Explanation:
Okay, so I need to find the minimum value of this expression:

$$
\sqrt{x^2 + 400} + \sqrt{y^2 + 900} + \sqrt{x^2 + y^2 - 80x - 100y + 4100}
$$

where $0 \le x \le 40$ and $0 \le y \le 50$. Hmm, let me try to break this down step by step.

First, let me look at each term individually. The first term is $\sqrt{x^2 + 400}$. That looks like the distance from the point $(x, 0)$ to the point $(0, 20)$, because if I think of the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, it's $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. So if I set $(x_1, y_1) = (x, 0)$ and $(x_2, y_2) = (0, 20)$, then the distance would be $\sqrt{(x - 0)^2 + (0 - 20)^2} = \sqrt{x^2 + 400}$. So that term is the distance from $(x, 0)$ to $(0, 20)$.

Similarly, the second term is $\sqrt{y^2 + 900}$. Let me check that. If I consider $(x_1, y_1) = (0, y)$ and $(x_2, y_2) = (0, 30)$, then the distance would be $\sqrt{(0 - 0)^2 + (y - 30)^2} = \sqrt{y^2 - 60y + 900}$. Wait, that's not exactly the same as $\sqrt{y^2 + 900}$. Hmm, maybe I need to adjust my coordinates.

Wait, maybe the second term is the distance from $(0, y)$ to some other point. Let me think. If I have $\sqrt{y^2 + 900}$, that's $\sqrt{y^2 + 30^2}$. So if I consider a point $(0, y)$ and another point $(a, b)$, then the distance squared is $(0 - a)^2 + (y - b)^2 = a^2 + (y - b)^2$. If I want this to be $y^2 + 900$, then $a^2 + (y - b)^2 = y^2 + 900$. Expanding the left side: $a^2 + y^2 - 2by + b^2 = y^2 + 900$. So simplifying, $a^2 - 2by + b^2 = 900$. For this to hold for all y, the coefficients of y must be zero. So -2b = 0 => b = 0, and then a^2 + b^2 = 900 => a^2 = 900 => a = 30 or -30. Therefore, the second term is the distance from $(0, y)$ to $(30, 0)$. Wait, but if b = 0, then the point is (a, 0). So the distance from (0, y) to (30, 0) is $\sqrt{(0 - 30)^2 + (y - 0)^2} = \sqrt{900 + y^2}$. Yes! So that term is the distance from (0, y) to (30, 0). So that's interesting.

So the first term is distance from (x, 0) to (0, 20), the second term is distance from (0, y) to (30, 0). Wait, but how does that relate to the third term?

Third term is $\sqrt{x^2 + y^2 - 80x - 100y + 4100}$. Let me try to simplify that expression under the square root. Let me complete the squares for x and y terms.

Starting with $x^2 - 80x + y^2 - 100y + 4100$. Let me group the x terms and y terms:

$x^2 - 80x + y^2 - 100y + 4100$.

For the x terms: $x^2 - 80x$. Completing the square: take half of -80, which is -40, square it: 1600. So $x^2 - 80x = (x - 40)^2 - 1600$.

For the y terms: $y^2 - 100y$. Completing the square: half of -100 is -50, square is 2500. So $y^2 - 100y = (y - 50)^2 - 2500$.

Putting it all back into the expression:

$(x - 40)^2 - 1600 + (y - 50)^2 - 2500 + 4100$.

Simplify constants: -1600 -2500 + 4100 = (-4100) + 4100 = 0. So the entire expression simplifies to $(x - 40)^2 + (y - 50)^2$. Therefore, the third term is $\sqrt{(x - 40)^2 + (y - 50)^2}$. That is the distance from the point (x, y) to the point (40, 50).

So putting it all together, the original expression is:

Distance from (x, 0) to (0, 20) + Distance from (0, y) to (30, 0) + Distance from (x, y) to (40, 50).

Wait, but how do these points relate? Let me think. Maybe there's a geometric interpretation here. Let me try to visualize.

Let me denote points:

Let me consider points A(0, 20), B(30, 0), and C(40, 50). Then, the expression is:

Distance from (x, 0) to A + Distance from (0, y) to B + Distance from (x, y) to C.

But (x, 0) is a point on the x-axis between 0 and 40, and (0, y) is a point on the y-axis between 0 and 50. However, the third term is the distance from (x, y) to C(40, 50). But (x, y) is a point in the rectangle [0,40] x [0,50]. Wait, but how do these terms relate?

Alternatively, maybe there's a way to interpret this as a path or something. Let me think. Suppose we have to go from some point to another via certain points. Wait, maybe if I think of (x, 0) as a point on the x-axis, (0, y) as a point on the y-axis, and (x, y) as a point in the plane. But how do these terms combine?

Alternatively, maybe there's a way to combine these terms. Let me think. Let me consider that the first term is distance from (x, 0) to A(0, 20), the second term is distance from (0, y) to B(30, 0), and the third term is distance from (x, y) to C(40, 50). But how do these relate? Maybe if I think of (x, y) as a point connected to (x, 0) and (0, y). Wait, but (x, 0) and (0, y) are points on the axes. Maybe there's a way to think of this as a path from A to B via (x, 0) and (0, y) and then to C? Not sure.

Alternatively, maybe there's a way to use reflection or something. Sometimes in optimization problems involving distances, reflecting points helps find minimal paths.

Alternatively, maybe I can think of variables x and y as coordinates of a point (x, y) in the rectangle [0,40] x [0,50], and then the expression is:

Distance from (x, 0) to A(0, 20) + Distance from (0, y) to B(30, 0) + Distance from (x, y) to C(40, 50).

But how can I relate these? Maybe if I consider that (x, 0) and (0, y) are points on the axes, and (x, y) is a point in the plane. Maybe there's a way to combine these terms. Let me think.

Alternatively, maybe I can think of the entire expression as the sum of three distances. Let me try to visualize:

Suppose I have a point P(x, 0) on the x-axis, a point Q(0, y) on the y-axis, and a point R(x, y) in the plane. Then the expression is |PA| + |QB| + |RC|, where A is (0, 20), B is (30, 0), and C is (40, 50). But how do these points relate? Maybe there's a way to connect them?

Alternatively, maybe I can think of the expression as |PA| + |QB| + |RC|, but with P and Q related to R? Since R is (x, y), and P is (x, 0), Q is (0, y). So P is the projection of R onto the x-axis, and Q is the projection of R onto the y-axis. So maybe there's a way to think of this as a path from A to P to R to C, and from B to Q to R? Not sure.

Alternatively, maybe I can think of the entire expression as |PA| + |QB| + |RC|, and try to find the minimum over x and y. But how?

Alternatively, maybe I can consider that for fixed x and y, the expression is the sum of three distances. But how can I minimize this?

Alternatively, maybe I can use calculus. Take partial derivatives with respect to x and y, set them to zero, and solve. But that might be complicated. Let me try.

First, let me denote the function as:

f(x, y) = sqrt(x² + 400) + sqrt(y² + 900) + sqrt((x - 40)² + (y - 50)²)

So, to find the minimum, take partial derivatives ∂f/∂x and ∂f/∂y, set them to zero.

Let me compute ∂f/∂x:

First term: derivative of sqrt(x² + 400) with respect to x is (1/(2*sqrt(x² + 400)))*2x = x / sqrt(x² + 400)

Second term: derivative of sqrt(y² + 900) with respect to x is 0.

Third term: derivative of sqrt((x - 40)² + (y - 50)²) with respect to x is [2(x - 40)] / [2*sqrt((x - 40)² + (y - 50)²)] = (x - 40)/sqrt((x - 40)² + (y - 50)²)

Therefore, ∂f/∂x = x / sqrt(x² + 400) + (x - 40)/sqrt((x - 40)² + (y - 50)²)

Similarly, ∂f/∂y:

First term: derivative of sqrt(x² + 400) with respect to y is 0.

Second term: derivative of sqrt(y² + 900) with respect to y is y / sqrt(y² + 900)

Third term: derivative of sqrt((x - 40)² + (y - 50)²) with respect to y is [2(y - 50)] / [2*sqrt(...)] = (y - 50)/sqrt(...)

Therefore, ∂f/∂y = y / sqrt(y² + 900) + (y - 50)/sqrt((x - 40)² + (y - 50)²)

To find critical points, set both partial derivatives to zero:

1. x / sqrt(x² + 400) + (x - 40)/sqrt((x - 40)² + (y - 50)²) = 0

2. y / sqrt(y² + 900) + (y - 50)/sqrt((x - 40)² + (y - 50)²) = 0

This seems complicated. Let me denote D = sqrt((x - 40)² + (y - 50)²). Then equations become:

1. x / sqrt(x² + 400) + (x - 40)/D = 0

2. y / sqrt(y² + 900) + (y - 50)/D = 0

Let me rearrange equation 1:

x / sqrt(x² + 400) = - (x - 40)/D

Similarly, equation 2:

y / sqrt(y² + 900) = - (y - 50)/D

Let me square both equations to eliminate D, but I need to be careful with signs.

From equation 1:

[x / sqrt(x² + 400)]² = [(x - 40)/D]^2

Which gives:

x² / (x² + 400) = (x - 40)² / D²

Similarly, from equation 2:

y² / (y² + 900) = (y - 50)² / D²

Therefore, both expressions equal to (x - 40)² / D² and (y - 50)² / D². Therefore, we can set them equal to each other:

x² / (x² + 400) = (x - 40)² / D²

and

y² / (y² + 900) = (y - 50)² / D²

Therefore, equate the two expressions:

x² / (x² + 400) = (x - 40)² / D² = (y - 50)² / D² * [D² / (y - 50)²] * [x² / (x² + 400)] ?

Wait, maybe better to take the ratio of the two equations. Let me think.

From the first equation:

x² / (x² + 400) = (x - 40)^2 / D^2

From the second equation:

y² / (y² + 900) = (y - 50)^2 / D^2

Therefore, if I take the ratio of the left sides and the ratio of the right sides:

[x² / (x² + 400)] / [y² / (y² + 900)] = [(x - 40)^2 / D^2] / [(y - 50)^2 / D^2] = [(x - 40)^2] / [(y - 50)^2]

Therefore:

[x² (y² + 900)] / [y² (x² + 400)] = [(x - 40)^2] / [(y - 50)^2]

Cross multiplying:

x² (y² + 900) (y - 50)^2 = y² (x² + 400) (x - 40)^2

This seems very complicated. Maybe there's a better way. Alternatively, maybe there's a relationship between x and y?

Alternatively, maybe we can assume that the minimal occurs when the point (x, y) lies along some line or has some relation to the points A, B, C?

Alternatively, maybe think of the problem as a combination of paths. Let me think again about the geometric interpretation.

Original expression:

Distance from (x, 0) to A(0, 20) + Distance from (0, y) to B(30, 0) + Distance from (x, y) to C(40, 50)

But (x, 0) and (0, y) are points on the axes. Maybe if I think of (x, 0) as a point on the x-axis and (0, y) as a point on the y-axis, and (x, y) as a point in the plane. Maybe there's a way to combine these terms.

Alternatively, think of the entire expression as the sum of three distances. Maybe if I can find a path that connects these points with minimal total length. For example, if I imagine starting at A(0, 20), going to (x, 0), then to (x, y), then to C(40, 50), and also going from B(30, 0) to (0, y). But this seems too vague.

Alternatively, maybe think of the expression as |PA| + |QB| + |RC|, where P is (x, 0), Q is (0, y), R is (x, y). Maybe there's a way to relate these points. For example, if I consider that P and Q are projections of R onto the axes, then maybe there's a way to express the total distance in terms of R?

Alternatively, maybe think of the expression as |PA| + |QB| + |RC|, and try to find the minimum over all possible P, Q, R with P on x-axis, Q on y-axis, R in the rectangle. But how?

Alternatively, maybe think of the problem as optimizing over x and y, so maybe there's a way to parametrize x and y?

Alternatively, maybe consider that the third term is the distance from (x, y) to (40, 50). If I can express the first two terms in terms of (x, y), maybe?

Wait, the first term is distance from (x, 0) to (0, 20). Let me think of that as distance from (x, 0) to A(0, 20). Similarly, the second term is distance from (0, y) to B(30, 0). Maybe if I can find a relationship between these points?

Alternatively, maybe think of the entire expression as |PA| + |QB| + |RC|, and try to find a way to combine these terms. For example, if I can find a path that goes from A to P to R to C and from B to Q to R, but I don't see the connection.

Alternatively, maybe think of the problem as minimizing |PA| + |QB| + |RC|, where P is (x, 0), Q is (0, y), R is (x, y). Maybe if I fix R, then P and Q are determined? Wait, no. For a given R(x, y), P is (x, 0) and Q is (0, y). So for each R, P and Q are fixed. Therefore, the expression is |PA| + |QB| + |RC|, where P and Q are projections of R onto the axes.

Alternatively, maybe think of this as |PA| + |QB| + |RC|, and try to find the minimum over R. But how?

