point at W = 200/π.

To confirm it's a maximum, check second derivative. The second derivative of A with respect to W is -π, which is negative, so it's a maximum.

Therefore, the maximum area is 20000/π.

But let me think again. The problem says the track consists of two edges of the field and two semicircles. Wait, maybe I misinterpreted the track's length. Maybe the track is the entire perimeter of the field? But the Asymptote code shows that the track is the rectangle with semicircular ends. So, the total length of the track would be the perimeter of the rectangle plus the circumference of the circle? Wait, no. Wait, the track is the two straight sides and the two semicircles. So, the total length of the track is 2L + 2*(π*(W/2)) = 2L + πW. Which is what I had before.

Alternatively, if the track was the entire perimeter of the rectangle, that would be 2L + 2W, but since there are semicircular ends, maybe the track is designed such that when you run around it, you have the two straight sides and the two semicircular ends. So, the total length of the track is 2L + πW.

But let me check with standard athletic tracks. In reality, a standard running track has two straight sides and two semicircular ends. The length of one lap is usually 400 meters. The formula for the length is 2L + 2*(πr), where r is the radius of the semicircular ends. But since the diameter is equal to the width of the rectangle, which is 2r. So, if the width is W, then radius is W/2. Therefore, the length of the track is 2L + 2*(π*(W/2)) = 2L + πW. Which matches my previous equation.

Therefore, my initial setup is correct. Therefore, the maximum area is 20000/π.

But let me check if there's another way to think about this. Maybe using the concept that for a given perimeter, the maximum area is a circle. But here, the shape is a rectangle with semicircular ends. However, the constraint is not on the perimeter of the rectangle, but on the length of the track, which is different.

Alternatively, if we think of the track as a closed loop, the total length is 400 meters. But the field is the rectangle, so maybe the area is the rectangle.

Alternatively, maybe the problem is similar to optimizing the area of a rectangle with semicircular ends, given the total length of the track. Which is exactly what I did.

Alternatively, maybe I can use the method of Lagrange multipliers. Let me try that.

We need to maximize A = L*W, subject to the constraint 2L + πW = 400.

Set up the Lagrangian: L = L*W - λ(2L + πW - 400).

Taking partial derivatives:

∂L/∂L = W - 2λ = 0 => W = 2λ

∂L/∂W = L - πλ = 0 => L = πλ

∂L/∂λ = -(2L + πW - 400) = 0 => 2L + πW = 400

From the first equation, W = 2λ. From the second equation, L = πλ. Therefore, substituting into the third equation:

2*(πλ) + π*(2λ) = 400 => 2πλ + 2πλ = 400 => 4πλ = 400 => λ = 400/(4π) = 100/π

Therefore, W = 2λ = 200/π, and L = πλ = π*(100/π) = 100.

So, same result. Therefore, area is 100*(200/π) = 20000/π.

Therefore, the maximum area is 20000 divided by π.

But let me check if the problem says "the largest possible area of the field". The field is the rectangle, so the area is L*W. So, yes, that's what I calculated.

Alternatively, if the field was the entire track, but the problem says "a rectangular field is enclosed in an athletic track", so the field is the rectangle, and the track is around it. So, the area of the field is the rectangle.

Therefore, the answer is 20000/π square meters.

But let me check if there's a different interpretation. For example, maybe the track is the entire perimeter of the field, but the field is a rectangle with semicircular ends? Wait, no. The Asymptote code shows a rectangle with semicircular ends, so the field is the rectangle, and the track is the path around it.

Alternatively, maybe the track is the rectangle itself? But the Asymptote code shows that the track is drawn with the rectangle and the semicircles.

Alternatively, maybe the problem is similar to a rectangle with semicircular ends, and the total length of the track is 400 meters. So, the track is the perimeter of the rectangle plus the circumference of the circle? Wait, no. If you have a rectangle with semicircular ends, the total length of the track (the perimeter) would be 2L + 2*(π*(W/2)) = 2L + πW. Which is what I had.

Alternatively, if the track was a full circle, but no, the Asymptote code shows a rectangle.

Alternatively, maybe the problem is that the track is the rectangle, and the two semicircles are part of the track? Wait, the problem says "the track consists of two edges of the field, and two semicircles". So, the track is made up of two edges of the field (so the two straight sides of the rectangle) and two semicircles. Therefore, the total length of the track is 2L + πW = 400.

Therefore, my original interpretation is correct.

Therefore, the maximum area is 20000/π.

But let me compute that value numerically to check if it's reasonable.

π is approximately 3.1416, so 20000 / 3.1416 ≈ 6366.2 square meters.

Is that a reasonable maximum area? Let me think. If the track was a circle with circumference 400 meters, the radius would be 400/(2π) ≈ 63.66 meters, and area would be πr² ≈ π*(63.66)^2 ≈ 12732.4 square meters. But here, the field is a rectangle with semicircular ends. Wait, but the area of the field is only the rectangle, not the entire track.

Wait, if the track is 400 meters, and the field is the rectangle, then the area of the field is L*W. If we had a circular track, the area would be πr², but here, the field is only the rectangle. So, 6366 is less than 12732, which makes sense because the field is only the rectangle.

Alternatively, if we had a square field with semicircular ends, but that's not necessarily the case.

Alternatively, maybe there's a different way to maximize the area. Let me think.

Suppose we consider the track as a closed loop, which is a rectangle with semicircular ends. The total length is 400 meters. The area of the field is the area of the rectangle.

Alternatively, if we think of the entire track as a shape, but the problem says the field is enclosed in the track. So, the field is the rectangle, and the track is around it.

Alternatively, maybe the problem is similar to optimizing the area of a rectangle with a fixed perimeter, but here the perimeter is not fixed, but the length of the track is fixed.

But in our case, the track's length is fixed, and we need to maximize the area of the rectangle.

Alternatively, maybe I can think of the problem as follows: the track is a closed loop with length 400 meters, which is made up of two straight sides of length L and two semicircular sides with diameter W. Therefore, the total length is 2L + πW = 400. The area of the field is the area of the rectangle, which is L*W.

Therefore, the problem reduces to maximizing L*W with 2L + πW = 400. Which is exactly what I did.

Therefore, the answer is 20000/π.

But let me check if there's a different approach. For example, using the AM-GM inequality.

We have 2L + πW = 400. Let me write this as 2L + πW = 400.

We need to maximize L*W.

Let me denote variables. Let me set variables such that I can apply AM-GM.

But AM-GM is for positive numbers. Let me think.

Let me consider 2L and πW as two terms.

But the product is L*W.

Alternatively, express the product in terms of 2L and πW.

Let me set x = 2L and y = πW. Then, x + y = 400.

We need to maximize (x/2)*(y/π) = (x*y)/(2π).

Therefore, maximizing x*y given that x + y = 400.

The maximum of x*y occurs when x = y = 200.

Therefore, maximum x*y = 200*200 = 40000.

Therefore, maximum area is 40000/(2π) = 20000/π.

Same result!

Therefore, this confirms the answer.

Therefore, the largest possible area of the field is 20000 divided by π square meters.

So, the answer is \boxed{\dfrac{20000}{\pi}}.

**Final Answer**
The largest possible area of the field is \boxed{\dfrac{20000}{\pi}} square meters.

To determine the **largest possible area of the rectangular field** enclosed in an athletic track, we begin by analyzing the structure of the track and the given constraint.

---

### **Understanding the Track Structure**

The track consists of:

- **Two straight sides** of the rectangle, each of length $ L $,
- **Two semicircular ends**, each with diameter $ W $ (the width of the rectangle).

The **total length of the track** is the sum of the two straight sides and the circumference of the full circle formed by the two semicircles:

$$
\text{Total length} = 2L + \pi W = 400 \text{ meters}
$$

---

### **Objective: Maximize the Area of the Field**

The **area of the rectangular field** is:

$$
A = L \cdot W
$$

We are to **maximize** this area under the constraint:

$$
2L + \pi W = 400
$$

---

### **Step 1: Express One Variable in Terms of the Other**

From the constraint:

$$
2L = 400 - \pi W \Rightarrow L = \frac{400 - \pi W}{2}
$$

Substitute into the area formula:

$$
A = L \cdot W = \left( \frac{400 - \pi W}{2} \right) \cdot W = \frac{400W - \pi W^2}{2}
$$

---

### **Step 2: Maximize the Area Function**

Let’s define the area as a function of $ W $:

$$
A(W) = \frac{400W - \pi W^2}{2}
$$

Take the derivative with respect to $ W $:

$$
\frac{dA}{dW} = \frac{400 - 2\pi W}{2} = 200 - \pi W
$$

Set the derivative equal to zero to find the critical point:

$$
200 - \pi W = 0 \Rightarrow W = \frac{200}{\pi}
$$

---

### **Step 3: Find the Corresponding $ L $ and Compute the Area**

Substitute $ W = \frac{200}{\pi} $ into the expression for $ L $:

$$
L = \frac{400 - \pi \cdot \frac{200}{\pi}}{2} = \frac{400 - 200}{2} = 100
$$

Now compute the area:

$$
A = L \cdot W = 100 \cdot \frac{200}{\pi} = \frac{20000}{\pi}
$$

---

### **Step 4: Confirm Maximum Using Second Derivative**

$$
\frac{d^2A}{dW^2} = -\pi < 0
$$

This confirms that the critical point is a **maximum**.

---

### **Final Answer**

$$
\boxed{\dfrac{20000}{\pi}}
$$

This is the **largest possible area** of the rectangular field, in square meters.

Solution:
Let the width of the rectangle be $w,$ and let the radius of each semicircle be $r.$

[asy]
unitsize(1 cm);

filldraw((0,0)--(3,0)--(3,2)--(0,2)--cycle,lightgreen);
draw((0,0)--(3,0),linewidth(2*bp));
draw((0,2)--(3,2),linewidth(2*bp));
draw(arc((3,1),1,-90,90),linewidth(2*bp));
draw(arc((0,1),1,90,270),linewidth(2*bp));

label("$w$", (1.5,0), S);
label("$r$", (3,1/2), E);
dot((3,1));
[/asy]

Then the length of the track is $2w + 2 \pi r = 400,$ so $w + \pi r = 200.$  By AM-GM,
\[200 = w + \pi r \ge 2 \sqrt{w \pi r},\]so $\sqrt{w \pi r} \le 100.$  Then $w \pi r \le 10000,$ so
\[wr \le \frac{10000}{\pi}.\]Then the area of the field, $2wr,$ must satisfy
\[2wr \le \frac{20000}{\pi}.\]Equality occurs when $w = 100$ and $r = \frac{100}{\pi},$ so the largest possible area is $\boxed{\frac{20000}{\pi}}.$


Explanation:
Okay, let's see. Kathy has $300 in the bank and she wants to withdraw half of it in British pounds and half in euros. Then we need to find out how many more euros than pounds she will have. Hmm, okay. Let me break this down step by step.

First, she wants to withdraw half of her $300 in pounds and half in euros. So, half of $300 is $150. That means she wants to convert $150 into pounds and $150 into euros. Wait, but the question says "withdraw half of it in British pounds and half of it in euros." So does that mean she's withdrawing half the amount in pounds and half in euros? Or is it half the amount in terms of the currency? Wait, maybe I need to clarify.

Wait, the problem says she has 300 USD in the bank. She wants to withdraw half of it in British pounds and half of it in euros. So, does that mean she's withdrawing half of the total amount (which is $150) in pounds and the other half ($150) in euros? But how does that work? Because if she's withdrawing in different currencies, the amount in pounds and euros would depend on the exchange rates.

Wait, maybe she wants to convert half of her USD into pounds and half into euros. So, half of $300 is $150. So she converts $150 to pounds and $150 to euros. Then, we need to find out how much that is in pounds and euros, and then find the difference between euros and pounds.

