Explanation:
Okay, so I need to figure out the measure of the largest angle in this convex quadrilateral. Let me start by recalling some basic properties of quadrilaterals.

First off, I remember that the sum of the interior angles in any quadrilateral is 360 degrees. That's a key point. So if I can figure out the measures of the other angles, I can find the largest one.

The problem says that the quadrilateral is convex. That means all its interior angles are less than 180 degrees. Also, it mentions that the largest angle is twice the measure of the smallest angle, and the other two angles are both right angles. Right angles are 90 degrees each.

Let me try to visualize this quadrilateral. Since two of the angles are right angles, maybe it's something like a rectangle but with one angle modified? But wait, in a rectangle, all angles are right angles. But here, there's a largest angle that's twice the smallest. So maybe it's not a rectangle.

Let me denote the angles. Let me call the smallest angle as 'x' degrees. Then the largest angle would be '2x' degrees. The other two angles are both right angles, so they are 90 degrees each.

So, adding up all the angles: x + 2x + 90 + 90 = 360 degrees. Let me write that equation down:

x + 2x + 90 + 90 = 360

Simplifying that:

3x + 180 = 360

Subtract 180 from both sides:

3x = 180

Divide both sides by 3:

x = 60

So the smallest angle is 60 degrees, and the largest angle is 2x, which is 120 degrees.

Wait, but hold on. Let me check if this makes sense. If the angles are 60, 120, 90, and 90, then adding them up: 60 + 120 + 90 + 90 = 360. Yes, that adds up.

But let me make sure that the quadrilateral is convex. All angles must be less than 180 degrees. The angles here are 60, 90, 90, and 120. All of these are less than 180, so that's good.

But wait, is there another possibility? For example, could the largest angle be more than 120? Let me think. Suppose the smallest angle is x, the largest is 2x, and the other two are 90. But maybe there's another configuration where the angles are different?

Wait, the problem says "the other two angles are both right angles." So that means there are exactly two right angles, and the other two angles are the smallest and the largest. So the angles are x, 2x, 90, 90.

So the equation is correct. Therefore, solving gives x = 60, so the largest angle is 120 degrees.

But let me check again. Maybe I made a wrong assumption? For example, maybe the two right angles are not the ones that are not the smallest and largest? Wait, the problem says "the measure of the largest angle is twice the measure of the smallest angle, and the other two angles are both right angles." So that means that among the four angles, one is the smallest (x), one is the largest (2x), and the other two are right angles. So yes, that's exactly the setup.

Therefore, the calculation seems correct.

But let me think again. Suppose someone might think that maybe the two right angles are the ones that are not the smallest and largest. But that's exactly what the problem says. The other two angles are right angles. So the two angles that are not the smallest and largest are right angles.

Alternatively, could there be a case where one of the right angles is the smallest or the largest? But if a right angle is 90 degrees, then if it's the smallest angle, then the largest angle would be 180 degrees. But that's impossible in a convex quadrilateral because all angles must be less than 180. Therefore, the right angles can't be the smallest or the largest. Therefore, the smallest angle must be less than 90, and the largest angle must be more than 90.

So in this case, the smallest angle is 60, which is less than 90, and the largest is 120, which is more than 90. So that works.

Alternatively, if someone thought that maybe the two right angles are the largest and smallest? But that can't be because the largest angle is twice the smallest. If one of the right angles is the smallest, then the largest would be 180, which is not allowed. So that's not possible.

Therefore, the answer should be 120 degrees.

But let me check once more. Let me think of a specific quadrilateral. For example, imagine a trapezoid. If it's a right trapezoid with two right angles, and the other two angles are 60 and 120. Wait, but in a trapezoid, consecutive angles between the bases are supplementary. Wait, but if it's not necessarily a trapezoid.

But maybe it's not a trapezoid. Maybe it's some other convex quadrilateral. But regardless, the sum of angles is 360. So as long as the angles add up to 360, and each is less than 180, it's okay.

So 60, 90, 90, 120. All less than 180. Sum to 360. So that works.

Therefore, I think the answer is 120 degrees.

But let me check if there's another way this could be set up. For example, maybe the two right angles are adjacent or opposite? But regardless, the sum is still the same.

Alternatively, maybe the problem is referring to the angles in a different way? For example, maybe the largest angle is twice the smallest, and the other two angles are right angles. So maybe there are two right angles, one smallest angle, and one largest angle.

Yes, that's exactly what we considered.

Alternatively, could there be more than two right angles? But the problem says "the other two angles are both right angles," which implies that there are exactly two right angles.

So, if there are two right angles, then the other two angles are x and 2x.

So, the equation is correct.

Therefore, I think my answer is correct.

**Final Answer**
The measure of the largest angle is \boxed{120} degrees.

To determine the measure of the largest angle in the given convex quadrilateral, we begin by recalling a fundamental geometric property:

> The sum of the interior angles of any quadrilateral is always **360 degrees**.

---

### Step 1: Define the Angles

Let’s denote the smallest angle as $ x $ degrees. According to the problem:

- The **largest angle** is **twice** the smallest angle, so it is $ 2x $.
- The **other two angles** are both **right angles**, which means they are each $ 90^\circ $.

So, the four angles of the quadrilateral are:
- $ x $ (smallest)
- $ 2x $ (largest)
- $ 90^\circ $
- $ 90^\circ $

---

### Step 2: Set Up the Equation

Using the angle sum property of quadrilaterals:

$$
x + 2x + 90 + 90 = 360
$$

Simplify:

$$
3x + 180 = 360
$$

Subtract 180 from both sides:

$$
3x = 180
$$

Divide both sides by 3:

$$
x = 60
$$

---

### Step 3: Find the Largest Angle

Since the largest angle is $ 2x $, we substitute $ x = 60 $:

$$
2x = 2 \times 60 = 120
$$

---

### Step 4: Verify Convexity

All angles must be **less than 180 degrees** for the quadrilateral to be convex. The angles are:

- $ 60^\circ $
- $ 120^\circ $
- $ 90^\circ $
- $ 90^\circ $

All are less than 180, so the quadrilateral is indeed convex.