Alternatively, maybe think of |PA| + |RC| as something. Let me think. If I fix R, then |PA| is distance from (x, 0) to A(0, 20). If I think of (x, 0) as a point on the x-axis, then |PA| is the distance from A to the x-axis at point (x, 0). Similarly, |QB| is the distance from B to the y-axis at (0, y). But how does that help?

Alternatively, maybe think of |PA| + |RC| as the sum of two distances. If I can find a way to relate them. For example, if I consider reflecting points?

Wait, reflection is a common technique in shortest path problems. For example, the shortest path that reflects off a line can be found by reflecting the point across the line.

But here, we have three terms. Maybe if I can find a way to combine them into a single path?

Alternatively, think of the entire expression as |PA| + |QB| + |RC|. If I can find a way to express this as a path from A to B to C via some points?

Wait, maybe not. Alternatively, think of |PA| + |RC| as |PA| + |RC|, and then add |QB|. Maybe if I can find a relationship between these terms.

Alternatively, maybe think of |PA| + |RC| as |PA| + |RC|, and then consider that for fixed R, |PA| + |RC| is minimized when...? Not sure.

Alternatively, maybe think of the problem as follows: Let me consider that (x, y) is a point in the rectangle [0,40] x [0,50]. Then, the expression is:

Distance from (x, 0) to A(0, 20) + Distance from (0, y) to B(30, 0) + Distance from (x, y) to C(40, 50)

But maybe if I can find coordinates for A, B, C such that there's a relation. Let me plot these points mentally:

A(0, 20): on the y-axis, 20 units up.

B(30, 0): on the x-axis, 30 units right.

C(40, 50): in the first quadrant, 40 right, 50 up.

So, if I imagine a path that goes from A to P(x, 0) to R(x, y) to C, and also from B to Q(0, y) to R. But how does that help?

Alternatively, maybe think of the entire expression as |PA| + |RC| + |QB|. If I can find a way to combine |PA| + |RC| and |QB|.

Alternatively, maybe think of |PA| + |RC| as |PA| + |RC|, and then |QB| is another term. Maybe if I can find a way to express |PA| + |RC| in terms of something else.

Alternatively, think of |PA| + |RC| as |PA| + |RC|. Let me consider that for fixed R(x, y), |PA| is the distance from A(0, 20) to (x, 0), and |RC| is the distance from (x, y) to C(40, 50). Maybe if I can find a relationship between these two distances?

Alternatively, maybe think of |PA| + |RC| as the sum of two distances from A to P and from R to C. If I can find a path from A to P to R to C, but not sure.

Alternatively, maybe think of the entire expression as |PA| + |QB| + |RC|, and try to find a way to minimize this by considering some geometric relations.

Alternatively, maybe think of the problem as optimizing over x and y, and try to find critical points. Earlier, I had the equations:

x / sqrt(x² + 400) + (x - 40)/D = 0

and

y / sqrt(y² + 900) + (y - 50)/D = 0

Where D = sqrt((x - 40)^2 + (y - 50)^2)

Let me denote the first equation as:

x / sqrt(x² + 400) = - (x - 40)/D

Similarly, the second equation:

y / sqrt(y² + 900) = - (y - 50)/D

Let me denote the left-hand sides as some values. Let me think of them as vectors or something. Alternatively, maybe think of the ratios.

Let me consider the first equation:

Let me denote vector from A to P: from A(0, 20) to P(x, 0). The vector is (x - 0, 0 - 20) = (x, -20). The length of this vector is sqrt(x² + 400). Similarly, the vector from R(x, y) to C(40, 50) is (40 - x, 50 - y), with length D.

But in the first equation, we have x / sqrt(x² + 400) = - (x - 40)/D. Let me note that x / sqrt(x² + 400) is the x-component of the unit vector from A to P. Similarly, -(x - 40)/D is the x-component of the unit vector from R to C multiplied by -1? Wait:

The unit vector from R to C is ( (40 - x)/D, (50 - y)/D ). So the x-component is (40 - x)/D. But in the equation, we have -(x - 40)/D = (40 - x)/D. So the first equation says that the x-component of the unit vector from A to P equals the x-component of the unit vector from R to C.

Similarly, the second equation says that the y-component of the unit vector from B to Q equals the y-component of the unit vector from R to C.

Wait, let me check:

For the second equation, y / sqrt(y² + 900) is the y-component of the unit vector from B to Q. Wait, point B is (30, 0), and Q is (0, y). So vector from B to Q is (0 - 30, y - 0) = (-30, y). The length is sqrt(900 + y²). The y-component of the unit vector is y / sqrt(900 + y²). The equation says that this equals - (y - 50)/D. Which is - (y - 50)/D = (50 - y)/D. Which is the y-component of the unit vector from R to C. So the second equation says that the y-component of the unit vector from B to Q equals the y-component of the unit vector from R to C.

So, putting it together, the first equation says that the x-component of the unit vector from A to P equals the x-component of the unit vector from R to C. Similarly, the second equation says that the y-component of the unit vector from B to Q equals the y-component of the unit vector from R to C.

Therefore, this suggests that the unit vector from R to C is equal to the unit vector from A to P and also equal to the unit vector from B to Q. Wait, but the unit vector from R to C is ( (40 - x)/D, (50 - y)/D ). The unit vector from A to P is (x / sqrt(x² + 400), -20 / sqrt(x² + 400)). Wait, earlier I thought the x-component was x / sqrt(x² + 400), but actually, the unit vector from A to P is (x, -20) normalized. So the x-component is x / sqrt(x² + 400), and the y-component is -20 / sqrt(x² + 400). Similarly, the unit vector from B to Q is (-30 / sqrt(900 + y²), y / sqrt(900 + y²)).

But according to the equations, the x-component of the unit vector from A to P equals the x-component of the unit vector from R to C, and the y-component of the unit vector from B to Q equals the y-component of the unit vector from R to C. Therefore, this suggests that the unit vector from R to C is equal to the unit vector from A to P in the x-component and equal to the unit vector from B to Q in the y-component. Therefore, the unit vector from R to C has components:

x-component: x / sqrt(x² + 400) = (40 - x)/D

y-component: (50 - y)/D = y / sqrt(y² + 900)

Wait, no. Wait, the second equation says that the y-component of the unit vector from B to Q equals the y-component of the unit vector from R to C. The y-component of the unit vector from B to Q is y / sqrt(y² + 900). The y-component of the unit vector from R to C is (50 - y)/D. Therefore, according to the second equation:

y / sqrt(y² + 900) = (50 - y)/D

Similarly, from the first equation:

x / sqrt(x² + 400) = (40 - x)/D

Therefore, we have:

From first equation:

x / sqrt(x² + 400) = (40 - x)/D => D = (40 - x) * sqrt(x² + 400) / x

From second equation:

y / sqrt(y² + 900) = (50 - y)/D => D = (50 - y) * sqrt(y² + 900) / y

Therefore, equate the two expressions for D:

(40 - x) * sqrt(x² + 400) / x = (50 - y) * sqrt(y² + 900) / y

This is a relation between x and y. Let me write that:

[(40 - x)/x] * sqrt(x² + 400) = [(50 - y)/y] * sqrt(y² + 900)

This seems complicated, but maybe there's a way to find x and y that satisfy this equation. Also, remember that D is also equal to sqrt((x - 40)^2 + (y - 50)^2). So we can write:

D = sqrt((x - 40)^2 + (y - 50)^2)

But from the first equation, D = (40 - x) * sqrt(x² + 400) / x. Therefore:

sqrt((x - 40)^2 + (y - 50)^2) = (40 - x) * sqrt(x² + 400) / x

Similarly, sqrt((x - 40)^2 + (y - 50)^2) = (40 - x) * sqrt(x² + 400) / x

Note that (x - 40)^2 = (40 - x)^2, so:

sqrt((40 - x)^2 + (y - 50)^2) = (40 - x) * sqrt(x² + 400) / x

Let me square both sides to eliminate the square roots:

(40 - x)^2 + (y - 50)^2 = [(40 - x)^2 * (x² + 400)] / x²

Multiply both sides by x²:

x²[(40 - x)^2 + (y - 50)^2] = (40 - x)^2 (x² + 400)

Expand the left side:

x²(40 - x)^2 + x²(y - 50)^2 = (40 - x)^2 x² + (40 - x)^2 * 400

Subtract x²(40 - x)^2 from both sides:

x²(y - 50)^2 = (40 - x)^2 * 400

Take square roots? Or solve for y?

Let me write:

x²(y - 50)^2 = 400(40 - x)^2

Take square roots (assuming positive quantities):

|x(y - 50)| = 20|40 - x|

Since x is between 0 and 40, 40 - x is non-negative, so |40 - x| = 40 - x. Also, x is non-negative, so |x(y - 50)| = x|y - 50|. Therefore:

x|y - 50| = 20(40 - x)

Similarly, from the second equation, we can derive a similar relation. Let me check.

From the second equation, we had:

D = (50 - y) * sqrt(y² + 900) / y

And D is also sqrt((x - 40)^2 + (y - 50)^2). Therefore:

sqrt((x - 40)^2 + (y - 50)^2) = (50 - y) * sqrt(y² + 900) / y

Square both sides:

(x - 40)^2 + (y - 50)^2 = [(50 - y)^2 (y² + 900)] / y²

Multiply both sides by y²:

y²[(x - 40)^2 + (y - 50)^2] = (50 - y)^2 (y² + 900)

Expand left side:

y²(x - 40)^2 + y²(y - 50)^2 = (50 - y)^2 y² + (50 - y)^2 * 900

Subtract y²(y - 50)^2 from both sides:

y²(x - 40)^2 = (50 - y)^2 * 900

Take square roots:

|y(x - 40)| = 30|50 - y|

Since y is between 0 and 50, 50 - y is non-negative, so |50 - y| = 50 - y. Also, y is non-negative, so |y(x - 40)| = y|x - 40|. Therefore:

y|x - 40| = 30(50 - y)

But since x is between 0 and 40, x - 40 is negative or zero, so |x - 40| = 40 - x. Therefore:

y(40 - x) = 30(50 - y)

So now we have two equations:

1. x|y - 50| = 20(40 - x)

2. y(40 - x) = 30(50 - y)

Let me focus on equation 2 first. Let me solve equation 2 for y in terms of x or vice versa.

Equation 2: y(40 - x) = 30(50 - y)

Expand:

40y - xy = 1500 - 30y

Bring all terms to left side:

40y - xy - 1500 + 30y = 0 => (40y + 30y) - xy - 1500 = 0 => 70y - xy - 1500 = 0

Factor y:

y(70 - x) = 1500 => y = 1500 / (70 - x)

So y is expressed in terms of x. Now, let's substitute this into equation 1.

Equation 1: x|y - 50| = 20(40 - x)

But since y = 1500 / (70 - x), let's compute y - 50:

y - 50 = [1500 / (70 - x)] - 50 = [1500 - 50(70 - x)] / (70 - x) = [1500 - 3500 + 50x] / (70 - x) = [50x - 2000] / (70 - x) = 50(x - 40) / (70 - x)

Therefore, |y - 50| = |50(x - 40)| / |70 - x| = 50|40 - x| / |70 - x|

Since x is between 0 and 40, 40 - x is non-negative, and 70 - x is positive (since x ≤ 40 < 70). Therefore, |y - 50| = 50(40 - x) / (70 - x)

Therefore, equation 1 becomes:

x * [50(40 - x) / (70 - x)] = 20(40 - x)

Assuming 40 - x ≠ 0 (we can check x = 40 later), we can divide both sides by (40 - x):

x * [50 / (70 - x)] = 20

Multiply both sides by (70 - x):

50x = 20(70 - x)

Expand:

50x = 1400 - 20x

Add 20x to both sides:

70x = 1400 => x = 1400 / 70 = 20

So x = 20. Then, from equation 2, y = 1500 / (70 - x) = 1500 / (70 - 20) = 1500 / 50 = 30.

So we have a critical point at x = 20, y = 30. Now, we need to check if this is within the given domain. Since 0 ≤ x ≤ 40 and 0 ≤ y ≤ 50, x = 20 and y = 30 are within the domain.

Now, we need to check if this critical point is indeed a minimum. Also, we should check the boundaries, because sometimes minima occur on the edges.

But first, let's compute the value of the function at this critical point.

Compute f(20, 30):

First term: sqrt(20² + 400) = sqrt(400 + 400) = sqrt(800) = 20√2 ≈ 28.28

Second term: sqrt(30² + 900) = sqrt(900 + 900) = sqrt(1800) = 30√2 ≈ 42.426

Third term: sqrt((20 - 40)^2 + (30 - 50)^2) = sqrt((-20)^2 + (-20)^2) = sqrt(400 + 400) = sqrt(800) = 20√2 ≈ 28.28

Total: 20√2 + 30√2 + 20√2 = 70√2 ≈ 98.99

So approximately 98.99. Now, we need to check if this is indeed the minimum, or if there's a lower value on the boundaries.

But before that, let me check if x = 40 is a possible case. Earlier, when solving equation 1, we assumed 40 - x ≠ 0. Let me check x = 40.

If x = 40, then from equation 2: y = 1500 / (70 - 40) = 1500 / 30 = 50. So y = 50. Let me check if this satisfies equation 1.