Yes, that seems to make sense. So let me confirm. The problem says "withdraw half of it in British pounds and half of it in euros." So "it" refers to the 300 USD. So half of it is $150. So she's converting $150 to pounds and $150 to euros. Then, the question is how many more euros than pounds will she have. So, after converting, we need to calculate the amount in euros and pounds, then subtract pounds from euros to find the difference.

Okay, so let's start with converting $150 to pounds. The exchange rate is 1 pound = 1.64 USD. So, to find out how many pounds she gets for $150, we divide the USD amount by the exchange rate. So pounds = USD / 1.64. So that would be 150 / 1.64.

Similarly, for euros, the exchange rate is 1 euro = 1.32 USD. So euros = USD / 1.32. So that would be 150 / 1.32.

Once we have both amounts, we subtract pounds from euros to find the difference. Then round to the nearest whole number.

Let me compute these values step by step.

First, converting $150 to pounds:

150 divided by 1.64. Let me calculate that. Let me get a calculator or do it manually. Let me try to compute 150 / 1.64.

Well, 1.64 times 90 is 147.6, because 1.64 * 90 = 1.64 * 9 * 10 = 14.76 * 10 = 147.6. Then 150 - 147.6 = 2.4. So 2.4 / 1.64 is approximately 1.463. So total pounds would be approximately 90 + 1.463 = 91.463 pounds. Let me check with a calculator approach.

Alternatively, 1.64 * 91 = 1.64 * 90 + 1.64 * 1 = 147.6 + 1.64 = 149.24. Then 150 - 149.24 = 0.76. So 0.76 / 1.64 ≈ 0.463. So total pounds ≈ 91.463. So approximately 91.46 pounds. Let me check with decimal division.

Alternatively, 150 / 1.64. Let me write this as 150 ÷ 1.64. Multiply numerator and denominator by 100 to eliminate decimals: 15000 ÷ 164. Let me compute that.

Divide 15000 by 164. Let me see how many times 164 goes into 15000.

First, 164 * 90 = 14760. Subtract that from 15000: 15000 - 14760 = 240. Then, 164 goes into 240 once (164), remainder 76. Then bring down a zero: 760. 164 * 4 = 656. Subtract from 760: 104. Bring down a zero: 1040. 164 * 6 = 984. Subtract: 1040 - 984 = 56. Bring down a zero: 560. 164 * 3 = 492. Subtract: 560 - 492 = 68. Bring down a zero: 680. 164 * 4 = 656. Subtract: 680 - 656 = 24. So so far, we have 91.463... So approximately 91.46 pounds. So rounding to two decimal places, 91.46 pounds. But since we need to round to the nearest whole number at the end, maybe we can keep more decimals for accuracy.

Alternatively, maybe I can use calculator-like steps. Let me check with approximate decimal values.

Alternatively, maybe I can use fractions. But perhaps it's easier to just proceed with decimal approximations.

So pounds ≈ 91.463.

Now, converting $150 to euros. The exchange rate is 1 euro = 1.32 USD. So euros = 150 / 1.32.

Again, let me compute that. Let me do 150 divided by 1.32.

First, note that 1.32 * 100 = 132. So 150 / 1.32 = (150 * 100) / 132 = 15000 / 132. Let me compute that.

Divide 15000 by 132. Let me see how many times 132 goes into 15000.

First, 132 * 100 = 13200. Subtract that from 15000: 15000 - 13200 = 1800.

Now, 132 * 13 = 1716. Subtract that from 1800: 1800 - 1716 = 84.

Bring down a zero: 840. 132 * 6 = 792. Subtract: 840 - 792 = 48.

Bring down a zero: 480. 132 * 3 = 396. Subtract: 480 - 396 = 84.

Bring down a zero: 840. Wait, this is repeating. So the decimal is 113.6363..., because 100 + 13.6363... So total is 113.6363... euros.

Wait, let me check again. Wait, 132 * 113 = ?

Wait, maybe I should do this step by step. Let me try again.

Original division: 15000 / 132.

First, divide numerator and denominator by 12 to simplify? Let me see. 132 divided by 12 is 11. 15000 divided by 12 is 1250. So now it's 1250 / 11. Which is approximately 113.6363... So yes, that's correct. So 1250 divided by 11 is 113.6363...

So euros ≈ 113.6363...

So approximately 113.64 euros.

So now, we have approximately 91.46 pounds and 113.64 euros.

The question asks how many more euros than pounds she will have. So we need to compute euros - pounds.

So 113.64 - 91.46 = ?

Let me compute that. 113.64 - 91.46. Subtract the whole numbers: 113 - 91 = 22. Subtract the decimals: 0.64 - 0.46 = 0.18. So total is 22.18. So approximately 22.18 euros more.

But we need to round to the nearest whole number. So 22.18 is approximately 22. So the answer would be 22.

Wait, but let me check if my approximated values for pounds and euros are accurate enough. Because if I rounded too early, maybe there's a slight error.

Let me recast the problem with more precise numbers.

First, pounds: 150 / 1.64.

Let me compute 150 / 1.64 exactly. Let me use decimal division.

1.64 × 91 = 1.64 × 90 + 1.64 × 1 = 147.6 + 1.64 = 149.24

Then, 150 - 149.24 = 0.76

So, 0.76 / 1.64 = 0.463...

So total pounds is 91.463...

Similarly, euros: 150 / 1.32 = 113.636...

So euros - pounds = 113.636... - 91.463... = ?

Let me compute that:

113.636... - 91.463... = (113 - 91) + (0.636... - 0.463...) = 22 + 0.173... ≈ 22.173...

So approximately 22.173, which rounds to 22 when rounded to the nearest whole number.

But let me check if there's a more precise way. Maybe I should carry out the exact calculation with more decimal places.

Alternatively, perhaps I can compute the exact difference without rounding intermediate steps.

Let me try that.

Let me compute euros - pounds = (150 / 1.32) - (150 / 1.64)

Factor out 150: 150*(1/1.32 - 1/1.64)

Compute 1/1.32 - 1/1.64.

First, compute 1/1.32. Let me calculate that.

1 / 1.32 ≈ 0.757575...

Similarly, 1 / 1.64 ≈ 0.609756...

So subtracting them: 0.757575... - 0.609756... ≈ 0.147819...

Then multiply by 150: 150 * 0.147819 ≈ ?

Compute 150 * 0.147819:

First, 100 * 0.147819 = 14.7819

50 * 0.147819 = 7.39095

Adding together: 14.7819 + 7.39095 = 22.17285

So approximately 22.17285, which is about 22.17, so rounds to 22. So that's consistent with the previous calculation.

Therefore, the answer is 22.

But let me check once more with exact fractions to see if there's any discrepancy.

Alternatively, let me compute 1/1.32 - 1/1.64.

First, convert 1.32 and 1.64 into fractions.

1.32 is 132/100 = 33/25.

Similarly, 1.64 is 164/100 = 41/25. Wait, 164 divided by 4 is 41, 100 divided by 4 is 25. Yes, so 1.64 = 41/25.

Therefore, 1/1.32 = 25/33, and 1/1.64 = 25/41.

Therefore, the difference is 25/33 - 25/41 = 25*(1/33 - 1/41) = 25*( (41 - 33)/(33*41) ) = 25*(8/(1353)) = 200 / 1353 ≈ ?

Compute 200 divided by 1353.

Well, 1353 × 0.147 ≈ 1353 × 0.1 + 1353 × 0.04 + 1353 × 0.007

= 135.3 + 54.12 + 9.471 ≈ 135.3 + 54.12 = 189.42 + 9.471 ≈ 198.891

Which is very close to 200. So 0.147 gives approximately 198.89, so 200 / 1353 ≈ 0.1478...

Therefore, multiplying by 150:

150 * (200 / 1353) = (150 * 200) / 1353 = 30000 / 1353 ≈ ?

Divide 30000 by 1353.

Well, 1353 × 22 = ?

1353 × 20 = 27060

1353 × 2 = 2706

Total: 27060 + 2706 = 29766

Subtract that from 30000: 30000 - 29766 = 234

So 22 + 234 / 1353 ≈ 22 + 0.173 ≈ 22.173, which is the same as before. So 22.173, which rounds to 22.

Therefore, the answer is 22.

But let me check once again with exact decimal calculations.

Alternatively, maybe I can use calculator-like computations for the initial conversions.

For pounds: 150 / 1.64.

Let me compute 1.64 × 91 = 149.24 as before. Then 150 - 149.24 = 0.76. So 0.76 / 1.64.

Let me compute 0.76 / 1.64. Multiply numerator and denominator by 100: 76 / 164. Simplify by dividing numerator and denominator by 4: 19 / 41 ≈ 0.4634...

So total pounds is 91 + 0.4634 ≈ 91.4634.

For euros: 150 / 1.32. Let me compute 1.32 × 113 = ?

1.32 × 100 = 132

1.32 × 13 = 17.16

Total: 132 + 17.16 = 149.16

Subtract from 150: 150 - 149.16 = 0.84

So 0.84 / 1.32 = 0.6363...

So total euros is 113 + 0.6363 ≈ 113.6363.

Therefore, euros - pounds = 113.6363 - 91.4634 = ?

Let me compute 113.6363 - 91.4634:

Subtract 91 from 113: 22.

Subtract 0.4634 from 0.6363: 0.6363 - 0.4634 = 0.1729.

So total difference is 22 + 0.1729 ≈ 22.1729, which is approximately 22.17, rounds to 22.

Therefore, regardless of the method, the answer is 22.

But let me check once more with another approach. Maybe using exact decimal values.

Alternatively, maybe I can use the exact values of the exchange rates and compute the difference.

But I think all methods lead to approximately 22.17, which rounds to 22. Therefore, the answer is 22.

But just to be thorough, let me check if there's any chance that the question is interpreted differently. For example, maybe "withdraw half of it in British pounds and half of it in euros" could mean that she withdraws half the amount in pounds and half in euros, but considering the exchange rates? Wait, but that would be a different interpretation. Let me think.

Suppose she wants to withdraw half of her money in pounds and half in euros. But since the currencies have different values, how would that work? For example, if she wants to withdraw half in pounds and half in euros, but the total amount in USD would be different? Wait, but the problem says she has 300 USD in the bank. So if she wants to withdraw half of it in pounds and half in euros, that would mean she is converting half of the USD into pounds and half into euros. Which is the same as the original interpretation. So I think that's the correct way.

Alternatively, if she wanted to withdraw half the amount in pounds and half in euros, but in terms of the value in USD? For example, if she wants to have half the value in pounds and half in euros, but that would be different. But the problem says "withdraw half of it in British pounds and half of it in euros". Since "it" refers to the 300 USD, so half of it is $150. So converting $150 to pounds and $150 to euros. So that's the correct interpretation.

Therefore, the answer is 22.

But just to make sure, let me check with exact numbers using calculator-like precision.

Let me compute 150 / 1.64:

Using a calculator, 150 divided by 1.64 equals approximately 91.463414634...

Similarly, 150 divided by 1.32 equals approximately 113.636363636...

Subtracting these two: 113.636363636... - 91.463414634... = ?

Let me compute 113.636363636 - 91.463414634:

First, subtract the whole numbers: 113 - 91 = 22.

Then subtract the decimals: 0.636363636 - 0.463414634.

Compute 0.636363636 - 0.463414634:

0.636363636 - 0.463414634 = 0.172949002.

So total difference is 22 + 0.172949002 ≈ 22.172949002.

Which is approximately 22.1729, so when rounded to the nearest whole number, it's 22, since 0.1729 is less than 0.5.

Therefore, the answer is 22.

I think that's solid. All methods converge to 22.17, which rounds to 22. Therefore, the answer is 22.