Equation 1: x|y - 50| = 20(40 - x). If x = 40, then left side is 40 * |50 - 50| = 0. Right side is 20*(40 - 40) = 0. So it's satisfied. Therefore, (40, 50) is also a critical point? Wait, but when x = 40, y = 50. Let me check if this is a valid point. Since x = 40 is allowed (upper bound), and y = 50 is allowed (upper bound). So this is another critical point. Let me compute the value of the function at (40, 50):

First term: sqrt(40² + 400) = sqrt(1600 + 400) = sqrt(2000) = 10√20 ≈ 44.72

Second term: sqrt(50² + 900) = sqrt(2500 + 900) = sqrt(3400) ≈ 58.31

Third term: sqrt((40 - 40)^2 + (50 - 50)^2) = sqrt(0 + 0) = 0

Total: 10√20 + sqrt(3400) + 0 ≈ 44.72 + 58.31 ≈ 103.03, which is larger than 70√2 ≈ 98.99. So this is worse. Therefore, the critical point at (20, 30) is better.

Similarly, we should check other boundaries. The boundaries occur when x = 0, x = 40, y = 0, y = 50, or combinations. Let me check these.

First, check x = 0:

If x = 0, then the first term is sqrt(0 + 400) = 20. The third term becomes sqrt((0 - 40)^2 + (y - 50)^2) = sqrt(1600 + (y - 50)^2). The second term is sqrt(y² + 900). So total function is 20 + sqrt(y² + 900) + sqrt(1600 + (y - 50)^2). Let me find the minimum over y in [0, 50].

Let me denote this as f(y) = 20 + sqrt(y² + 900) + sqrt((y - 50)^2 + 1600). Let me try to find the minimum of this function.

Alternatively, maybe there's a geometric interpretation here. The expression sqrt(y² + 900) is distance from (0, y) to (30, 0), and sqrt((y - 50)^2 + 1600) is distance from (0, y) to (-40, 50)? Wait, sqrt((y - 50)^2 + (x - (-40))^2). Wait, if x = 0, then sqrt((0 - 40)^2 + (y - 50)^2) = distance from (0, y) to (40, 50). Wait, but earlier, the third term when x = 0 is distance from (0, y) to (40, 50). So the total expression is 20 + distance from (0, y) to (30, 0) + distance from (0, y) to (40, 50). So for x = 0, the function is 20 + [distance from (0, y) to (30, 0) + distance from (0, y) to (40, 50)]. To minimize this, we can think of it as finding a point (0, y) on the y-axis that minimizes the sum of distances to (30, 0) and (40, 50), then adding 20.

But the minimal sum of distances from a point on the y-axis to two fixed points can be found by reflecting one of the points over the y-axis and finding the straight line distance. Let me recall that technique.

To minimize the sum of distances from a point on a line to two fixed points, you can reflect one point across the line and then the minimal path is the straight line between the reflected point and the other fixed point, intersecting the line.

So, let me reflect point (30, 0) over the y-axis. The reflection of (30, 0) over the y-axis is (-30, 0). Then, the minimal sum of distances from (0, y) to (30, 0) and (40, 50) is the distance from (-30, 0) to (40, 50), which is sqrt((40 - (-30))² + (50 - 0)^2) = sqrt(70² + 50²) = sqrt(4900 + 2500) = sqrt(7400) = 10*sqrt(74) ≈ 86.02. Therefore, the minimal value of the sum is 10√74, so the total function at x = 0 would be 20 + 10√74 ≈ 20 + 86.02 ≈ 106.02, which is larger than 70√2 ≈ 98.99. Therefore, the minimum on x = 0 is worse.

Similarly, check y = 0. If y = 0, then the second term is sqrt(0 + 900) = 30. The third term becomes sqrt((x - 40)^2 + (0 - 50)^2) = sqrt((x - 40)^2 + 2500). The first term is sqrt(x² + 400). So total function is sqrt(x² + 400) + 30 + sqrt((x - 40)^2 + 2500). Let me denote this as f(x) = sqrt(x² + 400) + sqrt((x - 40)^2 + 2500) + 30. Again, similar to before, maybe we can find the minimum.

This is similar to the previous case. The expression sqrt(x² + 400) + sqrt((x - 40)^2 + 2500) is the sum of distances from (x, 0) to (0, 20) and from (x, 0) to (40, 50). To minimize this sum, we can reflect one point over the x-axis. Let me reflect (40, 50) over the x-axis to get (40, -50). Then the minimal sum is the distance from (0, 20) to (40, -50), which is sqrt((40 - 0)^2 + (-50 - 20)^2) = sqrt(1600 + 4900) = sqrt(6500) = 10*sqrt(65) ≈ 80.62. Therefore, the minimal value of the sum is 10√65, so total function is 10√65 + 30 ≈ 80.62 + 30 ≈ 110.62, which is worse than 70√2.

Now check x = 40, y = 50, which we already did, gives ≈ 103.03.

Now check other boundaries. For example, when y = 50. Let me check y = 50. Then the second term is sqrt(50² + 900) = sqrt(2500 + 900) = sqrt(3400) ≈ 58.31. The third term becomes sqrt((x - 40)^2 + (50 - 50)^2) = sqrt((x - 40)^2) = |x - 40|. Since x is between 0 and 40, this is 40 - x. The first term is sqrt(x² + 400). Therefore, total function is sqrt(x² + 400) + sqrt(3400) + (40 - x). Let me denote this as f(x) = sqrt(x² + 400) + 40 - x + sqrt(3400). To minimize this, we can minimize sqrt(x² + 400) - x + 40 + sqrt(3400). The term sqrt(x² + 400) - x can be minimized. Let me compute derivative of sqrt(x² + 400) - x with respect to x:

d/dx [sqrt(x² + 400) - x] = [x / sqrt(x² + 400)] - 1. Setting to zero:

x / sqrt(x² + 400) - 1 = 0 => x / sqrt(x² + 400) = 1 => x = sqrt(x² + 400) => x² = x² + 400 => 0 = 400, which is impossible. Therefore, the function sqrt(x² + 400) - x is decreasing? Let me check the derivative. Since [x / sqrt(x² + 400)] - 1 is always negative because x / sqrt(x² + 400) < 1 for all real x. Therefore, the derivative is negative, so the function is decreasing. Therefore, the minimum occurs at x = 40. At x = 40, sqrt(40² + 400) - 40 = sqrt(1600 + 400) - 40 = sqrt(2000) - 40 ≈ 44.72 - 40 = 4.72. Therefore, total function is ≈ 4.72 + 40 + 58.31 ≈ 103.03, same as before. So worse than 70√2.

Similarly, check y = 0, which we did earlier.

Now check other boundaries, like when x = 0 and y = 0, but that's already covered.

What about when x = 0 and y = 50? Wait, that's covered in y = 50 case.

What about when x = 40 and y = 0? Let me check. If x = 40, y = 0, then first term is sqrt(40² + 400) = sqrt(2000) ≈ 44.72, second term is sqrt(0 + 900) = 30, third term is sqrt((40 - 40)^2 + (0 - 50)^2) = 50. Total ≈ 44.72 + 30 + 50 ≈ 124.72, which is worse.

Similarly, other corners: (0, 0): first term sqrt(0 + 400) = 20, second term sqrt(0 + 900) = 30, third term sqrt( (0 - 40)^2 + (0 - 50)^2 ) = sqrt(1600 + 2500) = sqrt(4100) ≈ 64.03. Total ≈ 20 + 30 + 64.03 ≈ 114.03, worse.

So, so far, the critical point (20, 30) gives the lowest value of approximately 98.99, which is 70√2. Let me check if there are other critical points on the boundaries. For example, when x is between 0 and 40, y between 0 and 50, but maybe the minimal occurs on some edge? Wait, we checked the boundaries where x or y is at their extremes, and found that the minimal is at (20, 30). However, to be thorough, maybe check if there are other critical points on the edges.

For example, suppose we consider the case when y = 50. Wait, we did that. Similarly for x = 0, etc. What about when x and y are not at the extremes but on some edge? For example, suppose we consider the case when y = 50, but we already checked that. Similarly for other edges.

Alternatively, maybe check if the critical point (20, 30) is indeed the minimum. Let me check another point near (20, 30). For example, take x = 20, y = 30. Let me perturb x slightly. Let me take x = 20 + h, y = 30 + k, and see if the function increases.

But maybe it's easier to check if the critical point satisfies the original equations. Earlier, we found that at (20, 30), the equations are satisfied. Also, since we derived it through the critical point equations, and the boundaries give higher values, it's likely that this is the minimum.

But to be thorough, let me check another point. For example, take x = 20, y = 30. Let me compute the value again:

First term: sqrt(20² + 20²) = sqrt(800) = 20√2 ≈ 28.28

Second term: sqrt(30² + 30²) = sqrt(1800) = 30√2 ≈ 42.426

Third term: sqrt( (20 - 40)^2 + (30 - 50)^2 ) = sqrt( (-20)^2 + (-20)^2 ) = sqrt(800) = 20√2 ≈ 28.28

Total: 20√2 + 30√2 + 20√2 = 70√2 ≈ 98.99

Now, let me check another point, say x = 10, y = 20. Let me compute:

First term: sqrt(10² + 400) = sqrt(100 + 400) = sqrt(500) ≈ 22.36

Second term: sqrt(20² + 900) = sqrt(400 + 900) = sqrt(1300) ≈ 36.06

Third term: sqrt( (10 - 40)^2 + (20 - 50)^2 ) = sqrt( (-30)^2 + (-30)^2 ) = sqrt(1800) ≈ 42.43

Total ≈ 22.36 + 36.06 + 42.43 ≈ 100.85, which is higher than 70√2 ≈ 98.99.

Another point: x = 25, y = 35.

First term: sqrt(25² + 400) = sqrt(625 + 400) = sqrt(1025) ≈ 32.02

Second term: sqrt(35² + 900) = sqrt(1225 + 900) = sqrt(2125) ≈ 46.09

Third term: sqrt( (25 - 40)^2 + (35 - 50)^2 ) = sqrt( (-15)^2 + (-15)^2 ) = sqrt(450) ≈ 21.21

Total ≈ 32.02 + 46.09 + 21.21 ≈ 99.32, which is still higher than 70√2.

Another point: x = 15, y = 25.

First term: sqrt(15² + 400) = sqrt(225 + 400) = sqrt(625) = 25

Second term: sqrt(25² + 900) = sqrt(625 + 900) = sqrt(1525) ≈ 39.05

Third term: sqrt( (15 - 40)^2 + (25 - 50)^2 ) = sqrt( (-25)^2 + (-25)^2 ) = sqrt(1250) ≈ 35.36

Total ≈ 25 + 39.05 + 35.36 ≈ 99.41, still higher.

Another point: x = 20, y = 30. Let me check if there's a point where the value is lower. Suppose I take x = 20, y = 30. What if I take x = 20, y = 30. Let me check if there's a way to get lower. Maybe if I take x = 20, y = 30, which is the critical point, and since the function is convex? Not sure, but given that the value is lower than nearby points, and the critical point equations are satisfied, it's likely the minimum.

Alternatively, think geometrically. The critical point (20, 30) is such that the unit vectors from the respective points are aligned. Let me think of the geometric meaning. The unit vector from A(0, 20) to P(20, 0) is (20, -20) normalized. The vector is (20, -20), which has length sqrt(800) = 20√2. The unit vector is (20 / 20√2, -20 / 20√2) = (1/√2, -1/√2). Similarly, the unit vector from R(20, 30) to C(40, 50) is (20, 20) normalized, which is (1/√2, 1/√2). Wait, but according to the earlier equations, the x-component of the unit vector from A to P equals the x-component of the unit vector from R to C. However, here, the x-component of the unit vector from A to P is 1/√2, and the x-component of the unit vector from R to C is 1/√2. Similarly, the y-component of the unit vector from B to Q is y / sqrt(y² + 900). For y = 30, that is 30 / sqrt(900 + 900) = 30 / (30√2) = 1/√2. The y-component of the unit vector from R to C is (50 - 30)/sqrt( (20 - 40)^2 + (30 - 50)^2 ) = 20 / sqrt(400 + 400) = 20 / (20√2) = 1/√2. Therefore, both components match. Therefore, the unit vectors are aligned in a way that the direction from R to C is the same as the direction from A to P and from B to Q. This suggests that the point R(20, 30) is such that the direction from R to C is the same as the direction from A to P and from B to Q. This seems like a balance point where the directions are aligned, which is likely the minimum.

Therefore, it's reasonable to conclude that the minimum occurs at (20, 30) with value 70√2. However, let me check if there's another way to see this.

Alternatively, think of the entire expression as the sum of three distances. If I imagine that the minimal path goes through points A, B, and C with some reflections. Wait, but how?

Alternatively, think of the problem as follows: The expression is |PA| + |QB| + |RC|, where P is (x, 0), Q is (0, y), R is (x, y). If I can find a way to relate these points. For example, if I consider that P and Q are projections of R onto the axes, then maybe there's a way to express |PA| + |QB| + |RC| in terms of R.