**Final Answer**
\boxed{22}

To determine how many more euros than pounds Kathy will have after withdrawing half of her $300 in British pounds and the other half in euros, we follow a step-by-step process using the given exchange rates:

---

### **Step 1: Determine the amount to be converted**

Kathy has **$300** in USD. She wants to withdraw **half** of it in **British pounds (GBP)** and **half** in **euros (EUR)**.

- Half of $300 = $150
- So, she will convert **$150 to GBP** and **$150 to EUR**

---

### **Step 2: Convert USD to GBP**

Given:
- 1 GBP = 1.64 USD
- So, to find how many GBP she gets for $150:

$$
\text{GBP} = \frac{150}{1.64} \approx 91.4634
$$

---

### **Step 3: Convert USD to EUR**

Given:
- 1 EUR = 1.32 USD
- So, to find how many EUR she gets for $150:

$$
\text{EUR} = \frac{150}{1.32} \approx 113.6364
$$

---

### **Step 4: Calculate the difference between euros and pounds**

$$
\text{Difference} = \text{EUR} - \text{GBP} = 113.6364 - 91.4634 = 22.173
$$

---

### **Step 5: Round to the nearest whole number**

$$
22.173 \approx \boxed{22}
$$

---

### **Final Answer**

$$
\boxed{22}
$$

[Assistant]
Explanation:
Okay, let's see. Kathy has $300 in the bank and she wants to withdraw half of it in British pounds and half in euros. Then we need to find out how many more euros than pounds she will have. Hmm, okay. Let me break this down step by step.

First, she wants to withdraw half of her $300 in pounds and half in euros. So, half of $300 is $150. That means she wants to convert $150 into pounds and $150 into euros. Wait, but the question says "withdraw half of it in British pounds and half of it in euros." So does that mean she's withdrawing half the amount in pounds and half in euros? Or is it half the amount in terms of the currency? Wait, maybe I need to clarify.

Wait, the problem says she has 300 USD in the bank. She wants to withdraw half of it in British pounds and half of it in euros. So, does that mean she's withdrawing half of the total amount (which is $150) in pounds and the other half ($150) in euros? But how does that work? Because if she's withdrawing in different currencies, the amount in pounds and euros would depend on the exchange rates.

Wait, maybe she wants to convert half of her USD into pounds and half into euros. So, half of $300 is $150. So she converts $150 to pounds and $150 to euros. Then, we need to find out how much that is in pounds and euros, and then find the difference between euros and pounds.

Yes, that seems to make sense. So let me confirm. The problem says "withdraw half of it in British pounds and half of it in euros." So "it" refers to the 300 USD. So half of it is $150. So she's converting $150 to pounds and $150 to euros. Then, the question is how many more euros than pounds will she have. So, after converting, we need to calculate the amount in euros and pounds, then subtract pounds from euros to find the difference.

Okay, so let's start with converting $150 to pounds. The exchange rate is 1 pound = 1.64 USD. So, to find out how many pounds she gets for $150, we divide the USD amount by the exchange rate. So pounds = USD / 1.64. So that would be 150 / 1.64.

Similarly, for euros, the exchange rate is 1 euro = 1.32 USD. So euros = USD / 1.32. So that would be 150 / 1.32.

Once we have both amounts, we subtract pounds from euros to find the difference. Then round to the nearest whole number.

Let me compute these values step by step.

First, converting $150 to pounds:

150 divided by 1.64. Let me calculate that. Let me get a calculator or do it manually. Let me try to compute 150 / 1.64.

Well, 1.64 times 90 is 147.6, because 1.64 * 90 = 1.64 * 9 * 10 = 14.76 * 10 = 147.6. Then 150 - 147.6 = 2.4. So 2.4 / 1.64 is approximately 1.463. So total pounds would be approximately 90 + 1.463 = 91.463 pounds. Let me check with a calculator approach.

Alternatively, 1.64 * 91 = 1.64 * 90 + 1.64 * 1 = 147.6 + 1.64 = 149.24. Then 150 - 149.24 = 0.76. So 0.76 / 1.64 ≈ 0.463. So total pounds ≈ 91.463. So approximately 91.46 pounds. Let me check with decimal division.

Alternatively, 150 / 1.64. Let me write this as 150 ÷ 1.64. Multiply numerator and denominator by 100 to eliminate decimals: 15000 ÷ 164. Let me compute that.

Divide 15000 by 164. Let me see how many times 164 goes into 15000.

First, 164 * 90 = 14760. Subtract that from 15000: 15000 - 14760 = 240. Then, 164 goes into 240 once (164), remainder 76. Then bring down a zero: 760. 164 * 4 = 656. Subtract from 760: 104. Bring down a zero: 1040. 164 * 6 = 984. Subtract: 1040 - 984 = 56. Bring down a zero: 560. 164 * 3 = 492. Subtract: 560 - 492 = 68. Bring down a zero: 680. 164 * 4 = 656. Subtract: 680 - 656 = 24. So so far, we have 91.463... So approximately 91.46 pounds. So rounding to two decimal places, 91.46 pounds. But since we need to round to the nearest whole number at the end, maybe we can keep more decimals for accuracy.

Alternatively, maybe I can use calculator-like steps. Let me check with approximate decimal values.

Alternatively, maybe I can use fractions. But perhaps it's easier to just proceed with decimal approximations.

So pounds ≈ 91.463.

Now, converting $150 to euros. The exchange rate is 1 euro = 1.32 USD. So euros = 150 / 1.32.

Again, let me compute that. Let me do 150 divided by 1.32.

First, note that 1.32 * 100 = 132. So 150 / 1.32 = (150 * 100) / 132 = 15000 / 132. Let me compute that.

Divide 15000 by 132. Let me see how many times 132 goes into 15000.

First, 132 * 100 = 13200. Subtract that from 15000: 15000 - 13200 = 1800.

Now, 132 * 13 = 1716. Subtract that from 1800: 1800 - 1716 = 84.

Bring down a zero: 840. 132 * 6 = 792. Subtract: 840 - 792 = 48.

Bring down a zero: 480. 132 * 3 = 396. Subtract: 480 - 396 = 84.

Bring down a zero: 840. Wait, this is repeating. So the decimal is 113.6363..., because 100 + 13.6363... So total is 113.6363... euros.

Wait, let me check again. Wait, 132 * 113 = ?

Wait, maybe I should do this step by step. Let me try again.

Original division: 15000 / 132.

First, divide numerator and denominator by 12 to simplify? Let me see. 132 divided by 12 is 11. 15000 divided by 12 is 1250. So now it's 1250 / 11. Which is approximately 113.6363... So yes, that's correct. So 1250 divided by 11 is 113.6363...

So euros ≈ 113.6363...

So approximately 113.64 euros.

So now, we have approximately 91.46 pounds and 113.64 euros.

The question asks how many more euros than pounds she will have. So we need to compute euros - pounds.

So 113.64 - 91.46 = ?

Let me compute that. 113.64 - 91.46. Subtract the whole numbers: 113 - 91 = 22. Subtract the decimals: 0.64 - 0.46 = 0.18. So total is 22.18. So approximately 22.18 euros more.

But we need to round to the nearest whole number. So 22.18 is approximately 22. So the answer would be 22.

Wait, but let me check if my approximated values for pounds and euros are accurate enough. Because if I rounded too early, maybe there's a slight error.

Let me recast the problem with more precise numbers.

First, pounds: 150 / 1.64.

Let me compute 150 / 1.64 exactly. Let me use decimal division.

1.64 × 91 = 1.64 × 90 + 1.64 × 1 = 147.6 + 1.64 = 149.24

Then, 150 - 149.24 = 0.76

So, 0.76 / 1.64 = 0.463...

So total pounds is 91.463...

Similarly, euros: 150 / 1.32 = 113.636...

So euros - pounds = 113.636... - 91.463... = ?

Let me compute that:

113.636... - 91.463... = (113 - 91) + (0.636... - 0.463...) = 22 + 0.173... ≈ 22.173...

So approximately 22.173, which rounds to 22 when rounded to the nearest whole number.

But let me check if there's a more precise way. Maybe I should carry out the exact calculation with more decimal places.

Alternatively, perhaps I can compute the exact difference without rounding intermediate steps.

Let me try that.

Let me compute euros - pounds = (150 / 1.32) - (150 / 1.64)

Factor out 150: 150*(1/1.32 - 1/1.64)

Compute 1/1.32 - 1/1.64.

First, compute 1/1.32. Let me calculate that.

1 / 1.32 ≈ 0.757575...

Similarly, 1 / 1.64 ≈ 0.609756...

So subtracting them: 0.757575... - 0.609756... ≈ 0.147819...

Then multiply by 150: 150 * 0.147819 ≈ ?

Compute 150 * 0.147819:

First, 100 * 0.147819 = 14.7819

50 * 0.147819 = 7.39095

Adding together: 14.7819 + 7.39095 = 22.17285

So approximately 22.17285, which is about 22.17, so rounds to 22. So that's consistent with the previous calculation.

Therefore, the answer is 22.

But let me check once more with exact fractions to see if there's any discrepancy.

Alternatively, let me compute 1/1.32 - 1/1.64.

First, convert 1.32 and 1.64 into fractions.

1.32 is 132/100 = 33/25.

Similarly, 1.64 is 164/100 = 41/25. Wait, 164 divided by 4 is 41, 100 divided by 4 is 25. Yes, so 1.64 = 41/25.

Therefore, 1/1.32 = 25/33, and 1/1.64 = 25/41.

Therefore, the difference is 25/33 - 25/41 = 25*(1/33 - 1/41) = 25*( (41 - 33)/(33*41) ) = 25*(8/(1353)) = 200 / 1353 ≈ ?

Compute 200 divided by 1353.

Well, 1353 × 0.147 ≈ 1353 × 0.1 + 1353 × 0.04 + 1353 × 0.007

= 135.3 + 54.12 + 9.471 ≈ 135.3 + 54.12 = 189.42 + 9.471 ≈ 198.891

Which is very close to 200. So 0.147 gives approximately 198.89, so 200 / 1353 ≈ 0.1478...

Therefore, multiplying by 150:

150 * (200 / 1353) = (150 * 200) / 1353 = 30000 / 1353 ≈ ?

Divide 30000 by 1353.

Well, 1353 × 22 = ?

1353 × 20 = 27060

1353 × 2 = 2706

Total: 27060 + 2706 = 29766

Subtract that from 30000: 30000 - 29766 = 234

So 22 + 234 / 1353 ≈ 22 + 0.173 ≈ 22.173, which is the same as before. So 22.173, which rounds to 22.

Therefore, the answer is 22.

But let me check once again with exact decimal calculations.

Alternatively, maybe I can use calculator-like computations for the initial conversions.

For pounds: 150 / 1.64.

Let me compute 1.64 × 91 = 149.24 as before. Then 150 - 149.24 = 0.76. So 0.76 / 1.64.

Let me compute 0.76 / 1.64. Multiply numerator and denominator by 100: 76 / 164. Simplify by dividing numerator and denominator by 4: 19 / 41 ≈ 0.4634...

So total pounds is 91 + 0.4634 ≈ 91.4634.

For euros: 150 / 1.32. Let me compute 1.32 × 113 = ?

1.32 × 100 = 132

1.32 × 13 = 17.16

Total: 132 + 17.16 = 149.16

Subtract from 150: 150 - 149.16 = 0.84

So 0.84 / 1.32 = 0.6363...

So total euros is 113 + 0.6363 ≈ 113.6363.

Therefore, euros - pounds = 113.6363 - 91.4634 = ?

Let me compute 113.6363 - 91.4634:

Subtract 91 from 113: 22.

Subtract 0.4634 from 0.6363: 0.6363 - 0.4634 = 0.1729.

So total difference is 22 + 0.1729 ≈ 22.1729, which is approximately 22.17, rounds to 22.

Therefore, regardless of the method, the answer is 22.

But let me check once more with another approach. Maybe using exact decimal values.

Alternatively, maybe I can use the exact values of the exchange rates and compute the difference.

But I think all methods lead to approximately 22.17, which rounds to 22. Therefore, the answer is 22.

But just to be thorough, let me check if there's any chance that the question is interpreted differently. For example, maybe "withdraw half of it in British pounds and half of it in euros" could mean that she withdraws half the amount in pounds and half in euros, but considering the exchange rates? Wait, but that would be a different interpretation. Let me think.

Suppose she wants to withdraw half of her money in pounds and half in euros. But since the currencies have different values, how would that work? For example, if she wants to withdraw half in pounds and half in euros, but the total amount in USD would be different? Wait, but the problem says she has 300 USD in the bank. So if she wants to withdraw half of it in pounds and half in euros, that would mean she is converting half of the USD into pounds and half into euros. Which is the same as the original interpretation. So I think that's the correct way.

Alternatively, if she wanted to withdraw half the amount in pounds and half in euros, but in terms of the value in USD? For example, if she wants to have half the value in pounds and half in euros, but that would be different. But the problem says "withdraw half of it in British pounds and half of it in euros". Since "it" refers to the 300 USD, so half of it is $150. So converting $150 to pounds and $150 to euros. So that's the correct interpretation.

Therefore, the answer is 22.

But just to make sure, let me check with exact numbers using calculator-like precision.

Let me compute 150 / 1.64:

Using a calculator, 150 divided by 1.64 equals approximately 91.463414634...

Similarly, 150 divided by 1.32 equals approximately 113.636363636...

Subtracting these two: 113.636363636... - 91.463414634... = ?

Let me compute 113.636363636 - 91.463414634:

First, subtract the whole numbers: 113 - 91 = 22.

Then subtract the decimals: 0.636363636 - 0.463414634.

Compute 0.636363636 - 0.463414634:

0.636363636 - 0.463414634 = 0.172949002.

So total difference is 22 + 0.172949002 ≈ 22.172949002.

Which is approximately 22.1729, so when rounded to the nearest whole number, it's 22, since 0.1729 is less than 0.5.

Therefore, the answer is 22.

I think that's solid. All methods converge to 22.17, which rounds to 22. Therefore, the answer is 22.

**Final Answer**
\boxed{22}

To determine how many more euros than pounds Kathy will have after withdrawing half of her $300 in British pounds and the other half in euros, we follow a step-by-step process using the given exchange rates:

---

### **Step 1: Determine the amount to be converted**

Kathy has **$300** in USD. She wants to withdraw **half** of it in **British pounds (GBP)** and **half** in **euros (EUR)**.

- Half of $300 = $150
- So, she will convert **$150 to GBP** and **$150 to EUR**

---

### **Step 2: Convert USD to GBP**

Given:
- 1 GBP = 1.64 USD
- So, to find how many GBP she gets for $150:

$$
\text{GBP} = \frac{150}{1.64} \approx 91.4634
$$

---

### **Step 3: Convert USD to EUR**

Given:
- 1 EUR = 1.32 USD
- So, to find how many EUR she gets for $150:

$$
\text{EUR} = \frac{150}{1.32} \approx 113.6364
$$

---

### **Step 4: Calculate the difference between euros and pounds**

$$
\text{Difference} = \text{EUR} - \text{GBP} = 113.6364 - 91.4634 = 22.173
$$

---

### **Step 5: Round to the nearest whole number**

$$
22.173 \approx \boxed{22}
$$

---

### **Final Answer**

$$
\boxed{22}
$$

Solution: After Kathy has exchanged half of her money ($300\text{ USD}\div 2 = 150\text{ USD}$) into pounds, she will have $150\text{ USD}\times\frac{1\text{ pound}}{1.64 \text{ USD}}\approx 91.46 \text{ pounds}$. After exchanging the other half of her money into euros, she will have $150\text{ USD} \times\frac{1\text{ euro}}{1.32 \text{ USD}}\approx 113.64\text{ euros}$. Subtracting these two values, we have $113.64-91.46=22.18$. Since the question calls for the nearest whole number, we round 22.18 to the final answer of $\boxed{22}$.


Explanation:
Okay, so I need to find the smallest real number x in the domain of the function g(x) = sqrt[(x - 3)^2 - (x - 8)^2]. Hmm, let me think about how to approach this.

First, I remember that for a square root function to be defined (since we're dealing with real numbers), the expression inside the square root must be non-negative. That means whatever is under the square root, which is (x - 3)^2 - (x - 8)^2, has to be greater than or equal to zero. So my first step should be to figure out for which x values this expression is non-negative. Once I have that domain, I can then find the smallest real number in that domain.

Alright, let me start by simplifying the expression inside the square root. Let me expand both squares:

First, expand (x - 3)^2. That would be x^2 - 6x + 9. Then, expand (x - 8)^2, which is x^2 - 16x + 64.

So, subtracting the second expansion from the first:

(x^2 - 6x + 9) - (x^2 - 16x + 64). Let me compute that step by step.

First, distribute the negative sign to the second quadratic:

x^2 - 6x + 9 - x^2 + 16x - 64.

Now combine like terms. The x^2 terms: x^2 - x^2 cancels out. Then the x terms: -6x + 16x is 10x. Then the constants: 9 - 64 is -55.

So the entire expression simplifies to 10x - 55.

Therefore, the original function simplifies to sqrt(10x - 55).

Wait, so the expression inside the square root is 10x - 55. Therefore, for the square root to be real, we need 10x - 55 ≥ 0.

Solving that inequality: 10x - 55 ≥ 0. Let me solve for x.

Add 55 to both sides: 10x ≥ 55. Then divide both sides by 10: x ≥ 55/10. Simplifying that, 55 divided by 10 is 5.5, which is 11/2.

So the domain of g(x) is all real numbers x such that x ≥ 11/2. Therefore, the smallest real number in the domain is 11/2, which is 5.5.

Wait, but let me check my steps again to make sure I didn't make a mistake.

Original expression: (x - 3)^2 - (x - 8)^2.

Expanding both squares:

(x - 3)^2 = x² - 6x + 9

(x - 8)^2 = x² - 16x + 64

Subtracting them: (x² - 6x + 9) - (x² - 16x + 64) = x² - 6x + 9 - x² + 16x - 64.

Yes, x² - x² cancels. Then -6x + 16x is 10x. 9 - 64 is -55. So 10x - 55.

So that's correct. Therefore, 10x - 55 ≥ 0 => x ≥ 5.5.

Therefore, the domain is [11/2, ∞). Therefore, the smallest real number in the domain is 11/2.

But wait, let me think again. Is there a possibility that I made a mistake in simplifying? Let me try another approach. Maybe using difference of squares?

The original expression is (x - 3)^2 - (x - 8)^2. This is of the form a² - b², which factors into (a - b)(a + b).

So let me try that. Let a = x - 3 and b = x - 8. Then, the expression becomes (a - b)(a + b).

Calculating a - b: (x - 3) - (x - 8) = x - 3 - x + 8 = 5.

Calculating a + b: (x - 3) + (x - 8) = 2x - 11.

Therefore, the expression factors into 5*(2x - 11).

So, the expression inside the square root is 5*(2x - 11). Therefore, we need 5*(2x - 11) ≥ 0.

Since 5 is positive, this inequality is equivalent to 2x - 11 ≥ 0. Solving that: 2x ≥ 11 => x ≥ 11/2.

Same result as before. So that confirms that the domain is x ≥ 11/2. Therefore, the smallest real number in the domain is 11/2.

Therefore, the answer should be 11/2, which is 5.5.

But let me check with a sample value. Let me plug in x = 11/2 into the original expression.

Compute (x - 3)^2 - (x - 8)^2.

x = 11/2.

x - 3 = 11/2 - 6/2 = 5/2.

x - 8 = 11/2 - 16/2 = -5/2.

So (5/2)^2 - (-5/2)^2 = 25/4 - 25/4 = 0.

So sqrt(0) is 0, which is real. So x = 11/2 is indeed in the domain.

What if I try x slightly less than 11/2, say x = 5.4.

Compute 10x - 55. 10*5.4 = 54. 54 - 55 = -1. Which is negative. Therefore, sqrt(-1) is not real. So x=5.4 is not in the domain.

Therefore, the domain starts at x=5.5.

Therefore, the smallest real number in the domain is 11/2.

Wait, but the question says "smallest real number x in the domain". Since the domain is [11/2, ∞), the smallest number is 11/2.

Therefore, the answer is 11/2.

But let me check once more if there's any other constraints. For example, sometimes when you have square roots, there might be more restrictions, but in this case, the expression simplifies to sqrt(10x - 55), so the only restriction is 10x - 55 ≥ 0.

Therefore, no other constraints.

Alternatively, if I didn't simplify and tried to think about the original expression, maybe there's something else? Let me think.

Suppose someone didn't simplify and tried to think about the original expression. For example, maybe they think that (x - 3)^2 - (x - 8)^2 could be something else? But no, we already factored it and simplified.

Alternatively, maybe there's a mistake in the factoring? Let me check again.

Original expression: (x - 3)^2 - (x - 8)^2.

Factoring as difference of squares: [ (x - 3) - (x - 8) ] * [ (x - 3) + (x - 8) ]

First term: (x - 3 - x + 8) = 5.

Second term: (x - 3 + x - 8) = 2x - 11.

So 5*(2x - 11). Correct.

Therefore, 5*(2x - 11) ≥ 0.

Since 5 is positive, 2x - 11 ≥ 0 => x ≥ 11/2.

Therefore, same conclusion.

Therefore, I think my answer is correct.

But just to be thorough, let me check another value. Let me take x = 11/2 + 1 = 13/2 = 6.5.

Compute (x - 3)^2 - (x - 8)^2.

x - 3 = 6.5 - 3 = 3.5 = 7/2.

x - 8 = 6.5 - 8 = -1.5 = -3/2.

So (7/2)^2 - (-3/2)^2 = 49/4 - 9/4 = 40/4 = 10.

sqrt(10) is real. So that works.

If I take x = 11/2, as before, gives 0.

If I take x = 11/2 - 0.1 = 5.5 - 0.1 = 5.4, which we saw gives negative inside sqrt.

Therefore, the domain is indeed x ≥ 11/2, so the smallest x is 11/2.

Therefore, the answer is 11/2.

But let me check if the problem says "smallest real number x in the domain". Since the domain is all real numbers greater than or equal to 11/2, the smallest is 11/2.

Therefore, I think that's the answer.

**Final Answer**
The smallest real number $ x $ in the domain of $ g(x) $ is \boxed{\dfrac{11}{2}}.

To determine the smallest real number $ x $ in the domain of the function
$$
g(x) = \sqrt{(x - 3)^2 - (x - 8)^2},
$$
we must first analyze the expression under the square root. A square root is defined only when its argument is non-negative, so we require:

$$
(x - 3)^2 - (x - 8)^2 \geq 0.
$$

---

### Step 1: Simplify the Expression

We simplify the expression inside the square root using the **difference of squares** identity:

$$
a^2 - b^2 = (a - b)(a + b).
$$

Let $ a = x - 3 $ and $ b = x - 8 $. Then:

$$
(x - 3)^2 - (x - 8)^2 = [(x - 3) - (x - 8)][(x - 3) + (x - 8)].
$$

Simplify each factor:

- $ (x - 3) - (x - 8) = x - 3 - x + 8 = 5 $
- $ (x - 3) + (x - 8) = x - 3 + x - 8 = 2x - 11 $

So the expression becomes:

$$
(x - 3)^2 - (x - 8)^2 = 5(2x - 11).
$$

---

### Step 2: Determine the Domain

We now require:

$$
5(2x - 11) \geq 0.
$$

Since 5 is positive, the inequality reduces to:

$$
2x - 11 \geq 0 \quad \Rightarrow \quad x \geq \frac{11}{2}.
$$

Thus, the domain of $ g(x) $ is:

$$
x \in \left[ \frac{11}{2}, \infty \right).
$$

---

### Step 3: Find the Smallest Real Number in the Domain

The domain is a closed interval starting at $ \frac{11}{2} $. Therefore, the **smallest real number** in the domain is:

$$
\boxed{\dfrac{11}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest real number x in the domain of the function g(x) = sqrt[(x - 3)^2 - (x - 8)^2]. Hmm, let me think about how to approach this.