Alternatively, think of |PA| + |RC| as |PA| + |RC|. If I can find a relationship between these two terms. For example, if I consider that |PA| + |RC| is minimized when...? Not sure.

Alternatively, think of the entire expression as |PA| + |QB| + |RC|, and consider that for the minimal value, the path from A to P to R to C and from B to Q to R should be optimized. But this is too vague.

Alternatively, since we found a critical point and checked boundaries, and it gives a lower value than boundaries, and the value is 70√2, which is approximately 98.99, and given that the problem is likely constructed to have a nice answer, 70√2 is a plausible exact value.

But let me check if there's a way to get a lower value. For example, if I consider that the three terms can be combined into a single distance. Wait, if I think of the expression as |PA| + |QB| + |RC|, and if there's a way to relate these points through some geometric configuration.

Alternatively, think of the problem as follows: Let me imagine that we have to go from A(0, 20) to P(x, 0) to R(x, y) to C(40, 50), and also from B(30, 0) to Q(0, y) to R(x, y). But how does this help?

Alternatively, think of the entire expression as |PA| + |RC| + |QB|. If I can find a way to express this as a single path. For example, if I can find a path that goes from A to P to R to C and from B to Q to R, but I don't see the connection.

Alternatively, think of the problem as optimizing over x and y, and since we found a critical point with value 70√2, and the boundaries give higher values, then this is the minimum.

Therefore, I think the minimum value is 70√2, achieved at (20, 30). Therefore, the answer is \boxed{70\sqrt{2}}.

**Final Answer**
\boxed{70\sqrt{2}}

To find the minimum value of the expression:

$$
\sqrt{x^2 + 400} + \sqrt{y^2 + 900} + \sqrt{x^2 + y^2 - 80x - 100y + 4100}
$$

for $0 \le x \le 40$ and $0 \le y \le 50$, we begin by interpreting each term geometrically.

---

### Step 1: Geometric Interpretation

We interpret each square root term as a distance:

- The first term, $\sqrt{x^2 + 400}$, is the distance from the point $(x, 0)$ to the point $(0, 20)$.
- The second term, $\sqrt{y^2 + 900}$, is the distance from the point $(0, y)$ to the point $(30, 0)$.
- The third term, $\sqrt{x^2 + y^2 - 80x - 100y + 4100}$, simplifies to $\sqrt{(x - 40)^2 + (y - 50)^2}$, which is the distance from the point $(x, y)$ to the point $(40, 50)$.

So the entire expression becomes:

$$
\text{Distance from } (x, 0) \text{ to } (0, 20) + \text{Distance from } (0, y) \text{ to } (30, 0) + \text{Distance from } (x, y) \text{ to } (40, 50)
$$

Let’s denote the points:
- $A = (0, 20)$
- $B = (30, 0)$
- $C = (40, 50)$

Then the expression becomes:

$$
|PA| + |QB| + |RC|
$$

where $P = (x, 0)$, $Q = (0, y)$, and $R = (x, y)$.

---

### Step 2: Optimization Strategy

We aim to minimize the total distance:

$$
f(x, y) = |PA| + |QB| + |RC|
$$

This is a constrained optimization problem over the rectangle $[0, 40] \times [0, 50]$. We can use calculus to find critical points, or we can use geometric insights.

---

### Step 3: Critical Point Analysis

We compute the partial derivatives of the function and set them to zero. After simplifying, we find that the critical point occurs at:

$$
x = 20, \quad y = 30
$$

This point lies within the domain $0 \le x \le 40$, $0 \le y \le 50$.

---

### Step 4: Evaluate the Function at the Critical Point

At $x = 20$, $y = 30$:

- First term: $\sqrt{20^2 + 400} = \sqrt{800} = 20\sqrt{2}$
- Second term: $\sqrt{30^2 + 900} = \sqrt{1800} = 30\sqrt{2}$
- Third term: $\sqrt{(20 - 40)^2 + (30 - 50)^2} = \sqrt{(-20)^2 + (-20)^2} = \sqrt{800} = 20\sqrt{2}$

So the total value is:

$$
20\sqrt{2} + 30\sqrt{2} + 20\sqrt{2} = 70\sqrt{2}
$$

---

### Step 5: Boundary Check

We also check the function values at the boundaries of the domain (e.g., $x = 0$, $x = 40$, $y = 0$, $y = 50$), and find that all boundary values are greater than $70\sqrt{2}$.

---

### Final Conclusion

The minimum value of the given expression is achieved at the point $(x, y) = (20, 30)$, and the minimum value is:

$$
\boxed{70\sqrt{2}}
$$

[Assistant]
Explanation:
Okay, so I need to find the minimum value of this expression:

$$
\sqrt{x^2 + 400} + \sqrt{y^2 + 900} + \sqrt{x^2 + y^2 - 80x - 100y + 4100}
$$

where $0 \le x \le 40$ and $0 \le y \le 50$. Hmm, let me try to break this down step by step.

First, let me look at each term individually. The first term is $\sqrt{x^2 + 400}$. That looks like the distance from the point $(x, 0)$ to the point $(0, 20)$, because if I think of the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, it's $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. So if I set $(x_1, y_1) = (x, 0)$ and $(x_2, y_2) = (0, 20)$, then the distance would be $\sqrt{(x - 0)^2 + (0 - 20)^2} = \sqrt{x^2 + 400}$. So that term is the distance from $(x, 0)$ to $(0, 20)$.

Similarly, the second term is $\sqrt{y^2 + 900}$. Let me check that. If I consider $(x_1, y_1) = (0, y)$ and $(x_2, y_2) = (0, 30)$, then the distance would be $\sqrt{(0 - 0)^2 + (y - 30)^2} = \sqrt{y^2 - 60y + 900}$. Wait, that's not exactly the same as $\sqrt{y^2 + 900}$. Hmm, maybe I need to adjust my coordinates.

Wait, maybe the second term is the distance from $(0, y)$ to some other point. Let me think. If I have $\sqrt{y^2 + 900}$, that's $\sqrt{y^2 + 30^2}$. So if I consider a point $(0, y)$ and another point $(a, b)$, then the distance squared is $(0 - a)^2 + (y - b)^2 = a^2 + (y - b)^2$. If I want this to be $y^2 + 900$, then $a^2 + (y - b)^2 = y^2 + 900$. Expanding the left side: $a^2 + y^2 - 2by + b^2 = y^2 + 900$. So simplifying, $a^2 - 2by + b^2 = 900$. For this to hold for all y, the coefficients of y must be zero. So -2b = 0 => b = 0, and then a^2 + b^2 = 900 => a^2 = 900 => a = 30 or -30. Therefore, the second term is the distance from $(0, y)$ to $(30, 0)$. Wait, but if b = 0, then the point is (a, 0). So the distance from (0, y) to (30, 0) is $\sqrt{(0 - 30)^2 + (y - 0)^2} = \sqrt{900 + y^2}$. Yes! So that term is the distance from (0, y) to (30, 0). So that's interesting.

So the first term is distance from (x, 0) to (0, 20), the second term is distance from (0, y) to (30, 0). Wait, but how does that relate to the third term?

Third term is $\sqrt{x^2 + y^2 - 80x - 100y + 4100}$. Let me try to simplify that expression under the square root. Let me complete the squares for x and y terms.

Starting with $x^2 - 80x + y^2 - 100y + 4100$. Let me group the x terms and y terms:

$x^2 - 80x + y^2 - 100y + 4100$.

For the x terms: $x^2 - 80x$. Completing the square: take half of -80, which is -40, square it: 1600. So $x^2 - 80x = (x - 40)^2 - 1600$.

For the y terms: $y^2 - 100y$. Completing the square: half of -100 is -50, square is 2500. So $y^2 - 100y = (y - 50)^2 - 2500$.

Putting it all back into the expression:

$(x - 40)^2 - 1600 + (y - 50)^2 - 2500 + 4100$.

Simplify constants: -1600 -2500 + 4100 = (-4100) + 4100 = 0. So the entire expression simplifies to $(x - 40)^2 + (y - 50)^2$. Therefore, the third term is $\sqrt{(x - 40)^2 + (y - 50)^2}$. That is the distance from the point (x, y) to the point (40, 50).

So putting it all together, the original expression is:

Distance from (x, 0) to (0, 20) + Distance from (0, y) to (30, 0) + Distance from (x, y) to (40, 50).

Wait, but how do these points relate? Let me think. Maybe there's a geometric interpretation here. Let me try to visualize.

Let me denote points:

Let me consider points A(0, 20), B(30, 0), and C(40, 50). Then, the expression is:

Distance from (x, 0) to A + Distance from (0, y) to B + Distance from (x, y) to C.

But (x, 0) is a point on the x-axis between 0 and 40, and (0, y) is a point on the y-axis between 0 and 50. However, the third term is the distance from (x, y) to C(40, 50). But (x, y) is a point in the rectangle [0,40] x [0,50]. Wait, but how do these terms relate?

Alternatively, maybe there's a way to interpret this as a path or something. Let me think. Suppose we have to go from some point to another via certain points. Wait, maybe if I think of (x, 0) as a point on the x-axis, (0, y) as a point on the y-axis, and (x, y) as a point in the plane. But how do these terms combine?

Alternatively, maybe there's a way to combine these terms. Let me think. Let me consider that the first term is distance from (x, 0) to A(0, 20), the second term is distance from (0, y) to B(30, 0), and the third term is distance from (x, y) to C(40, 50). But how do these relate? Maybe if I think of (x, y) as a point connected to (x, 0) and (0, y). Wait, but (x, 0) and (0, y) are points on the axes. Maybe there's a way to think of this as a path from A to B via (x, 0) and (0, y) and then to C? Not sure.

Alternatively, maybe there's a way to use reflection or something. Sometimes in optimization problems involving distances, reflecting points helps find minimal paths.

Alternatively, maybe I can think of variables x and y as coordinates of a point (x, y) in the rectangle [0,40] x [0,50], and then the expression is:

Distance from (x, 0) to A(0, 20) + Distance from (0, y) to B(30, 0) + Distance from (x, y) to C(40, 50).

But how can I relate these? Maybe if I consider that (x, 0) and (0, y) are points on the axes, and (x, y) is a point in the plane. Maybe there's a way to combine these terms. Let me think.

Alternatively, maybe I can think of the entire expression as the sum of three distances. Let me try to visualize:

Suppose I have a point P(x, 0) on the x-axis, a point Q(0, y) on the y-axis, and a point R(x, y) in the plane. Then the expression is |PA| + |QB| + |RC|, where A is (0, 20), B is (30, 0), and C is (40, 50). But how do these points relate? Maybe there's a way to connect them?

Alternatively, maybe I can think of the expression as |PA| + |QB| + |RC|, but with P and Q related to R? Since R is (x, y), and P is (x, 0), Q is (0, y). So P is the projection of R onto the x-axis, and Q is the projection of R onto the y-axis. So maybe there's a way to think of this as a path from A to P to R to C, and from B to Q to R? Not sure.

Alternatively, maybe I can think of the entire expression as |PA| + |QB| + |RC|, and try to find the minimum over x and y. But how?

Alternatively, maybe I can consider that for fixed x and y, the expression is the sum of three distances. But how can I minimize this?

Alternatively, maybe I can use calculus. Take partial derivatives with respect to x and y, set them to zero, and solve. But that might be complicated. Let me try.

First, let me denote the function as:

f(x, y) = sqrt(x² + 400) + sqrt(y² + 900) + sqrt((x - 40)² + (y - 50)²)

So, to find the minimum, take partial derivatives ∂f/∂x and ∂f/∂y, set them to zero.

Let me compute ∂f/∂x:

First term: derivative of sqrt(x² + 400) with respect to x is (1/(2*sqrt(x² + 400)))*2x = x / sqrt(x² + 400)

Second term: derivative of sqrt(y² + 900) with respect to x is 0.

Third term: derivative of sqrt((x - 40)² + (y - 50)²) with respect to x is [2(x - 40)] / [2*sqrt((x - 40)² + (y - 50)²)] = (x - 40)/sqrt((x - 40)² + (y - 50)²)

Therefore, ∂f/∂x = x / sqrt(x² + 400) + (x - 40)/sqrt((x - 40)² + (y - 50)²)

Similarly, ∂f/∂y:

First term: derivative of sqrt(x² + 400) with respect to y is 0.

Second term: derivative of sqrt(y² + 900) with respect to y is y / sqrt(y² + 900)

Third term: derivative of sqrt((x - 40)² + (y - 50)²) with respect to y is [2(y - 50)] / [2*sqrt(...)] = (y - 50)/sqrt(...)

Therefore, ∂f/∂y = y / sqrt(y² + 900) + (y - 50)/sqrt((x - 40)² + (y - 50)²)

To find critical points, set both partial derivatives to zero:

1. x / sqrt(x² + 400) + (x - 40)/sqrt((x - 40)² + (y - 50)²) = 0

2. y / sqrt(y² + 900) + (y - 50)/sqrt((x - 40)² + (y - 50)²) = 0

This seems complicated. Let me denote D = sqrt((x - 40)² + (y - 50)²). Then equations become:

1. x / sqrt(x² + 400) + (x - 40)/D = 0

2. y / sqrt(y² + 900) + (y - 50)/D = 0

Let me rearrange equation 1:

x / sqrt(x² + 400) = - (x - 40)/D

Similarly, equation 2:

y / sqrt(y² + 900) = - (y - 50)/D

Let me square both equations to eliminate D, but I need to be careful with signs.

From equation 1:

[x / sqrt(x² + 400)]² = [(x - 40)/D]^2

Which gives:

x² / (x² + 400) = (x - 40)² / D²

Similarly, from equation 2:

y² / (y² + 900) = (y - 50)² / D²

Therefore, both expressions equal to (x - 40)² / D² and (y - 50)² / D². Therefore, we can set them equal to each other:

x² / (x² + 400) = (x - 40)² / D²

and

y² / (y² + 900) = (y - 50)² / D²

Therefore, equate the two expressions:

x² / (x² + 400) = (x - 40)² / D² = (y - 50)² / D² * [D² / (y - 50)²] * [x² / (x² + 400)] ?

Wait, maybe better to take the ratio of the two equations. Let me think.

From the first equation:

x² / (x² + 400) = (x - 40)^2 / D^2

From the second equation:

y² / (y² + 900) = (y - 50)^2 / D^2

Therefore, if I take the ratio of the left sides and the ratio of the right sides:

[x² / (x² + 400)] / [y² / (y² + 900)] = [(x - 40)^2 / D^2] / [(y - 50)^2 / D^2] = [(x - 40)^2] / [(y - 50)^2]

Therefore:

[x² (y² + 900)] / [y² (x² + 400)] = [(x - 40)^2] / [(y - 50)^2]

Cross multiplying:

x² (y² + 900) (y - 50)^2 = y² (x² + 400) (x - 40)^2

This seems very complicated. Maybe there's a better way. Alternatively, maybe there's a relationship between x and y?

Alternatively, maybe we can assume that the minimal occurs when the point (x, y) lies along some line or has some relation to the points A, B, C?

Alternatively, maybe think of the problem as a combination of paths. Let me think again about the geometric interpretation.

Original expression:

Distance from (x, 0) to A(0, 20) + Distance from (0, y) to B(30, 0) + Distance from (x, y) to C(40, 50)

But (x, 0) and (0, y) are points on the axes. Maybe if I think of (x, 0) as a point on the x-axis and (0, y) as a point on the y-axis, and (x, y) as a point in the plane. Maybe there's a way to combine these terms.

Alternatively, think of the entire expression as the sum of three distances. Maybe if I can find a path that connects these points with minimal total length. For example, if I imagine starting at A(0, 20), going to (x, 0), then to (x, y), then to C(40, 50), and also going from B(30, 0) to (0, y). But this seems too vague.

Alternatively, maybe think of the expression as |PA| + |QB| + |RC|, where P is (x, 0), Q is (0, y), R is (x, y). Maybe there's a way to relate these points. For example, if I consider that P and Q are projections of R onto the axes, then maybe there's a way to express the total distance in terms of R?

Alternatively, maybe think of the expression as |PA| + |QB| + |RC|, and try to find the minimum over all possible P, Q, R with P on x-axis, Q on y-axis, R in the rectangle. But how?

Alternatively, maybe think of the problem as optimizing over x and y, so maybe there's a way to parametrize x and y?

Alternatively, maybe consider that the third term is the distance from (x, y) to (40, 50). If I can express the first two terms in terms of (x, y), maybe?

Wait, the first term is distance from (x, 0) to (0, 20). Let me think of that as distance from (x, 0) to A(0, 20). Similarly, the second term is distance from (0, y) to B(30, 0). Maybe if I can find a relationship between these points?

Alternatively, maybe think of the entire expression as |PA| + |QB| + |RC|, and try to find a way to combine these terms. For example, if I can find a path that goes from A to P to R to C and from B to Q to R, but I don't see the connection.

Alternatively, maybe think of the problem as minimizing |PA| + |QB| + |RC|, where P is (x, 0), Q is (0, y), R is (x, y). Maybe if I fix R, then P and Q are determined? Wait, no. For a given R(x, y), P is (x, 0) and Q is (0, y). So for each R, P and Q are fixed. Therefore, the expression is |PA| + |QB| + |RC|, where P and Q are projections of R onto the axes.

Alternatively, maybe think of this as |PA| + |QB| + |RC|, and try to find the minimum over R. But how?

Alternatively, maybe think of |PA| + |RC| as something. Let me think. If I fix R, then |PA| is distance from (x, 0) to A(0, 20). If I think of (x, 0) as a point on the x-axis, then |PA| is the distance from A to the x-axis at point (x, 0). Similarly, |QB| is the distance from B to the y-axis at (0, y). But how does that help?

Alternatively, maybe think of |PA| + |RC| as the sum of two distances. If I can find a way to relate them. For example, if I consider reflecting points?

Wait, reflection is a common technique in shortest path problems. For example, the shortest path that reflects off a line can be found by reflecting the point across the line.

But here, we have three terms. Maybe if I can find a way to combine them into a single path?

Alternatively, think of the entire expression as |PA| + |QB| + |RC|. If I can find a way to express this as a path from A to B to C via some points?

Wait, maybe not. Alternatively, think of |PA| + |RC| as |PA| + |RC|, and then add |QB|. Maybe if I can find a relationship between these terms.

Alternatively, maybe think of |PA| + |RC| as |PA| + |RC|, and then consider that for fixed R, |PA| + |RC| is minimized when...? Not sure.

Alternatively, maybe think of the problem as follows: Let me consider that (x, y) is a point in the rectangle [0,40] x [0,50]. Then, the expression is:

Distance from (x, 0) to A(0, 20) + Distance from (0, y) to B(30, 0) + Distance from (x, y) to C(40, 50)

But maybe if I can find coordinates for A, B, C such that there's a relation. Let me plot these points mentally:

A(0, 20): on the y-axis, 20 units up.

B(30, 0): on the x-axis, 30 units right.

C(40, 50): in the first quadrant, 40 right, 50 up.

So, if I imagine a path that goes from A to P(x, 0) to R(x, y) to C, and also from B to Q(0, y) to R. But how does that help?

Alternatively, maybe think of the entire expression as |PA| + |RC| + |QB|. If I can find a way to combine |PA| + |RC| and |QB|.

Alternatively, maybe think of |PA| + |RC| as |PA| + |RC|, and then |QB| is another term. Maybe if I can find a way to express |PA| + |RC| in terms of something else.

Alternatively, think of |PA| + |RC| as |PA| + |RC|. Let me consider that for fixed R(x, y), |PA| is the distance from A(0, 20) to (x, 0), and |RC| is the distance from (x, y) to C(40, 50). Maybe if I can find a relationship between these two distances?

Alternatively, maybe think of |PA| + |RC| as the sum of two distances from A to P and from R to C. If I can find a path from A to P to R to C, but not sure.

Alternatively, maybe think of the entire expression as |PA| + |QB| + |RC|, and try to find a way to minimize this by considering some geometric relations.

Alternatively, maybe think of the problem as optimizing over x and y, and try to find critical points. Earlier, I had the equations:

x / sqrt(x² + 400) + (x - 40)/D = 0

and

y / sqrt(y² + 900) + (y - 50)/D = 0

Where D = sqrt((x - 40)^2 + (y - 50)^2)

Let me denote the first equation as:

x / sqrt(x² + 400) = - (x - 40)/D

Similarly, the second equation:

y / sqrt(y² + 900) = - (y - 50)/D

Let me denote the left-hand sides as some values. Let me think of them as vectors or something. Alternatively, maybe think of the ratios.

Let me consider the first equation:

Let me denote vector from A to P: from A(0, 20) to P(x, 0). The vector is (x - 0, 0 - 20) = (x, -20). The length of this vector is sqrt(x² + 400). Similarly, the vector from R(x, y) to C(40, 50) is (40 - x, 50 - y), with length D.

But in the first equation, we have x / sqrt(x² + 400) = - (x - 40)/D. Let me note that x / sqrt(x² + 400) is the x-component of the unit vector from A to P. Similarly, -(x - 40)/D is the x-component of the unit vector from R to C multiplied by -1? Wait:

The unit vector from R to C is ( (40 - x)/D, (50 - y)/D ). So the x-component is (40 - x)/D. But in the equation, we have -(x - 40)/D = (40 - x)/D. So the first equation says that the x-component of the unit vector from A to P equals the x-component of the unit vector from R to C.

Similarly, the second equation says that the y-component of the unit vector from B to Q equals the y-component of the unit vector from R to C.

Wait, let me check:

For the second equation, y / sqrt(y² + 900) is the y-component of the unit vector from B to Q. Wait, point B is (30, 0), and Q is (0, y). So vector from B to Q is (0 - 30, y - 0) = (-30, y). The length is sqrt(900 + y²). The y-component of the unit vector is y / sqrt(900 + y²). The equation says that this equals - (y - 50)/D. Which is - (y - 50)/D = (50 - y)/D. Which is the y-component of the unit vector from R to C. So the second equation says that the y-component of the unit vector from B to Q equals the y-component of the unit vector from R to C.

So, putting it together, the first equation says that the x-component of the unit vector from A to P equals the x-component of the unit vector from R to C. Similarly, the second equation says that the y-component of the unit vector from B to Q equals the y-component of the unit vector from R to C.

Therefore, this suggests that the unit vector from R to C is equal to the unit vector from A to P and also equal to the unit vector from B to Q. Wait, but the unit vector from R to C is ( (40 - x)/D, (50 - y)/D ). The unit vector from A to P is (x / sqrt(x² + 400), -20 / sqrt(x² + 400)). Wait, earlier I thought the x-component was x / sqrt(x² + 400), but actually, the unit vector from A to P is (x, -20) normalized. So the x-component is x / sqrt(x² + 400), and the y-component is -20 / sqrt(x² + 400). Similarly, the unit vector from B to Q is (-30 / sqrt(900 + y²), y / sqrt(900 + y²)).

But according to the equations, the x-component of the unit vector from A to P equals the x-component of the unit vector from R to C, and the y-component of the unit vector from B to Q equals the y-component of the unit vector from R to C. Therefore, this suggests that the unit vector from R to C is equal to the unit vector from A to P in the x-component and equal to the unit vector from B to Q in the y-component. Therefore, the unit vector from R to C has components:

x-component: x / sqrt(x² + 400) = (40 - x)/D

y-component: (50 - y)/D = y / sqrt(y² + 900)

Wait, no. Wait, the second equation says that the y-component of the unit vector from B to Q equals the y-component of the unit vector from R to C. The y-component of the unit vector from B to Q is y / sqrt(y² + 900). The y-component of the unit vector from R to C is (50 - y)/D. Therefore, according to the second equation:

y / sqrt(y² + 900) = (50 - y)/D

Similarly, from the first equation:

x / sqrt(x² + 400) = (40 - x)/D

Therefore, we have:

From first equation:

x / sqrt(x² + 400) = (40 - x)/D => D = (40 - x) * sqrt(x² + 400) / x

From second equation:

y / sqrt(y² + 900) = (50 - y)/D => D = (50 - y) * sqrt(y² + 900) / y

Therefore, equate the two expressions for D:

(40 - x) * sqrt(x² + 400) / x = (50 - y) * sqrt(y² + 900) / y

This is a relation between x and y. Let me write that:

[(40 - x)/x] * sqrt(x² + 400) = [(50 - y)/y] * sqrt(y² + 900)

This seems complicated, but maybe there's a way to find x and y that satisfy this equation. Also, remember that D is also equal to sqrt((x - 40)^2 + (y - 50)^2). So we can write:

D = sqrt((x - 40)^2 + (y - 50)^2)

But from the first equation, D = (40 - x) * sqrt(x² + 400) / x. Therefore:

sqrt((x - 40)^2 + (y - 50)^2) = (40 - x) * sqrt(x² + 400) / x

Similarly, sqrt((x - 40)^2 + (y - 50)^2) = (40 - x) * sqrt(x² + 400) / x

Note that (x - 40)^2 = (40 - x)^2, so:

sqrt((40 - x)^2 + (y - 50)^2) = (40 - x) * sqrt(x² + 400) / x

Let me square both sides to eliminate the square roots:

(40 - x)^2 + (y - 50)^2 = [(40 - x)^2 * (x² + 400)] / x²

Multiply both sides by x²:

x²[(40 - x)^2 + (y - 50)^2] = (40 - x)^2 (x² + 400)

Expand the left side:

x²(40 - x)^2 + x²(y - 50)^2 = (40 - x)^2 x² + (40 - x)^2 * 400

Subtract x²(40 - x)^2 from both sides:

x²(y - 50)^2 = (40 - x)^2 * 400

Take square roots? Or solve for y?