First, I remember that for a square root function to be defined (since we're dealing with real numbers), the expression inside the square root must be non-negative. That means whatever is under the square root, which is (x - 3)^2 - (x - 8)^2, has to be greater than or equal to zero. So my first step should be to figure out for which x values this expression is non-negative. Once I have that domain, I can then find the smallest real number in that domain.

Alright, let me start by simplifying the expression inside the square root. Let me expand both squares:

First, expand (x - 3)^2. That would be x^2 - 6x + 9. Then, expand (x - 8)^2, which is x^2 - 16x + 64.

So, subtracting the second expansion from the first:

(x^2 - 6x + 9) - (x^2 - 16x + 64). Let me compute that step by step.

First, distribute the negative sign to the second quadratic:

x^2 - 6x + 9 - x^2 + 16x - 64.

Now combine like terms. The x^2 terms: x^2 - x^2 cancels out. Then the x terms: -6x + 16x is 10x. Then the constants: 9 - 64 is -55.

So the entire expression simplifies to 10x - 55.

Therefore, the original function simplifies to sqrt(10x - 55).

Wait, so the expression inside the square root is 10x - 55. Therefore, for the square root to be real, we need 10x - 55 ≥ 0.

Solving that inequality: 10x - 55 ≥ 0. Let me solve for x.

Add 55 to both sides: 10x ≥ 55. Then divide both sides by 10: x ≥ 55/10. Simplifying that, 55 divided by 10 is 5.5, which is 11/2.

So the domain of g(x) is all real numbers x such that x ≥ 11/2. Therefore, the smallest real number in the domain is 11/2, which is 5.5.

Wait, but let me check my steps again to make sure I didn't make a mistake.

Original expression: (x - 3)^2 - (x - 8)^2.

Expanding both squares:

(x - 3)^2 = x² - 6x + 9

(x - 8)^2 = x² - 16x + 64

Subtracting them: (x² - 6x + 9) - (x² - 16x + 64) = x² - 6x + 9 - x² + 16x - 64.

Yes, x² - x² cancels. Then -6x + 16x is 10x. 9 - 64 is -55. So 10x - 55.

So that's correct. Therefore, 10x - 55 ≥ 0 => x ≥ 5.5.

Therefore, the domain is [11/2, ∞). Therefore, the smallest real number in the domain is 11/2.

But wait, let me think again. Is there a possibility that I made a mistake in simplifying? Let me try another approach. Maybe using difference of squares?

The original expression is (x - 3)^2 - (x - 8)^2. This is of the form a² - b², which factors into (a - b)(a + b).

So let me try that. Let a = x - 3 and b = x - 8. Then, the expression becomes (a - b)(a + b).

Calculating a - b: (x - 3) - (x - 8) = x - 3 - x + 8 = 5.

Calculating a + b: (x - 3) + (x - 8) = 2x - 11.

Therefore, the expression factors into 5*(2x - 11).

So, the expression inside the square root is 5*(2x - 11). Therefore, we need 5*(2x - 11) ≥ 0.

Since 5 is positive, this inequality is equivalent to 2x - 11 ≥ 0. Solving that: 2x ≥ 11 => x ≥ 11/2.

Same result as before. So that confirms that the domain is x ≥ 11/2. Therefore, the smallest real number in the domain is 11/2.

Therefore, the answer should be 11/2, which is 5.5.

But let me check with a sample value. Let me plug in x = 11/2 into the original expression.

Compute (x - 3)^2 - (x - 8)^2.

x = 11/2.

x - 3 = 11/2 - 6/2 = 5/2.

x - 8 = 11/2 - 16/2 = -5/2.

So (5/2)^2 - (-5/2)^2 = 25/4 - 25/4 = 0.

So sqrt(0) is 0, which is real. So x = 11/2 is indeed in the domain.

What if I try x slightly less than 11/2, say x = 5.4.

Compute 10x - 55. 10*5.4 = 54. 54 - 55 = -1. Which is negative. Therefore, sqrt(-1) is not real. So x=5.4 is not in the domain.

Therefore, the domain starts at x=5.5.

Therefore, the smallest real number in the domain is 11/2.

Wait, but the question says "smallest real number x in the domain". Since the domain is [11/2, ∞), the smallest number is 11/2.

Therefore, the answer is 11/2.

But let me check once more if there's any other constraints. For example, sometimes when you have square roots, there might be more restrictions, but in this case, the expression simplifies to sqrt(10x - 55), so the only restriction is 10x - 55 ≥ 0.

Therefore, no other constraints.

Alternatively, if I didn't simplify and tried to think about the original expression, maybe there's something else? Let me think.

Suppose someone didn't simplify and tried to think about the original expression. For example, maybe they think that (x - 3)^2 - (x - 8)^2 could be something else? But no, we already factored it and simplified.

Alternatively, maybe there's a mistake in the factoring? Let me check again.

Original expression: (x - 3)^2 - (x - 8)^2.

Factoring as difference of squares: [ (x - 3) - (x - 8) ] * [ (x - 3) + (x - 8) ]

First term: (x - 3 - x + 8) = 5.

Second term: (x - 3 + x - 8) = 2x - 11.

So 5*(2x - 11). Correct.

Therefore, 5*(2x - 11) ≥ 0.

Since 5 is positive, 2x - 11 ≥ 0 => x ≥ 11/2.

Therefore, same conclusion.

Therefore, I think my answer is correct.

But just to be thorough, let me check another value. Let me take x = 11/2 + 1 = 13/2 = 6.5.

Compute (x - 3)^2 - (x - 8)^2.

x - 3 = 6.5 - 3 = 3.5 = 7/2.

x - 8 = 6.5 - 8 = -1.5 = -3/2.

So (7/2)^2 - (-3/2)^2 = 49/4 - 9/4 = 40/4 = 10.

sqrt(10) is real. So that works.

If I take x = 11/2, as before, gives 0.

If I take x = 11/2 - 0.1 = 5.5 - 0.1 = 5.4, which we saw gives negative inside sqrt.

Therefore, the domain is indeed x ≥ 11/2, so the smallest x is 11/2.

Therefore, the answer is 11/2.

But let me check if the problem says "smallest real number x in the domain". Since the domain is all real numbers greater than or equal to 11/2, the smallest is 11/2.

Therefore, I think that's the answer.

**Final Answer**
The smallest real number $ x $ in the domain of $ g(x) $ is \boxed{\dfrac{11}{2}}.

To determine the smallest real number $ x $ in the domain of the function
$$
g(x) = \sqrt{(x - 3)^2 - (x - 8)^2},
$$
we must first analyze the expression under the square root. A square root is defined only when its argument is non-negative, so we require:

$$
(x - 3)^2 - (x - 8)^2 \geq 0.
$$

---

### Step 1: Simplify the Expression

We simplify the expression inside the square root using the **difference of squares** identity:

$$
a^2 - b^2 = (a - b)(a + b).
$$

Let $ a = x - 3 $ and $ b = x - 8 $. Then:

$$
(x - 3)^2 - (x - 8)^2 = [(x - 3) - (x - 8)][(x - 3) + (x - 8)].
$$

Simplify each factor:

- $ (x - 3) - (x - 8) = x - 3 - x + 8 = 5 $
- $ (x - 3) + (x - 8) = x - 3 + x - 8 = 2x - 11 $

So the expression becomes:

$$
(x - 3)^2 - (x - 8)^2 = 5(2x - 11).
$$

---

### Step 2: Determine the Domain

We now require:

$$
5(2x - 11) \geq 0.
$$

Since 5 is positive, the inequality reduces to:

$$
2x - 11 \geq 0 \quad \Rightarrow \quad x \geq \frac{11}{2}.
$$

Thus, the domain of $ g(x) $ is:

$$
x \in \left[ \frac{11}{2}, \infty \right).
$$

---

### Step 3: Find the Smallest Real Number in the Domain

The domain is a closed interval starting at $ \frac{11}{2} $. Therefore, the **smallest real number** in the domain is:

$$
\boxed{\dfrac{11}{2}}
$$

Solution:
A real number $x$ is in the domain of $g$ if and only if $$(x-3)^2 - (x-8)^2 \ge 0.$$ Expanding this out and simplifying, we get $$10x - 55\ge 0;$$ the smallest solution is $x=\frac{55}{10}=\boxed{\frac{11}{2}}$.

Alternatively, once we have the quadratic equation $$(x-3)^2 - (x-8)^2 \ge 0,$$ instead of expanding it out, we can observe that $(x-3)^2$ is the square of the distance from $x$ to $3$ on the number line, while $(x-8)^2$ is the square of the distance from $x$ to $8$. Thus, $(x-3)^2-(x-8)^2\ge 0$ is true if $x$ is closer to $8$ than to $3$, which is true if and only if $x\ge \frac{8+3}{2} = \boxed{\frac{11}{2}}$.


Explanation:
Okay, so I need to find the smallest distance between the origin and a point on the graph of $ y = \frac{1}{\sqrt{2}}(x^2 - 3) $. The answer should be in the form $ \sqrt{a}/b $, where $ a $ and $ b $ are positive integers, and $ a $ is square-free. Then I have to find $ a + b $. Alright, let me think about how to approach this.

First, I remember that the distance between a point $ (x, y) $ and the origin $ (0, 0) $ is given by the distance formula: $ D = \sqrt{x^2 + y^2} $. Since we want the smallest distance, maybe I can minimize this distance function. However, dealing with square roots can be a bit messy, so maybe I can minimize the square of the distance instead, which should have its minimum at the same point. That might simplify calculations.

So, let me define $ D^2 = x^2 + y^2 $. Since $ y $ is given in terms of $ x $, I can substitute that into the equation. Let me do that:

$ D^2 = x^2 + \left( \frac{1}{\sqrt{2}}(x^2 - 3) \right)^2 $.

Let me simplify this expression. First, square the $ y $ term:

$ \left( \frac{1}{\sqrt{2}}(x^2 - 3) \right)^2 = \frac{1}{2}(x^2 - 3)^2 $.

So, substituting back into $ D^2 $:

$ D^2 = x^2 + \frac{1}{2}(x^2 - 3)^2 $.

Now, I need to find the minimum value of this function. To do that, I can take the derivative with respect to $ x $, set it equal to zero, and solve for $ x $. Then check if it's a minimum using the second derivative or some other method.

Let me compute the derivative. Let me denote $ f(x) = x^2 + \frac{1}{2}(x^2 - 3)^2 $.

First, expand $ (x^2 - 3)^2 $:

$ (x^2 - 3)^2 = x^4 - 6x^2 + 9 $.

Therefore, $ f(x) = x^2 + \frac{1}{2}(x^4 - 6x^2 + 9) $.

Let me simplify this:

$ f(x) = x^2 + \frac{1}{2}x^4 - 3x^2 + \frac{9}{2} $.

Combine like terms:

$ f(x) = \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} $.

Now, take the derivative $ f'(x) $:

$ f'(x) = 2x^3 - 4x $.