Let me write:

x²(y - 50)^2 = 400(40 - x)^2

Take square roots (assuming positive quantities):

|x(y - 50)| = 20|40 - x|

Since x is between 0 and 40, 40 - x is non-negative, so |40 - x| = 40 - x. Also, x is non-negative, so |x(y - 50)| = x|y - 50|. Therefore:

x|y - 50| = 20(40 - x)

Similarly, from the second equation, we can derive a similar relation. Let me check.

From the second equation, we had:

D = (50 - y) * sqrt(y² + 900) / y

And D is also sqrt((x - 40)^2 + (y - 50)^2). Therefore:

sqrt((x - 40)^2 + (y - 50)^2) = (50 - y) * sqrt(y² + 900) / y

Square both sides:

(x - 40)^2 + (y - 50)^2 = [(50 - y)^2 (y² + 900)] / y²

Multiply both sides by y²:

y²[(x - 40)^2 + (y - 50)^2] = (50 - y)^2 (y² + 900)

Expand left side:

y²(x - 40)^2 + y²(y - 50)^2 = (50 - y)^2 y² + (50 - y)^2 * 900

Subtract y²(y - 50)^2 from both sides:

y²(x - 40)^2 = (50 - y)^2 * 900

Take square roots:

|y(x - 40)| = 30|50 - y|

Since y is between 0 and 50, 50 - y is non-negative, so |50 - y| = 50 - y. Also, y is non-negative, so |y(x - 40)| = y|x - 40|. Therefore:

y|x - 40| = 30(50 - y)

But since x is between 0 and 40, x - 40 is negative or zero, so |x - 40| = 40 - x. Therefore:

y(40 - x) = 30(50 - y)

So now we have two equations:

1. x|y - 50| = 20(40 - x)

2. y(40 - x) = 30(50 - y)

Let me focus on equation 2 first. Let me solve equation 2 for y in terms of x or vice versa.

Equation 2: y(40 - x) = 30(50 - y)

Expand:

40y - xy = 1500 - 30y

Bring all terms to left side:

40y - xy - 1500 + 30y = 0 => (40y + 30y) - xy - 1500 = 0 => 70y - xy - 1500 = 0

Factor y:

y(70 - x) = 1500 => y = 1500 / (70 - x)

So y is expressed in terms of x. Now, let's substitute this into equation 1.

Equation 1: x|y - 50| = 20(40 - x)

But since y = 1500 / (70 - x), let's compute y - 50:

y - 50 = [1500 / (70 - x)] - 50 = [1500 - 50(70 - x)] / (70 - x) = [1500 - 3500 + 50x] / (70 - x) = [50x - 2000] / (70 - x) = 50(x - 40) / (70 - x)

Therefore, |y - 50| = |50(x - 40)| / |70 - x| = 50|40 - x| / |70 - x|

Since x is between 0 and 40, 40 - x is non-negative, and 70 - x is positive (since x ≤ 40 < 70). Therefore, |y - 50| = 50(40 - x) / (70 - x)

Therefore, equation 1 becomes:

x * [50(40 - x) / (70 - x)] = 20(40 - x)

Assuming 40 - x ≠ 0 (we can check x = 40 later), we can divide both sides by (40 - x):

x * [50 / (70 - x)] = 20

Multiply both sides by (70 - x):

50x = 20(70 - x)

Expand:

50x = 1400 - 20x

Add 20x to both sides:

70x = 1400 => x = 1400 / 70 = 20

So x = 20. Then, from equation 2, y = 1500 / (70 - x) = 1500 / (70 - 20) = 1500 / 50 = 30.

So we have a critical point at x = 20, y = 30. Now, we need to check if this is within the given domain. Since 0 ≤ x ≤ 40 and 0 ≤ y ≤ 50, x = 20 and y = 30 are within the domain.

Now, we need to check if this critical point is indeed a minimum. Also, we should check the boundaries, because sometimes minima occur on the edges.

But first, let's compute the value of the function at this critical point.

Compute f(20, 30):

First term: sqrt(20² + 400) = sqrt(400 + 400) = sqrt(800) = 20√2 ≈ 28.28

Second term: sqrt(30² + 900) = sqrt(900 + 900) = sqrt(1800) = 30√2 ≈ 42.426

Third term: sqrt((20 - 40)^2 + (30 - 50)^2) = sqrt((-20)^2 + (-20)^2) = sqrt(400 + 400) = sqrt(800) = 20√2 ≈ 28.28

Total: 20√2 + 30√2 + 20√2 = 70√2 ≈ 98.99

So approximately 98.99. Now, we need to check if this is indeed the minimum, or if there's a lower value on the boundaries.

But before that, let me check if x = 40 is a possible case. Earlier, when solving equation 1, we assumed 40 - x ≠ 0. Let me check x = 40.

If x = 40, then from equation 2: y = 1500 / (70 - 40) = 1500 / 30 = 50. So y = 50. Let me check if this satisfies equation 1.

Equation 1: x|y - 50| = 20(40 - x). If x = 40, then left side is 40 * |50 - 50| = 0. Right side is 20*(40 - 40) = 0. So it's satisfied. Therefore, (40, 50) is also a critical point? Wait, but when x = 40, y = 50. Let me check if this is a valid point. Since x = 40 is allowed (upper bound), and y = 50 is allowed (upper bound). So this is another critical point. Let me compute the value of the function at (40, 50):

First term: sqrt(40² + 400) = sqrt(1600 + 400) = sqrt(2000) = 10√20 ≈ 44.72

Second term: sqrt(50² + 900) = sqrt(2500 + 900) = sqrt(3400) ≈ 58.31

Third term: sqrt((40 - 40)^2 + (50 - 50)^2) = sqrt(0 + 0) = 0

Total: 10√20 + sqrt(3400) + 0 ≈ 44.72 + 58.31 ≈ 103.03, which is larger than 70√2 ≈ 98.99. So this is worse. Therefore, the critical point at (20, 30) is better.

Similarly, we should check other boundaries. The boundaries occur when x = 0, x = 40, y = 0, y = 50, or combinations. Let me check these.

First, check x = 0:

If x = 0, then the first term is sqrt(0 + 400) = 20. The third term becomes sqrt((0 - 40)^2 + (y - 50)^2) = sqrt(1600 + (y - 50)^2). The second term is sqrt(y² + 900). So total function is 20 + sqrt(y² + 900) + sqrt(1600 + (y - 50)^2). Let me find the minimum over y in [0, 50].

Let me denote this as f(y) = 20 + sqrt(y² + 900) + sqrt((y - 50)^2 + 1600). Let me try to find the minimum of this function.

Alternatively, maybe there's a geometric interpretation here. The expression sqrt(y² + 900) is distance from (0, y) to (30, 0), and sqrt((y - 50)^2 + 1600) is distance from (0, y) to (-40, 50)? Wait, sqrt((y - 50)^2 + (x - (-40))^2). Wait, if x = 0, then sqrt((0 - 40)^2 + (y - 50)^2) = distance from (0, y) to (40, 50). Wait, but earlier, the third term when x = 0 is distance from (0, y) to (40, 50). So the total expression is 20 + distance from (0, y) to (30, 0) + distance from (0, y) to (40, 50). So for x = 0, the function is 20 + [distance from (0, y) to (30, 0) + distance from (0, y) to (40, 50)]. To minimize this, we can think of it as finding a point (0, y) on the y-axis that minimizes the sum of distances to (30, 0) and (40, 50), then adding 20.

But the minimal sum of distances from a point on the y-axis to two fixed points can be found by reflecting one of the points over the y-axis and finding the straight line distance. Let me recall that technique.

To minimize the sum of distances from a point on a line to two fixed points, you can reflect one point across the line and then the minimal path is the straight line between the reflected point and the other fixed point, intersecting the line.

So, let me reflect point (30, 0) over the y-axis. The reflection of (30, 0) over the y-axis is (-30, 0). Then, the minimal sum of distances from (0, y) to (30, 0) and (40, 50) is the distance from (-30, 0) to (40, 50), which is sqrt((40 - (-30))² + (50 - 0)^2) = sqrt(70² + 50²) = sqrt(4900 + 2500) = sqrt(7400) = 10*sqrt(74) ≈ 86.02. Therefore, the minimal value of the sum is 10√74, so the total function at x = 0 would be 20 + 10√74 ≈ 20 + 86.02 ≈ 106.02, which is larger than 70√2 ≈ 98.99. Therefore, the minimum on x = 0 is worse.

Similarly, check y = 0. If y = 0, then the second term is sqrt(0 + 900) = 30. The third term becomes sqrt((x - 40)^2 + (0 - 50)^2) = sqrt((x - 40)^2 + 2500). The first term is sqrt(x² + 400). So total function is sqrt(x² + 400) + 30 + sqrt((x - 40)^2 + 2500). Let me denote this as f(x) = sqrt(x² + 400) + sqrt((x - 40)^2 + 2500) + 30. Again, similar to before, maybe we can find the minimum.

This is similar to the previous case. The expression sqrt(x² + 400) + sqrt((x - 40)^2 + 2500) is the sum of distances from (x, 0) to (0, 20) and from (x, 0) to (40, 50). To minimize this sum, we can reflect one point over the x-axis. Let me reflect (40, 50) over the x-axis to get (40, -50). Then the minimal sum is the distance from (0, 20) to (40, -50), which is sqrt((40 - 0)^2 + (-50 - 20)^2) = sqrt(1600 + 4900) = sqrt(6500) = 10*sqrt(65) ≈ 80.62. Therefore, the minimal value of the sum is 10√65, so total function is 10√65 + 30 ≈ 80.62 + 30 ≈ 110.62, which is worse than 70√2.

Now check x = 40, y = 50, which we already did, gives ≈ 103.03.

Now check other boundaries. For example, when y = 50. Let me check y = 50. Then the second term is sqrt(50² + 900) = sqrt(2500 + 900) = sqrt(3400) ≈ 58.31. The third term becomes sqrt((x - 40)^2 + (50 - 50)^2) = sqrt((x - 40)^2) = |x - 40|. Since x is between 0 and 40, this is 40 - x. The first term is sqrt(x² + 400). Therefore, total function is sqrt(x² + 400) + sqrt(3400) + (40 - x). Let me denote this as f(x) = sqrt(x² + 400) + 40 - x + sqrt(3400). To minimize this, we can minimize sqrt(x² + 400) - x + 40 + sqrt(3400). The term sqrt(x² + 400) - x can be minimized. Let me compute derivative of sqrt(x² + 400) - x with respect to x:

d/dx [sqrt(x² + 400) - x] = [x / sqrt(x² + 400)] - 1. Setting to zero:

x / sqrt(x² + 400) - 1 = 0 => x / sqrt(x² + 400) = 1 => x = sqrt(x² + 400) => x² = x² + 400 => 0 = 400, which is impossible. Therefore, the function sqrt(x² + 400) - x is decreasing? Let me check the derivative. Since [x / sqrt(x² + 400)] - 1 is always negative because x / sqrt(x² + 400) < 1 for all real x. Therefore, the derivative is negative, so the function is decreasing. Therefore, the minimum occurs at x = 40. At x = 40, sqrt(40² + 400) - 40 = sqrt(1600 + 400) - 40 = sqrt(2000) - 40 ≈ 44.72 - 40 = 4.72. Therefore, total function is ≈ 4.72 + 40 + 58.31 ≈ 103.03, same as before. So worse than 70√2.

Similarly, check y = 0, which we did earlier.

Now check other boundaries, like when x = 0 and y = 0, but that's already covered.

What about when x = 0 and y = 50? Wait, that's covered in y = 50 case.

What about when x = 40 and y = 0? Let me check. If x = 40, y = 0, then first term is sqrt(40² + 400) = sqrt(2000) ≈ 44.72, second term is sqrt(0 + 900) = 30, third term is sqrt((40 - 40)^2 + (0 - 50)^2) = 50. Total ≈ 44.72 + 30 + 50 ≈ 124.72, which is worse.

Similarly, other corners: (0, 0): first term sqrt(0 + 400) = 20, second term sqrt(0 + 900) = 30, third term sqrt( (0 - 40)^2 + (0 - 50)^2 ) = sqrt(1600 + 2500) = sqrt(4100) ≈ 64.03. Total ≈ 20 + 30 + 64.03 ≈ 114.03, worse.

So, so far, the critical point (20, 30) gives the lowest value of approximately 98.99, which is 70√2. Let me check if there are other critical points on the boundaries. For example, when x is between 0 and 40, y between 0 and 50, but maybe the minimal occurs on some edge? Wait, we checked the boundaries where x or y is at their extremes, and found that the minimal is at (20, 30). However, to be thorough, maybe check if there are other critical points on the edges.

For example, suppose we consider the case when y = 50. Wait, we did that. Similarly for x = 0, etc. What about when x and y are not at the extremes but on some edge? For example, suppose we consider the case when y = 50, but we already checked that. Similarly for other edges.

Alternatively, maybe check if the critical point (20, 30) is indeed the minimum. Let me check another point near (20, 30). For example, take x = 20, y = 30. Let me perturb x slightly. Let me take x = 20 + h, y = 30 + k, and see if the function increases.

But maybe it's easier to check if the critical point satisfies the original equations. Earlier, we found that at (20, 30), the equations are satisfied. Also, since we derived it through the critical point equations, and the boundaries give higher values, it's likely that this is the minimum.