Wait, let me check that again. The derivative of $ \frac{1}{2}x^4 $ is $ 2x^3 $, right? Because $ \frac{d}{dx} x^n = n x^{n-1} $, so $ \frac{1}{2} \times 4x^3 = 2x^3 $. Then the derivative of $ -2x^2 $ is $ -4x $. And the derivative of $ \frac{9}{2} $ is 0. So yes, $ f'(x) = 2x^3 - 4x $.

To find critical points, set $ f'(x) = 0 $:

$ 2x^3 - 4x = 0 $.

Factor out 2x:

$ 2x(x^2 - 2) = 0 $.

So, the critical points are when $ 2x = 0 $ or $ x^2 - 2 = 0 $.

Therefore, $ x = 0 $ or $ x = \sqrt{2} $ or $ x = -\sqrt{2} $.

Now, I need to check which of these critical points gives the minimum distance. Since the function $ f(x) $ is even? Wait, let me check. The original function $ y = \frac{1}{\sqrt{2}}(x^2 - 3) $ is even, because replacing $ x $ with $ -x $ doesn't change the value. Therefore, the graph is symmetric about the y-axis. Therefore, the distance from the origin should also be symmetric. Therefore, the minimum distance should occur at $ x = \sqrt{2} $ and $ x = -\sqrt{2} $, and maybe also at $ x = 0 $. Let me check the value of $ f(x) $ at these points.

First, at $ x = 0 $:

$ f(0) = \frac{1}{2}(0)^4 - 2(0)^2 + \frac{9}{2} = 0 - 0 + \frac{9}{2} = \frac{9}{2} $.

So, the distance squared is $ 9/2 $, so the distance is $ \sqrt{9/2} = 3/\sqrt{2} $, which is approximately 2.121.

Now, let's check at $ x = \sqrt{2} $:

First compute $ x^2 = (\sqrt{2})^2 = 2 $.

So, $ f(\sqrt{2}) = \frac{1}{2}(2)^2 - 2(2) + \frac{9}{2} $.

Wait, hold on. Wait, earlier I had simplified $ f(x) $ as $ \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} $. So, substituting $ x = \sqrt{2} $:

First, $ x^4 = (x^2)^2 = (2)^2 = 4 $.

Therefore, $ f(\sqrt{2}) = \frac{1}{2}(4) - 2(2) + \frac{9}{2} = 2 - 4 + 4.5 = 2.5 $.

Which is $ 5/2 $. Therefore, the distance squared is $ 5/2 $, so the distance is $ \sqrt{5/2} = \sqrt{10}/2 $. Wait, because $ \sqrt{5/2} = \sqrt{10}/2 $. Let me check that:

$ \sqrt{5/2} = \sqrt{5}/\sqrt{2} = (\sqrt{5} \times \sqrt{2}) / (\sqrt{2} \times \sqrt{2}) ) = \sqrt{10}/2 $. Yes, that's correct.

Similarly, at $ x = -\sqrt{2} $, the value will be the same because of the even function. So, the distance squared is 5/2, which is less than 9/2. Therefore, the minimum distance occurs at $ x = \pm \sqrt{2} $, and the minimal distance is $ \sqrt{10}/2 $.

Wait, but the problem says the answer can be expressed as $ \sqrt{a}/b $. So, $ \sqrt{10}/2 $ is already in that form, with $ a = 10 $ and $ b = 2 $. Then $ a + b = 12 $. But wait, let me check if 10 is square-free. 10 factors into 2 * 5, neither of which are squares, so yes, 10 is square-free. Therefore, the answer is 10 + 2 = 12. Is that correct?

Wait, but let me double-check my calculations because sometimes when you do calculus, you might make a mistake. Let me verify the steps again.

First, the distance squared is $ x^2 + y^2 $, with $ y = \frac{1}{\sqrt{2}}(x^2 - 3) $. Then substituting:

$ D^2 = x^2 + \frac{1}{2}(x^2 - 3)^2 $. Then expanding $ (x^2 - 3)^2 $ gives $ x^4 - 6x^2 + 9 $, so:

$ D^2 = x^2 + \frac{1}{2}x^4 - 3x^2 + \frac{9}{2} $, which simplifies to $ \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} $. Then derivative is $ 2x^3 - 4x $, set to zero gives critical points at x=0, x=±√2. Then evaluating D² at x=√2:

x² = 2, so x^4 = 4. Then D² = (1/2)(4) - 2*(2) + 9/2 = 2 - 4 + 4.5 = 2.5, which is 5/2. So D = sqrt(5/2) = sqrt(10)/2. So that seems correct.

But let me check if there's another way to approach this problem, maybe using geometry or something else, to confirm.

Alternatively, maybe using Lagrange multipliers? Let me try that.

We need to minimize the distance squared function $ f(x, y) = x^2 + y^2 $ subject to the constraint $ g(x, y) = y - \frac{1}{\sqrt{2}}(x^2 - 3) = 0 $.

The Lagrangian is $ \mathcal{L} = x^2 + y^2 - \lambda \left( y - \frac{1}{\sqrt{2}}(x^2 - 3) \right) $.

Taking partial derivatives:

∂L/∂x = 2x - λ*(-2x / sqrt(2)) = 0

Wait, let me compute that properly.

Wait, the constraint is $ g(x, y) = y - \frac{1}{\sqrt{2}}(x^2 - 3) = 0 $, so the gradient of g is (dg/dx, dg/dy) = (-2x / sqrt(2), 1).

Therefore, the Lagrangian equations are:

∇f = λ∇g.

So:

df/dx = 2x = λ * (-2x / sqrt(2))

df/dy = 2y = λ * 1

And the constraint equation: y = (1 / sqrt(2))(x² - 3)

So, from the first equation:

2x = λ * (-2x / sqrt(2)).

Let me solve for λ.

If x ≠ 0, we can divide both sides by 2x:

1 = λ * (-1 / sqrt(2)) => λ = -sqrt(2).

If x = 0, then the first equation becomes 0 = 0, so we have to check that case separately.

So, case 1: x ≠ 0, then λ = -sqrt(2).

From the second equation: 2y = λ => 2y = -sqrt(2) => y = -sqrt(2)/2.

Now, using the constraint equation: y = (1 / sqrt(2))(x² - 3). So:

-sqrt(2)/2 = (1 / sqrt(2))(x² - 3)

Multiply both sides by sqrt(2):

- (sqrt(2) * sqrt(2))/2 = x² - 3 => -2/2 = x² - 3 => -1 = x² - 3 => x² = 2 => x = ±sqrt(2). Which matches our previous result. So that's consistent.

Case 2: x = 0. Then from the constraint equation, y = (1 / sqrt(2))(0 - 3) = -3 / sqrt(2). Then the distance squared is 0² + y² = (9 / 2), which is the same as before. So, that's the point (0, -3 / sqrt(2)).

So, comparing the two cases, the minimal distance squared is 5/2, which is less than 9/2, so the minimal distance is sqrt(5/2) = sqrt(10)/2. Therefore, the answer is sqrt(10)/2, so a = 10, b = 2, so a + b = 12. Therefore, the answer is 12.

But wait, let me check once more. Maybe I made a mistake in simplifying sqrt(5/2). Let me confirm:

sqrt(5/2) can be written as sqrt(10)/sqrt(4) ? Wait, no. Wait sqrt(5/2) is equal to sqrt(10)/2 because:

sqrt(5)/sqrt(2) = (sqrt(5) * sqrt(2)) / (sqrt(2) * sqrt(2)) ) = sqrt(10)/2. Yes, that's correct. So sqrt(5/2) is indeed sqrt(10)/2. Therefore, the expression is in the form sqrt(a)/b where a = 10 and b = 2. Since 10 is square-free (its prime factors are 2 and 5, neither of which are squared), so that's acceptable.

But just to be thorough, let me check if there are any other critical points or if I missed something. For example, when I used calculus, I found critical points at x = 0, ±√2. The second derivative test could confirm if these are minima or maxima.

Let me compute the second derivative of f(x). Earlier, f'(x) = 2x³ - 4x. Then f''(x) = 6x² - 4.

At x = 0: f''(0) = 0 - 4 = -4 < 0. Therefore, x = 0 is a local maximum. Which makes sense, since the distance squared at x=0 is 9/2, which is larger than at x=±√2. So that's a maximum.

At x = ±√2: f''(x) = 6*( (√2)^2 ) - 4 = 6*(2) - 4 = 12 - 4 = 8 > 0. Therefore, these points are local minima. Therefore, the minimal distance squared is indeed 5/2, so the minimal distance is sqrt(5/2) = sqrt(10)/2. Therefore, the answer is 10 + 2 = 12.

But just to be 100% sure, let me check with another method. Maybe parametric equations or something else.

Alternatively, think geometrically. The graph of y = (1/√2)(x² - 3) is a parabola opening upwards. The origin is at (0, 0). The minimal distance from the origin to the parabola would be along the line connecting the origin to the closest point on the parabola. The line connecting the origin to the closest point should be perpendicular to the tangent of the parabola at that point.

So, let me try this approach. Let me suppose that the closest point is (x, y) on the parabola. Then the vector from the origin to (x, y) is (x, y). The tangent line at (x, y) has a slope equal to the derivative of the function at that point.

First, compute the derivative of y with respect to x. Given y = (1/√2)(x² - 3), so dy/dx = (2x)/√2 = √2 x.

Therefore, the slope of the tangent line at (x, y) is √2 x. Therefore, the slope of the normal line (perpendicular to tangent) is -1/(√2 x), provided that x ≠ 0.

But the line connecting the origin to (x, y) is the normal line. Therefore, the slope of this line is y/x. Therefore, we have:

y/x = -1/(√2 x)

Wait, let me write that equation:

The slope of the normal line is -1/(dy/dx) = -1/(√2 x). But also, the slope of the line from the origin to (x, y) is y/x. Therefore, equating these:

y/x = -1/(√2 x)

Multiply both sides by x (assuming x ≠ 0):

y = -1/√2

So, this gives us y = -1/√2. But since the point (x, y) lies on the parabola, we can substitute y into the equation of the parabola:

-1/√2 = (1/√2)(x² - 3)

Multiply both sides by √2:

-1 = x² - 3 => x² = 2 => x = ±√2. Which is the same result as before. Therefore, the closest points are at (±√2, -1/√2). Therefore, the distance from the origin is sqrt( (√2)^2 + (-1/√2)^2 ) = sqrt(2 + 1/2) = sqrt(5/2) = sqrt(10)/2. Same result. Therefore, this confirms the previous calculations.

Therefore, I can be confident that the minimal distance is sqrt(10)/2, so a = 10, b = 2, and a + b = 12.

But just to make sure, let me check if there's a possibility that the minimal distance occurs at another point. For example, maybe when the line from the origin is tangent to the parabola? Wait, but we already found that the normal line at the closest point passes through the origin, so that's the condition for the minimal distance. Therefore, that should be the case.

Alternatively, maybe I can parametrize the distance squared function and check its minimum. Let me try that again.

We had D² = x² + [ (x² - 3)/√2 ]² = x² + (x^4 - 6x² + 9)/2. Which simplifies to (2x² + x^4 - 6x² + 9)/2 = (x^4 - 4x² + 9)/2. Wait, hold on, earlier I had written f(x) as (1/2)x^4 - 2x² + 9/2, which is the same as (x^4 - 4x² + 9)/2. So, D² = (x^4 - 4x² + 9)/2. Then taking derivative, which is [4x³ - 8x]/2 = 2x³ - 4x, which is same as before. So, same critical points. So, same result.

Alternatively, maybe completing the square or something else? Let me see. Let me consider D² = (x^4 - 4x² + 9)/2. Let me set z = x², so z ≥ 0. Then D² = (z² - 4z + 9)/2. So, we can think of this as a quadratic in z: f(z) = (z² - 4z + 9)/2. To find the minimum of this quadratic function for z ≥ 0.

The quadratic function z² - 4z + 9 has its vertex at z = -b/(2a) = 4/(2*1) = 2. Since the coefficient of z² is positive, the minimum occurs at z = 2. Therefore, the minimum value is ( (2)^2 - 4*(2) + 9 ) / 2 = (4 - 8 + 9)/2 = 5/2. Therefore, D² = 5/2, so D = sqrt(5/2) = sqrt(10)/2. Therefore, same result. Therefore, this method also gives the same answer. Therefore, this confirms again that the minimal distance is sqrt(10)/2.

Therefore, I think this is solid. Therefore, the answer is 10 + 2 = 12.

But just to check once more, let me compute the numerical value. sqrt(10)/2 is approximately 3.1623 / 2 ≈ 1.581. Let me check the distance at x = sqrt(2). The point is (sqrt(2), y). Let me compute y: y = (1/sqrt(2)) * ( (sqrt(2))² - 3 ) = (1/sqrt(2))*(2 - 3) = (-1)/sqrt(2). Therefore, the point is (sqrt(2), -1/sqrt(2)). The distance from the origin is sqrt( (sqrt(2))² + (-1/sqrt(2))² ) = sqrt(2 + 1/2) = sqrt(5/2) ≈ 1.581, which is indeed less than the distance at x=0, which was 3/sqrt(2) ≈ 2.121. So that seems correct.

Alternatively, if I pick another point on the parabola, say x = 1. Then y = (1/sqrt(2))(1 - 3) = (-2)/sqrt(2) = -sqrt(2). The distance squared is 1² + (-sqrt(2))² = 1 + 2 = 3, which is larger than 5/2 = 2.5. So that's consistent. If I take x = 2, then y = (1/sqrt(2))(4 - 3) = 1/sqrt(2). The distance squared is 4 + (1/2) = 4.5, which is larger. So, yes, seems like the minimal distance is indeed at x = sqrt(2).

Therefore, I think I can confidently say that the answer is 12.

**Final Answer**
\boxed{12}

To find the smallest distance between the origin and a point on the graph of the function
$$
y = \frac{1}{\sqrt{2}}(x^2 - 3),
$$
we begin by using the **distance formula** between a point $(x, y)$ and the origin $(0, 0)$:

$$
D = \sqrt{x^2 + y^2}.
$$

To simplify the minimization process, we minimize the **square of the distance**, which is:

$$
D^2 = x^2 + y^2.
$$

Substituting the expression for $ y $ from the given function:

$$
D^2 = x^2 + \left( \frac{1}{\sqrt{2}}(x^2 - 3) \right)^2 = x^2 + \frac{1}{2}(x^2 - 3)^2.
$$

Expanding the square:

$$
(x^2 - 3)^2 = x^4 - 6x^2 + 9,
$$
so
$$
D^2 = x^2 + \frac{1}{2}(x^4 - 6x^2 + 9) = \frac{1}{2}x^4 - 2x^2 + \frac{9}{2}.
$$

Now, we find the critical points by taking the derivative of $ D^2 $ with respect to $ x $:

$$
\frac{d}{dx} \left( \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} \right) = 2x^3 - 4x.
$$

Setting the derivative equal to zero:

$$
2x^3 - 4x = 0 \Rightarrow 2x(x^2 - 2) = 0.
$$

This gives the critical points:

$$
x = 0, \quad x = \pm \sqrt{2}.
$$

We now evaluate $ D^2 $ at these points:

- At $ x = 0 $:
  $$
  D^2 = \frac{1}{2}(0)^4 - 2(0)^2 + \frac{9}{2} = \frac{9}{2}.
  $$

- At $ x = \pm \sqrt{2} $:
  $$
  x^2 = 2, \quad x^4 = 4, \quad \text{so} \quad D^2 = \frac{1}{2}(4) - 2(2) + \frac{9}{2} = 2 - 4 + \frac{9}{2} = \frac{5}{2}.
  $$

Thus, the **minimum distance squared** is $ \frac{5}{2} $, and the **minimum distance** is:

$$
D = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}.
$$

This is in the form $ \frac{\sqrt{a}}{b} $, where $ a = 10 $ and $ b = 2 $. Since 10 is square-free (its prime factorization is $ 2 \times 5 $), the conditions of the problem are satisfied.

---

### Final Answer:

$$
\boxed{12}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest distance between the origin and a point on the graph of $ y = \frac{1}{\sqrt{2}}(x^2 - 3) $. The answer should be in the form $ \sqrt{a}/b $, where $ a $ and $ b $ are positive integers, and $ a $ is square-free. Then I have to find $ a + b $. Alright, let me think about how to approach this.

First, I remember that the distance between a point $ (x, y) $ and the origin $ (0, 0) $ is given by the distance formula: $ D = \sqrt{x^2 + y^2} $. Since we want the smallest distance, maybe I can minimize this distance function. However, dealing with square roots can be a bit messy, so maybe I can minimize the square of the distance instead, which should have its minimum at the same point. That might simplify calculations.

So, let me define $ D^2 = x^2 + y^2 $. Since $ y $ is given in terms of $ x $, I can substitute that into the equation. Let me do that:

$ D^2 = x^2 + \left( \frac{1}{\sqrt{2}}(x^2 - 3) \right)^2 $.

Let me simplify this expression. First, square the $ y $ term:

$ \left( \frac{1}{\sqrt{2}}(x^2 - 3) \right)^2 = \frac{1}{2}(x^2 - 3)^2 $.

So, substituting back into $ D^2 $:

$ D^2 = x^2 + \frac{1}{2}(x^2 - 3)^2 $.

Now, I need to find the minimum value of this function. To do that, I can take the derivative with respect to $ x $, set it equal to zero, and solve for $ x $. Then check if it's a minimum using the second derivative or some other method.

Let me compute the derivative. Let me denote $ f(x) = x^2 + \frac{1}{2}(x^2 - 3)^2 $.

First, expand $ (x^2 - 3)^2 $:

$ (x^2 - 3)^2 = x^4 - 6x^2 + 9 $.

Therefore, $ f(x) = x^2 + \frac{1}{2}(x^4 - 6x^2 + 9) $.

Let me simplify this:

$ f(x) = x^2 + \frac{1}{2}x^4 - 3x^2 + \frac{9}{2} $.

Combine like terms:

$ f(x) = \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} $.

Now, take the derivative $ f'(x) $:

$ f'(x) = 2x^3 - 4x $.

Wait, let me check that again. The derivative of $ \frac{1}{2}x^4 $ is $ 2x^3 $, right? Because $ \frac{d}{dx} x^n = n x^{n-1} $, so $ \frac{1}{2} \times 4x^3 = 2x^3 $. Then the derivative of $ -2x^2 $ is $ -4x $. And the derivative of $ \frac{9}{2} $ is 0. So yes, $ f'(x) = 2x^3 - 4x $.

To find critical points, set $ f'(x) = 0 $:

$ 2x^3 - 4x = 0 $.

Factor out 2x:

$ 2x(x^2 - 2) = 0 $.

So, the critical points are when $ 2x = 0 $ or $ x^2 - 2 = 0 $.

Therefore, $ x = 0 $ or $ x = \sqrt{2} $ or $ x = -\sqrt{2} $.

Now, I need to check which of these critical points gives the minimum distance. Since the function $ f(x) $ is even? Wait, let me check. The original function $ y = \frac{1}{\sqrt{2}}(x^2 - 3) $ is even, because replacing $ x $ with $ -x $ doesn't change the value. Therefore, the graph is symmetric about the y-axis. Therefore, the distance from the origin should also be symmetric. Therefore, the minimum distance should occur at $ x = \sqrt{2} $ and $ x = -\sqrt{2} $, and maybe also at $ x = 0 $. Let me check the value of $ f(x) $ at these points.

First, at $ x = 0 $:

$ f(0) = \frac{1}{2}(0)^4 - 2(0)^2 + \frac{9}{2} = 0 - 0 + \frac{9}{2} = \frac{9}{2} $.

So, the distance squared is $ 9/2 $, so the distance is $ \sqrt{9/2} = 3/\sqrt{2} $, which is approximately 2.121.

Now, let's check at $ x = \sqrt{2} $:

First compute $ x^2 = (\sqrt{2})^2 = 2 $.

So, $ f(\sqrt{2}) = \frac{1}{2}(2)^2 - 2(2) + \frac{9}{2} $.

Wait, hold on. Wait, earlier I had simplified $ f(x) $ as $ \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} $. So, substituting $ x = \sqrt{2} $:

First, $ x^4 = (x^2)^2 = (2)^2 = 4 $.

Therefore, $ f(\sqrt{2}) = \frac{1}{2}(4) - 2(2) + \frac{9}{2} = 2 - 4 + 4.5 = 2.5 $.

Which is $ 5/2 $. Therefore, the distance squared is $ 5/2 $, so the distance is $ \sqrt{5/2} = \sqrt{10}/2 $. Wait, because $ \sqrt{5/2} = \sqrt{10}/2 $. Let me check that:

$ \sqrt{5/2} = \sqrt{5}/\sqrt{2} = (\sqrt{5} \times \sqrt{2}) / (\sqrt{2} \times \sqrt{2}) ) = \sqrt{10}/2 $. Yes, that's correct.

Similarly, at $ x = -\sqrt{2} $, the value will be the same because of the even function. So, the distance squared is 5/2, which is less than 9/2. Therefore, the minimum distance occurs at $ x = \pm \sqrt{2} $, and the minimal distance is $ \sqrt{10}/2 $.

Wait, but the problem says the answer can be expressed as $ \sqrt{a}/b $. So, $ \sqrt{10}/2 $ is already in that form, with $ a = 10 $ and $ b = 2 $. Then $ a + b = 12 $. But wait, let me check if 10 is square-free. 10 factors into 2 * 5, neither of which are squares, so yes, 10 is square-free. Therefore, the answer is 10 + 2 = 12. Is that correct?

Wait, but let me double-check my calculations because sometimes when you do calculus, you might make a mistake. Let me verify the steps again.

First, the distance squared is $ x^2 + y^2 $, with $ y = \frac{1}{\sqrt{2}}(x^2 - 3) $. Then substituting:

$ D^2 = x^2 + \frac{1}{2}(x^2 - 3)^2 $. Then expanding $ (x^2 - 3)^2 $ gives $ x^4 - 6x^2 + 9 $, so:

$ D^2 = x^2 + \frac{1}{2}x^4 - 3x^2 + \frac{9}{2} $, which simplifies to $ \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} $. Then derivative is $ 2x^3 - 4x $, set to zero gives critical points at x=0, x=±√2. Then evaluating D² at x=√2:

x² = 2, so x^4 = 4. Then D² = (1/2)(4) - 2*(2) + 9/2 = 2 - 4 + 4.5 = 2.5, which is 5/2. So D = sqrt(5/2) = sqrt(10)/2. So that seems correct.

But let me check if there's another way to approach this problem, maybe using geometry or something else, to confirm.

Alternatively, maybe using Lagrange multipliers? Let me try that.

We need to minimize the distance squared function $ f(x, y) = x^2 + y^2 $ subject to the constraint $ g(x, y) = y - \frac{1}{\sqrt{2}}(x^2 - 3) = 0 $.

The Lagrangian is $ \mathcal{L} = x^2 + y^2 - \lambda \left( y - \frac{1}{\sqrt{2}}(x^2 - 3) \right) $.

Taking partial derivatives:

∂L/∂x = 2x - λ*(-2x / sqrt(2)) = 0

Wait, let me compute that properly.

Wait, the constraint is $ g(x, y) = y - \frac{1}{\sqrt{2}}(x^2 - 3) = 0 $, so the gradient of g is (dg/dx, dg/dy) = (-2x / sqrt(2), 1).