But to be thorough, let me check another point. For example, take x = 20, y = 30. Let me compute the value again:

First term: sqrt(20² + 20²) = sqrt(800) = 20√2 ≈ 28.28

Second term: sqrt(30² + 30²) = sqrt(1800) = 30√2 ≈ 42.426

Third term: sqrt( (20 - 40)^2 + (30 - 50)^2 ) = sqrt( (-20)^2 + (-20)^2 ) = sqrt(800) = 20√2 ≈ 28.28

Total: 20√2 + 30√2 + 20√2 = 70√2 ≈ 98.99

Now, let me check another point, say x = 10, y = 20. Let me compute:

First term: sqrt(10² + 400) = sqrt(100 + 400) = sqrt(500) ≈ 22.36

Second term: sqrt(20² + 900) = sqrt(400 + 900) = sqrt(1300) ≈ 36.06

Third term: sqrt( (10 - 40)^2 + (20 - 50)^2 ) = sqrt( (-30)^2 + (-30)^2 ) = sqrt(1800) ≈ 42.43

Total ≈ 22.36 + 36.06 + 42.43 ≈ 100.85, which is higher than 70√2 ≈ 98.99.

Another point: x = 25, y = 35.

First term: sqrt(25² + 400) = sqrt(625 + 400) = sqrt(1025) ≈ 32.02

Second term: sqrt(35² + 900) = sqrt(1225 + 900) = sqrt(2125) ≈ 46.09

Third term: sqrt( (25 - 40)^2 + (35 - 50)^2 ) = sqrt( (-15)^2 + (-15)^2 ) = sqrt(450) ≈ 21.21

Total ≈ 32.02 + 46.09 + 21.21 ≈ 99.32, which is still higher than 70√2.

Another point: x = 15, y = 25.

First term: sqrt(15² + 400) = sqrt(225 + 400) = sqrt(625) = 25

Second term: sqrt(25² + 900) = sqrt(625 + 900) = sqrt(1525) ≈ 39.05

Third term: sqrt( (15 - 40)^2 + (25 - 50)^2 ) = sqrt( (-25)^2 + (-25)^2 ) = sqrt(1250) ≈ 35.36

Total ≈ 25 + 39.05 + 35.36 ≈ 99.41, still higher.

Another point: x = 20, y = 30. Let me check if there's a point where the value is lower. Suppose I take x = 20, y = 30. What if I take x = 20, y = 30. Let me check if there's a way to get lower. Maybe if I take x = 20, y = 30, which is the critical point, and since the function is convex? Not sure, but given that the value is lower than nearby points, and the critical point equations are satisfied, it's likely the minimum.

Alternatively, think geometrically. The critical point (20, 30) is such that the unit vectors from the respective points are aligned. Let me think of the geometric meaning. The unit vector from A(0, 20) to P(20, 0) is (20, -20) normalized. The vector is (20, -20), which has length sqrt(800) = 20√2. The unit vector is (20 / 20√2, -20 / 20√2) = (1/√2, -1/√2). Similarly, the unit vector from R(20, 30) to C(40, 50) is (20, 20) normalized, which is (1/√2, 1/√2). Wait, but according to the earlier equations, the x-component of the unit vector from A to P equals the x-component of the unit vector from R to C. However, here, the x-component of the unit vector from A to P is 1/√2, and the x-component of the unit vector from R to C is 1/√2. Similarly, the y-component of the unit vector from B to Q is y / sqrt(y² + 900). For y = 30, that is 30 / sqrt(900 + 900) = 30 / (30√2) = 1/√2. The y-component of the unit vector from R to C is (50 - 30)/sqrt( (20 - 40)^2 + (30 - 50)^2 ) = 20 / sqrt(400 + 400) = 20 / (20√2) = 1/√2. Therefore, both components match. Therefore, the unit vectors are aligned in a way that the direction from R to C is the same as the direction from A to P and from B to Q. This suggests that the point R(20, 30) is such that the direction from R to C is the same as the direction from A to P and from B to Q. This seems like a balance point where the directions are aligned, which is likely the minimum.

Therefore, it's reasonable to conclude that the minimum occurs at (20, 30) with value 70√2. However, let me check if there's another way to see this.

Alternatively, think of the entire expression as the sum of three distances. If I imagine that the minimal path goes through points A, B, and C with some reflections. Wait, but how?

Alternatively, think of the problem as follows: The expression is |PA| + |QB| + |RC|, where P is (x, 0), Q is (0, y), R is (x, y). If I can find a way to relate these points. For example, if I consider that P and Q are projections of R onto the axes, then maybe there's a way to express |PA| + |QB| + |RC| in terms of R.

Alternatively, think of |PA| + |RC| as |PA| + |RC|. If I can find a relationship between these two terms. For example, if I consider that |PA| + |RC| is minimized when...? Not sure.

Alternatively, think of the entire expression as |PA| + |QB| + |RC|, and consider that for the minimal value, the path from A to P to R to C and from B to Q to R should be optimized. But this is too vague.

Alternatively, since we found a critical point and checked boundaries, and it gives a lower value than boundaries, and the value is 70√2, which is approximately 98.99, and given that the problem is likely constructed to have a nice answer, 70√2 is a plausible exact value.

But let me check if there's a way to get a lower value. For example, if I consider that the three terms can be combined into a single distance. Wait, if I think of the expression as |PA| + |QB| + |RC|, and if there's a way to relate these points through some geometric configuration.

Alternatively, think of the problem as follows: Let me imagine that we have to go from A(0, 20) to P(x, 0) to R(x, y) to C(40, 50), and also from B(30, 0) to Q(0, y) to R(x, y). But how does this help?

Alternatively, think of the entire expression as |PA| + |RC| + |QB|. If I can find a way to express this as a single path. For example, if I can find a path that goes from A to P to R to C and from B to Q to R, but I don't see the connection.

Alternatively, think of the problem as optimizing over x and y, and since we found a critical point with value 70√2, and the boundaries give higher values, then this is the minimum.

Therefore, I think the minimum value is 70√2, achieved at (20, 30). Therefore, the answer is \boxed{70\sqrt{2}}.

**Final Answer**
\boxed{70\sqrt{2}}

To find the minimum value of the expression:

$$
\sqrt{x^2 + 400} + \sqrt{y^2 + 900} + \sqrt{x^2 + y^2 - 80x - 100y + 4100}
$$

for $0 \le x \le 40$ and $0 \le y \le 50$, we begin by interpreting each term geometrically.

---

### Step 1: Geometric Interpretation

We interpret each square root term as a distance:

- The first term, $\sqrt{x^2 + 400}$, is the distance from the point $(x, 0)$ to the point $(0, 20)$.
- The second term, $\sqrt{y^2 + 900}$, is the distance from the point $(0, y)$ to the point $(30, 0)$.
- The third term, $\sqrt{x^2 + y^2 - 80x - 100y + 4100}$, simplifies to $\sqrt{(x - 40)^2 + (y - 50)^2}$, which is the distance from the point $(x, y)$ to the point $(40, 50)$.

So the entire expression becomes:

$$
\text{Distance from } (x, 0) \text{ to } (0, 20) + \text{Distance from } (0, y) \text{ to } (30, 0) + \text{Distance from } (x, y) \text{ to } (40, 50)
$$

Let’s denote the points:
- $A = (0, 20)$
- $B = (30, 0)$
- $C = (40, 50)$

Then the expression becomes:

$$
|PA| + |QB| + |RC|
$$

where $P = (x, 0)$, $Q = (0, y)$, and $R = (x, y)$.

---

### Step 2: Optimization Strategy

We aim to minimize the total distance:

$$
f(x, y) = |PA| + |QB| + |RC|
$$

This is a constrained optimization problem over the rectangle $[0, 40] \times [0, 50]$. We can use calculus to find critical points, or we can use geometric insights.

---

### Step 3: Critical Point Analysis

We compute the partial derivatives of the function and set them to zero. After simplifying, we find that the critical point occurs at:

$$
x = 20, \quad y = 30
$$

This point lies within the domain $0 \le x \le 40$, $0 \le y \le 50$.

---

### Step 4: Evaluate the Function at the Critical Point

At $x = 20$, $y = 30$:

- First term: $\sqrt{20^2 + 400} = \sqrt{800} = 20\sqrt{2}$
- Second term: $\sqrt{30^2 + 900} = \sqrt{1800} = 30\sqrt{2}$
- Third term: $\sqrt{(20 - 40)^2 + (30 - 50)^2} = \sqrt{(-20)^2 + (-20)^2} = \sqrt{800} = 20\sqrt{2}$

So the total value is:

$$
20\sqrt{2} + 30\sqrt{2} + 20\sqrt{2} = 70\sqrt{2}
$$

---

### Step 5: Boundary Check

We also check the function values at the boundaries of the domain (e.g., $x = 0$, $x = 40$, $y = 0$, $y = 50$), and find that all boundary values are greater than $70\sqrt{2}$.

---

### Final Conclusion

The minimum value of the given expression is achieved at the point $(x, y) = (20, 30)$, and the minimum value is:

$$
\boxed{70\sqrt{2}}
$$

Solution:
Completing the square in $x$ and $y,$ the expression becomes
\[\sqrt{x^2 + 400} + \sqrt{y^2 + 900} + \sqrt{(x - 40)^2 + (y - 50)^2} = \sqrt{x^2 + 400} + \sqrt{y^2 + 900} + \sqrt{(40 - x)^2 + (50 - y)^2}.\]By QM-AM,
\begin{align*}
\sqrt{\frac{x^2 + 400}{2}} &\ge \frac{x + 20}{2}, \\
\sqrt{\frac{y^2 + 900}{2}} &\ge \frac{y + 30}{2}, \\
\sqrt{\frac{(40 - x)^2 + (50 - y)^2}{2}} &\ge \frac{(40 - x) + (50 - y)}{2},
\end{align*}so
\begin{align*}
&\sqrt{x^2 + 400} + \sqrt{y^2 + 900} + \sqrt{(40 - x)^2 + (50 - y)^2} \\
&\ge \sqrt{2} \cdot \frac{x + 20}{2} + \sqrt{2} \cdot \frac{y + 30}{2} + \sqrt{2} \cdot \frac{(40 - x) + (50 - y)}{2} \\
&= 70 \sqrt{2}.
\end{align*}Equality occurs when $x = 20$ and $y = 30,$ so the minimum value is $\boxed{70 \sqrt{2}}.$


Explanation:
Okay, so Bill is walking in different directions, and I need to figure out how far he is from his starting point in a straight line. Let me try to visualize this.

First, he walks 1/2 mile south. Then he goes 3/4 mile east, and then another 1/2 mile south. Hmm, so maybe I can use coordinates to track his position? Let me think. If I consider his starting point as the origin (0,0), then moving south would be in the negative y-direction, and east would be the positive x-direction.

Let me set up a coordinate system. Let's say the starting point is (0,0). When he walks south, his y-coordinate decreases. So, first, he walks 1/2 mile south. That would take him to (0, -1/2). Then he walks 3/4 mile east, which is along the positive x-axis. So adding 3/4 to his x-coordinate, making his position (3/4, -1/2). Then he walks another 1/2 mile south, so his y-coordinate decreases by another 1/2. That would make his final position (3/4, -1/2 - 1/2) = (3/4, -1).

Now, I need to find the straight-line distance from his starting point (0,0) to his final position (3/4, -1). To find the straight-line distance, I can use the distance formula. The distance between two points (x1, y1) and (x2, y2) is sqrt[(x2 - x1)^2 + (y2 - y1)^2].

Plugging in the values, the distance would be sqrt[(3/4 - 0)^2 + (-1 - 0)^2] = sqrt[(9/16) + 1]. Let me compute that.

First, 9/16 is 0.5625, and 1 is 1. So adding them together gives 0.5625 + 1 = 1.5625. Then the square root of 1.5625 is... Hmm, sqrt(1.5625). Let me think. I know that 1.25 squared is 1.5625 because 1.25 * 1.25 = 1.5625. So the distance is 1.25 miles.

Wait, but the question says to express the answer as a decimal to the nearest hundredth. 1.25 is already to the nearest hundredth, right? Because the hundredth place is the second decimal place. 1.25 is 1.25, so that's 1.25.

But let me double-check my steps to make sure I didn't make a mistake.

Starting at (0,0). South 1/2 mile: (0, -0.5). Then east 3/4 mile: (0.75, -0.5). Then south another 1/2 mile: (0.75, -1.0). Then distance from (0,0) is sqrt(0.75^2 + (-1.0)^2).

Calculating 0.75 squared: 0.75 * 0.75 = 0.5625. Then (-1.0)^2 is 1.0. Adding them gives 1.5625. Square root of that is indeed 1.25.

So 1.25 miles. But wait, 1.25 is exact, so as a decimal, it's already to the nearest hundredth. So the answer is 1.25.

But let me check if there's another way to think about this. Maybe using right triangles?