Therefore, the Lagrangian equations are:

∇f = λ∇g.

So:

df/dx = 2x = λ * (-2x / sqrt(2))

df/dy = 2y = λ * 1

And the constraint equation: y = (1 / sqrt(2))(x² - 3)

So, from the first equation:

2x = λ * (-2x / sqrt(2)).

Let me solve for λ.

If x ≠ 0, we can divide both sides by 2x:

1 = λ * (-1 / sqrt(2)) => λ = -sqrt(2).

If x = 0, then the first equation becomes 0 = 0, so we have to check that case separately.

So, case 1: x ≠ 0, then λ = -sqrt(2).

From the second equation: 2y = λ => 2y = -sqrt(2) => y = -sqrt(2)/2.

Now, using the constraint equation: y = (1 / sqrt(2))(x² - 3). So:

-sqrt(2)/2 = (1 / sqrt(2))(x² - 3)

Multiply both sides by sqrt(2):

- (sqrt(2) * sqrt(2))/2 = x² - 3 => -2/2 = x² - 3 => -1 = x² - 3 => x² = 2 => x = ±sqrt(2). Which matches our previous result. So that's consistent.

Case 2: x = 0. Then from the constraint equation, y = (1 / sqrt(2))(0 - 3) = -3 / sqrt(2). Then the distance squared is 0² + y² = (9 / 2), which is the same as before. So, that's the point (0, -3 / sqrt(2)).

So, comparing the two cases, the minimal distance squared is 5/2, which is less than 9/2, so the minimal distance is sqrt(5/2) = sqrt(10)/2. Therefore, the answer is sqrt(10)/2, so a = 10, b = 2, so a + b = 12. Therefore, the answer is 12.

But wait, let me check once more. Maybe I made a mistake in simplifying sqrt(5/2). Let me confirm:

sqrt(5/2) can be written as sqrt(10)/sqrt(4) ? Wait, no. Wait sqrt(5/2) is equal to sqrt(10)/2 because:

sqrt(5)/sqrt(2) = (sqrt(5) * sqrt(2)) / (sqrt(2) * sqrt(2)) ) = sqrt(10)/2. Yes, that's correct. So sqrt(5/2) is indeed sqrt(10)/2. Therefore, the expression is in the form sqrt(a)/b where a = 10 and b = 2. Since 10 is square-free (its prime factors are 2 and 5, neither of which are squared), so that's acceptable.

But just to be thorough, let me check if there are any other critical points or if I missed something. For example, when I used calculus, I found critical points at x = 0, ±√2. The second derivative test could confirm if these are minima or maxima.

Let me compute the second derivative of f(x). Earlier, f'(x) = 2x³ - 4x. Then f''(x) = 6x² - 4.

At x = 0: f''(0) = 0 - 4 = -4 < 0. Therefore, x = 0 is a local maximum. Which makes sense, since the distance squared at x=0 is 9/2, which is larger than at x=±√2. So that's a maximum.

At x = ±√2: f''(x) = 6*( (√2)^2 ) - 4 = 6*(2) - 4 = 12 - 4 = 8 > 0. Therefore, these points are local minima. Therefore, the minimal distance squared is indeed 5/2, so the minimal distance is sqrt(5/2) = sqrt(10)/2. Therefore, the answer is 10 + 2 = 12.

But just to be 100% sure, let me check with another method. Maybe parametric equations or something else.

Alternatively, think geometrically. The graph of y = (1/√2)(x² - 3) is a parabola opening upwards. The origin is at (0, 0). The minimal distance from the origin to the parabola would be along the line connecting the origin to the closest point on the parabola. The line connecting the origin to the closest point should be perpendicular to the tangent of the parabola at that point.

So, let me try this approach. Let me suppose that the closest point is (x, y) on the parabola. Then the vector from the origin to (x, y) is (x, y). The tangent line at (x, y) has a slope equal to the derivative of the function at that point.

First, compute the derivative of y with respect to x. Given y = (1/√2)(x² - 3), so dy/dx = (2x)/√2 = √2 x.

Therefore, the slope of the tangent line at (x, y) is √2 x. Therefore, the slope of the normal line (perpendicular to tangent) is -1/(√2 x), provided that x ≠ 0.

But the line connecting the origin to (x, y) is the normal line. Therefore, the slope of this line is y/x. Therefore, we have:

y/x = -1/(√2 x)

Wait, let me write that equation:

The slope of the normal line is -1/(dy/dx) = -1/(√2 x). But also, the slope of the line from the origin to (x, y) is y/x. Therefore, equating these:

y/x = -1/(√2 x)

Multiply both sides by x (assuming x ≠ 0):

y = -1/√2

So, this gives us y = -1/√2. But since the point (x, y) lies on the parabola, we can substitute y into the equation of the parabola:

-1/√2 = (1/√2)(x² - 3)

Multiply both sides by √2:

-1 = x² - 3 => x² = 2 => x = ±√2. Which is the same result as before. Therefore, the closest points are at (±√2, -1/√2). Therefore, the distance from the origin is sqrt( (√2)^2 + (-1/√2)^2 ) = sqrt(2 + 1/2) = sqrt(5/2) = sqrt(10)/2. Same result. Therefore, this confirms the previous calculations.

Therefore, I can be confident that the minimal distance is sqrt(10)/2, so a = 10, b = 2, and a + b = 12.

But just to make sure, let me check if there's a possibility that the minimal distance occurs at another point. For example, maybe when the line from the origin is tangent to the parabola? Wait, but we already found that the normal line at the closest point passes through the origin, so that's the condition for the minimal distance. Therefore, that should be the case.

Alternatively, maybe I can parametrize the distance squared function and check its minimum. Let me try that again.

We had D² = x² + [ (x² - 3)/√2 ]² = x² + (x^4 - 6x² + 9)/2. Which simplifies to (2x² + x^4 - 6x² + 9)/2 = (x^4 - 4x² + 9)/2. Wait, hold on, earlier I had written f(x) as (1/2)x^4 - 2x² + 9/2, which is the same as (x^4 - 4x² + 9)/2. So, D² = (x^4 - 4x² + 9)/2. Then taking derivative, which is [4x³ - 8x]/2 = 2x³ - 4x, which is same as before. So, same critical points. So, same result.

Alternatively, maybe completing the square or something else? Let me see. Let me consider D² = (x^4 - 4x² + 9)/2. Let me set z = x², so z ≥ 0. Then D² = (z² - 4z + 9)/2. So, we can think of this as a quadratic in z: f(z) = (z² - 4z + 9)/2. To find the minimum of this quadratic function for z ≥ 0.

The quadratic function z² - 4z + 9 has its vertex at z = -b/(2a) = 4/(2*1) = 2. Since the coefficient of z² is positive, the minimum occurs at z = 2. Therefore, the minimum value is ( (2)^2 - 4*(2) + 9 ) / 2 = (4 - 8 + 9)/2 = 5/2. Therefore, D² = 5/2, so D = sqrt(5/2) = sqrt(10)/2. Therefore, same result. Therefore, this method also gives the same answer. Therefore, this confirms again that the minimal distance is sqrt(10)/2.

Therefore, I think this is solid. Therefore, the answer is 10 + 2 = 12.

But just to check once more, let me compute the numerical value. sqrt(10)/2 is approximately 3.1623 / 2 ≈ 1.581. Let me check the distance at x = sqrt(2). The point is (sqrt(2), y). Let me compute y: y = (1/sqrt(2)) * ( (sqrt(2))² - 3 ) = (1/sqrt(2))*(2 - 3) = (-1)/sqrt(2). Therefore, the point is (sqrt(2), -1/sqrt(2)). The distance from the origin is sqrt( (sqrt(2))² + (-1/sqrt(2))² ) = sqrt(2 + 1/2) = sqrt(5/2) ≈ 1.581, which is indeed less than the distance at x=0, which was 3/sqrt(2) ≈ 2.121. So that seems correct.

Alternatively, if I pick another point on the parabola, say x = 1. Then y = (1/sqrt(2))(1 - 3) = (-2)/sqrt(2) = -sqrt(2). The distance squared is 1² + (-sqrt(2))² = 1 + 2 = 3, which is larger than 5/2 = 2.5. So that's consistent. If I take x = 2, then y = (1/sqrt(2))(4 - 3) = 1/sqrt(2). The distance squared is 4 + (1/2) = 4.5, which is larger. So, yes, seems like the minimal distance is indeed at x = sqrt(2).

Therefore, I think I can confidently say that the answer is 12.

**Final Answer**
\boxed{12}

To find the smallest distance between the origin and a point on the graph of the function
$$
y = \frac{1}{\sqrt{2}}(x^2 - 3),
$$
we begin by using the **distance formula** between a point $(x, y)$ and the origin $(0, 0)$:

$$
D = \sqrt{x^2 + y^2}.
$$

To simplify the minimization process, we minimize the **square of the distance**, which is:

$$
D^2 = x^2 + y^2.
$$

Substituting the expression for $ y $ from the given function:

$$
D^2 = x^2 + \left( \frac{1}{\sqrt{2}}(x^2 - 3) \right)^2 = x^2 + \frac{1}{2}(x^2 - 3)^2.
$$

Expanding the square:

$$
(x^2 - 3)^2 = x^4 - 6x^2 + 9,
$$
so
$$
D^2 = x^2 + \frac{1}{2}(x^4 - 6x^2 + 9) = \frac{1}{2}x^4 - 2x^2 + \frac{9}{2}.
$$

Now, we find the critical points by taking the derivative of $ D^2 $ with respect to $ x $:

$$
\frac{d}{dx} \left( \frac{1}{2}x^4 - 2x^2 + \frac{9}{2} \right) = 2x^3 - 4x.
$$

Setting the derivative equal to zero:

$$
2x^3 - 4x = 0 \Rightarrow 2x(x^2 - 2) = 0.
$$

This gives the critical points:

$$
x = 0, \quad x = \pm \sqrt{2}.
$$

We now evaluate $ D^2 $ at these points:

- At $ x = 0 $:
  $$
  D^2 = \frac{1}{2}(0)^4 - 2(0)^2 + \frac{9}{2} = \frac{9}{2}.
  $$

- At $ x = \pm \sqrt{2} $:
  $$
  x^2 = 2, \quad x^4 = 4, \quad \text{so} \quad D^2 = \frac{1}{2}(4) - 2(2) + \frac{9}{2} = 2 - 4 + \frac{9}{2} = \frac{5}{2}.
  $$

Thus, the **minimum distance squared** is $ \frac{5}{2} $, and the **minimum distance** is:

$$
D = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}.
$$

This is in the form $ \frac{\sqrt{a}}{b} $, where $ a = 10 $ and $ b = 2 $. Since 10 is square-free (its prime factorization is $ 2 \times 5 $), the conditions of the problem are satisfied.

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### Final Answer:

$$
\boxed{12}
$$

Solution:
By the distance formula, we are trying to minimize $\sqrt{x^2+y^2}=\sqrt{x^2+(1/2)(x^4-6x^2+9)}$. In general, minimization problems like this require calculus, but one optimization method that sometimes works is to try to complete the square.  Pulling out a factor of $1/2$ from under the radical, we have \begin{align*}
\frac{1}{\sqrt{2}}\sqrt{2x^2+x^4-6x^2+9}&=\frac{1}{\sqrt{2}}\sqrt{(x^4-4x^2+4)+5} \\
&= \frac{1}{\sqrt{2}}\sqrt{(x^2-2)^2+5}.
\end{align*}This last expression is minimized when the square equals $0$, i.e. when $x=\sqrt{2}$. Then the distance is $\sqrt{5}/\sqrt{2}=\sqrt{10}/2$.  Hence the desired answer is $\boxed{12}$.