He walked south twice, so total south distance is 1/2 + 1/2 = 1 mile south. Then east 3/4 mile. So if I imagine a right triangle where one leg is 1 mile south and the other leg is 3/4 mile east. Then the hypotenuse would be the straight-line distance.

So hypotenuse squared is (1)^2 + (3/4)^2 = 1 + 9/16 = 25/16. Wait, 1 is 16/16, so 16/16 + 9/16 = 25/16. Then square root of 25/16 is 5/4, which is 1.25. Yep, same answer.

So that confirms it. So the answer is 1.25.

But just to make sure, maybe I made a mistake in the coordinate system? Let me think again.

If he goes south first, then east, then south again. So the total displacement is 1 mile south and 3/4 mile east. So the straight line is the hypotenuse of a right triangle with legs 1 and 0.75. So yes, same as before.

Alternatively, if I had messed up the directions? For example, if I thought that going south twice would be different? But no, south is straight down, so adding both south movements.

Alternatively, maybe I should have considered the east movement as negative? No, because east is positive x-direction.

Wait, but in coordinates, if you go south, that's negative y, east is positive x. So the final position is (3/4, -1). So the distance is sqrt( (3/4)^2 + (-1)^2 ). Which is sqrt(9/16 + 1) = sqrt(25/16) = 5/4 = 1.25.

Yes, that's correct.

So I think the answer is 1.25. Therefore, expressed as a decimal to the nearest hundredth, it's 1.25.

But just to check, maybe the problem expects something else? Let me check the problem again.

"Bill walks 1/2 mile south, then 3/4 mile east, and finally 1/2 mile south. How many miles is he, in a direct line, from his starting point? Express your answer as a decimal to the nearest hundredth."

No, that's exactly what I did. So I think 1.25 is correct.

Alternatively, maybe I need to convert fractions to decimals first? Let me try that approach.

First leg: 1/2 mile south is 0.5 miles. Then 3/4 mile east is 0.75 miles. Then another 0.5 miles south. So total south is 0.5 + 0.5 = 1.0 miles. So displacement is 1.0 miles south and 0.75 miles east. Then distance is sqrt(1.0^2 + 0.75^2) = sqrt(1 + 0.5625) = sqrt(1.5625) = 1.25.

Same result.

Therefore, I think the answer is 1.25.

But just to make sure, maybe I can think of it as vectors.

First vector: south 1/2 mile. So in vector terms, that's (0, -1/2). Then east 3/4 mile is (3/4, 0). Then south 1/2 mile is (0, -1/2). Adding all vectors together:

(0 + 3/4 + 0, -1/2 + 0 -1/2) = (3/4, -1). So same as before.

Then magnitude of the vector is sqrt( (3/4)^2 + (-1)^2 ) = same as before.

So yeah, 1.25.

Therefore, I think the answer is 1.25.

But just to check if there's any possible mistake in calculation. Let me compute sqrt(1.5625).

Well, 1.25 squared is 1.5625. Because 1.2 squared is 1.44, 1.25 squared is (1 + 0.25)^2 = 1 + 0.5 + 0.0625 = 1.5625. So yes, sqrt(1.5625) is 1.25.

Therefore, the answer is 1.25.

I think that's solid. So I can confidently say the answer is 1.25.

**Final Answer**
\boxed{1.25}

To determine how far Bill is from his starting point after walking in different directions, we can model his movements using a coordinate system and apply the **distance formula**.

---

### Step 1: Assign Coordinates

Let’s place Bill’s starting point at the origin:
$$
(0, 0)
$$

- **First movement**: Bill walks **1/2 mile south**.
  In a coordinate system, moving south decreases the y-coordinate.
  So, his new position is:
  $$
  (0, -\frac{1}{2})
  $$

- **Second movement**: Bill walks **3/4 mile east**.
  Moving east increases the x-coordinate.
  So, his new position is:
  $$
  \left(\frac{3}{4}, -\frac{1}{2}\right)
  $$

- **Third movement**: Bill walks **1/2 mile south** again.
  This further decreases the y-coordinate.
  So, his final position is:
  $$
  \left(\frac{3}{4}, -\frac{1}{2} - \frac{1}{2}\right) = \left(\frac{3}{4}, -1\right)
  $$

---

### Step 2: Use the Distance Formula

We now compute the straight-line distance from the starting point $(0, 0)$ to the final position $\left(\frac{3}{4}, -1\right)$ using the distance formula:

$$
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

Substitute the coordinates:

$$
\text{Distance} = \sqrt{\left(\frac{3}{4} - 0\right)^2 + (-1 - 0)^2} = \sqrt{\left(\frac{3}{4}\right)^2 + (-1)^2}
$$

$$
= \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{25}{16}} = \frac{5}{4} = 1.25
$$

---

### Final Answer

$$
\boxed{1.25}
$$

[Assistant]
Explanation:
Okay, so Bill is walking in different directions, and I need to figure out how far he is from his starting point in a straight line. Let me try to visualize this.

First, he walks 1/2 mile south. Then he goes 3/4 mile east, and then another 1/2 mile south. Hmm, so maybe I can use coordinates to track his position? Let me think. If I consider his starting point as the origin (0,0), then moving south would be in the negative y-direction, and east would be the positive x-direction.

Let me set up a coordinate system. Let's say the starting point is (0,0). When he walks south, his y-coordinate decreases. So, first, he walks 1/2 mile south. That would take him to (0, -1/2). Then he walks 3/4 mile east, which is along the positive x-axis. So adding 3/4 to his x-coordinate, making his position (3/4, -1/2). Then he walks another 1/2 mile south, so his y-coordinate decreases by another 1/2. That would make his final position (3/4, -1/2 - 1/2) = (3/4, -1).

Now, I need to find the straight-line distance from his starting point (0,0) to his final position (3/4, -1). To find the straight-line distance, I can use the distance formula. The distance between two points (x1, y1) and (x2, y2) is sqrt[(x2 - x1)^2 + (y2 - y1)^2].

Plugging in the values, the distance would be sqrt[(3/4 - 0)^2 + (-1 - 0)^2] = sqrt[(9/16) + 1]. Let me compute that.

First, 9/16 is 0.5625, and 1 is 1. So adding them together gives 0.5625 + 1 = 1.5625. Then the square root of 1.5625 is... Hmm, sqrt(1.5625). Let me think. I know that 1.25 squared is 1.5625 because 1.25 * 1.25 = 1.5625. So the distance is 1.25 miles.

Wait, but the question says to express the answer as a decimal to the nearest hundredth. 1.25 is already to the nearest hundredth, right? Because the hundredth place is the second decimal place. 1.25 is 1.25, so that's 1.25.

But let me double-check my steps to make sure I didn't make a mistake.

Starting at (0,0). South 1/2 mile: (0, -0.5). Then east 3/4 mile: (0.75, -0.5). Then south another 1/2 mile: (0.75, -1.0). Then distance from (0,0) is sqrt(0.75^2 + (-1.0)^2).

Calculating 0.75 squared: 0.75 * 0.75 = 0.5625. Then (-1.0)^2 is 1.0. Adding them gives 1.5625. Square root of that is indeed 1.25.

So 1.25 miles. But wait, 1.25 is exact, so as a decimal, it's already to the nearest hundredth. So the answer is 1.25.

But let me check if there's another way to think about this. Maybe using right triangles?

He walked south twice, so total south distance is 1/2 + 1/2 = 1 mile south. Then east 3/4 mile. So if I imagine a right triangle where one leg is 1 mile south and the other leg is 3/4 mile east. Then the hypotenuse would be the straight-line distance.

So hypotenuse squared is (1)^2 + (3/4)^2 = 1 + 9/16 = 25/16. Wait, 1 is 16/16, so 16/16 + 9/16 = 25/16. Then square root of 25/16 is 5/4, which is 1.25. Yep, same answer.

So that confirms it. So the answer is 1.25.

But just to make sure, maybe I made a mistake in the coordinate system? Let me think again.

If he goes south first, then east, then south again. So the total displacement is 1 mile south and 3/4 mile east. So the straight line is the hypotenuse of a right triangle with legs 1 and 0.75. So yes, same as before.

Alternatively, if I had messed up the directions? For example, if I thought that going south twice would be different? But no, south is straight down, so adding both south movements.

Alternatively, maybe I should have considered the east movement as negative? No, because east is positive x-direction.

Wait, but in coordinates, if you go south, that's negative y, east is positive x. So the final position is (3/4, -1). So the distance is sqrt( (3/4)^2 + (-1)^2 ). Which is sqrt(9/16 + 1) = sqrt(25/16) = 5/4 = 1.25.

Yes, that's correct.

So I think the answer is 1.25. Therefore, expressed as a decimal to the nearest hundredth, it's 1.25.

But just to check, maybe the problem expects something else? Let me check the problem again.

"Bill walks 1/2 mile south, then 3/4 mile east, and finally 1/2 mile south. How many miles is he, in a direct line, from his starting point? Express your answer as a decimal to the nearest hundredth."

No, that's exactly what I did. So I think 1.25 is correct.

Alternatively, maybe I need to convert fractions to decimals first? Let me try that approach.

First leg: 1/2 mile south is 0.5 miles. Then 3/4 mile east is 0.75 miles. Then another 0.5 miles south. So total south is 0.5 + 0.5 = 1.0 miles. So displacement is 1.0 miles south and 0.75 miles east. Then distance is sqrt(1.0^2 + 0.75^2) = sqrt(1 + 0.5625) = sqrt(1.5625) = 1.25.

Same result.

Therefore, I think the answer is 1.25.

But just to make sure, maybe I can think of it as vectors.

First vector: south 1/2 mile. So in vector terms, that's (0, -1/2). Then east 3/4 mile is (3/4, 0). Then south 1/2 mile is (0, -1/2). Adding all vectors together:

(0 + 3/4 + 0, -1/2 + 0 -1/2) = (3/4, -1). So same as before.

Then magnitude of the vector is sqrt( (3/4)^2 + (-1)^2 ) = same as before.

So yeah, 1.25.

Therefore, I think the answer is 1.25.

But just to check if there's any possible mistake in calculation. Let me compute sqrt(1.5625).

Well, 1.25 squared is 1.5625. Because 1.2 squared is 1.44, 1.25 squared is (1 + 0.25)^2 = 1 + 0.5 + 0.0625 = 1.5625. So yes, sqrt(1.5625) is 1.25.

Therefore, the answer is 1.25.

I think that's solid. So I can confidently say the answer is 1.25.

**Final Answer**
\boxed{1.25}

To determine how far Bill is from his starting point after walking in different directions, we can model his movements using a coordinate system and apply the **distance formula**.

---

### Step 1: Assign Coordinates

Let’s place Bill’s starting point at the origin:
$$
(0, 0)
$$

- **First movement**: Bill walks **1/2 mile south**.
  In a coordinate system, moving south decreases the y-coordinate.
  So, his new position is:
  $$
  (0, -\frac{1}{2})
  $$

- **Second movement**: Bill walks **3/4 mile east**.
  Moving east increases the x-coordinate.
  So, his new position is:
  $$
  \left(\frac{3}{4}, -\frac{1}{2}\right)
  $$

- **Third movement**: Bill walks **1/2 mile south** again.
  This further decreases the y-coordinate.
  So, his final position is:
  $$
  \left(\frac{3}{4}, -\frac{1}{2} - \frac{1}{2}\right) = \left(\frac{3}{4}, -1\right)
  $$

---

### Step 2: Use the Distance Formula

We now compute the straight-line distance from the starting point $(0, 0)$ to the final position $\left(\frac{3}{4}, -1\right)$ using the distance formula:

$$
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
$$

Substitute the coordinates:

$$
\text{Distance} = \sqrt{\left(\frac{3}{4} - 0\right)^2 + (-1 - 0)^2} = \sqrt{\left(\frac{3}{4}\right)^2 + (-1)^2}
$$

$$
= \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{25}{16}} = \frac{5}{4} = 1.25
$$

---

### Final Answer

$$
\boxed{1.25}
$$

Solution:
The diagram on the left shows the path of Bill's walk. As the diagram on the right illustrates, he could also have walked from $A$ to $B$ by first walking 1 mile south then $\frac{3}{4}$ mile east. [asy]
pair a=(0,1), b=(.75, 0), c=(0,.5), d=(.75,.5), o=(0,0);
draw(a--b--d--c--cycle);
label("$A$", a, NW);
label("$B$", b, SE);
label("$\frac{1}{2}$", (0,0.75), W);
label("$\frac{3}{4}$", (.7, 0.66),W);
label("$\frac{1}{2}$", (.75, .25), E);

picture pic;
draw(pic, a--b--o--cycle);
label(pic, "$A$", a, NW);
label(pic, "$B$", b, SE);
label(pic, "$\frac{3}{4}$", (.375,0), S);
label(pic, "1", (0, .5), W);
add(shift(1.5,0)*pic);
[/asy] By the Pythagorean Theorem, \[(AB)^2=1^2+\left(\frac{3}{4}\right)^2=1+\frac{9}{16}=\frac{25}{16},\]so $AB=\frac{5}{4}=1\frac{1}{4}$, or $\boxed{1.25}$.