---

### Final Answer

$$
\boxed{120}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the measure of the largest angle in this convex quadrilateral. Let me start by recalling some basic properties of quadrilaterals.

First off, I remember that the sum of the interior angles in any quadrilateral is 360 degrees. That's a key point. So if I can figure out the measures of the other angles, I can find the largest one.

The problem says that the quadrilateral is convex. That means all its interior angles are less than 180 degrees. Also, it mentions that the largest angle is twice the measure of the smallest angle, and the other two angles are both right angles. Right angles are 90 degrees each.

Let me try to visualize this quadrilateral. Since two of the angles are right angles, maybe it's something like a rectangle but with one angle modified? But wait, in a rectangle, all angles are right angles. But here, there's a largest angle that's twice the smallest. So maybe it's not a rectangle.

Let me denote the angles. Let me call the smallest angle as 'x' degrees. Then the largest angle would be '2x' degrees. The other two angles are both right angles, so they are 90 degrees each.

So, adding up all the angles: x + 2x + 90 + 90 = 360 degrees. Let me write that equation down:

x + 2x + 90 + 90 = 360

Simplifying that:

3x + 180 = 360

Subtract 180 from both sides:

3x = 180

Divide both sides by 3:

x = 60

So the smallest angle is 60 degrees, and the largest angle is 2x, which is 120 degrees.

Wait, but hold on. Let me check if this makes sense. If the angles are 60, 120, 90, and 90, then adding them up: 60 + 120 + 90 + 90 = 360. Yes, that adds up.

But let me make sure that the quadrilateral is convex. All angles must be less than 180 degrees. The angles here are 60, 90, 90, and 120. All of these are less than 180, so that's good.

But wait, is there another possibility? For example, could the largest angle be more than 120? Let me think. Suppose the smallest angle is x, the largest is 2x, and the other two are 90. But maybe there's another configuration where the angles are different?

Wait, the problem says "the other two angles are both right angles." So that means there are exactly two right angles, and the other two angles are the smallest and the largest. So the angles are x, 2x, 90, 90.

So the equation is correct. Therefore, solving gives x = 60, so the largest angle is 120 degrees.

But let me check again. Maybe I made a wrong assumption? For example, maybe the two right angles are not the ones that are not the smallest and largest? Wait, the problem says "the measure of the largest angle is twice the measure of the smallest angle, and the other two angles are both right angles." So that means that among the four angles, one is the smallest (x), one is the largest (2x), and the other two are right angles. So yes, that's exactly the setup.

Therefore, the calculation seems correct.

But let me think again. Suppose someone might think that maybe the two right angles are the ones that are not the smallest and largest. But that's exactly what the problem says. The other two angles are right angles. So the two angles that are not the smallest and largest are right angles.

Alternatively, could there be a case where one of the right angles is the smallest or the largest? But if a right angle is 90 degrees, then if it's the smallest angle, then the largest angle would be 180 degrees. But that's impossible in a convex quadrilateral because all angles must be less than 180. Therefore, the right angles can't be the smallest or the largest. Therefore, the smallest angle must be less than 90, and the largest angle must be more than 90.

So in this case, the smallest angle is 60, which is less than 90, and the largest is 120, which is more than 90. So that works.

Alternatively, if someone thought that maybe the two right angles are the largest and smallest? But that can't be because the largest angle is twice the smallest. If one of the right angles is the smallest, then the largest would be 180, which is not allowed. So that's not possible.

Therefore, the answer should be 120 degrees.

But let me check once more. Let me think of a specific quadrilateral. For example, imagine a trapezoid. If it's a right trapezoid with two right angles, and the other two angles are 60 and 120. Wait, but in a trapezoid, consecutive angles between the bases are supplementary. Wait, but if it's not necessarily a trapezoid.

But maybe it's not a trapezoid. Maybe it's some other convex quadrilateral. But regardless, the sum of angles is 360. So as long as the angles add up to 360, and each is less than 180, it's okay.

So 60, 90, 90, 120. All less than 180. Sum to 360. So that works.

Therefore, I think the answer is 120 degrees.

But let me check if there's another way this could be set up. For example, maybe the two right angles are adjacent or opposite? But regardless, the sum is still the same.

Alternatively, maybe the problem is referring to the angles in a different way? For example, maybe the largest angle is twice the smallest, and the other two angles are right angles. So maybe there are two right angles, one smallest angle, and one largest angle.

Yes, that's exactly what we considered.

Alternatively, could there be more than two right angles? But the problem says "the other two angles are both right angles," which implies that there are exactly two right angles.

So, if there are two right angles, then the other two angles are x and 2x.

So, the equation is correct.

Therefore, I think my answer is correct.

**Final Answer**
The measure of the largest angle is \boxed{120} degrees.

To determine the measure of the largest angle in the given convex quadrilateral, we begin by recalling a fundamental geometric property:

> The sum of the interior angles of any quadrilateral is always **360 degrees**.

---

### Step 1: Define the Angles

Let’s denote the smallest angle as $ x $ degrees. According to the problem:

- The **largest angle** is **twice** the smallest angle, so it is $ 2x $.
- The **other two angles** are both **right angles**, which means they are each $ 90^\circ $.

So, the four angles of the quadrilateral are:
- $ x $ (smallest)
- $ 2x $ (largest)
- $ 90^\circ $
- $ 90^\circ $

---

### Step 2: Set Up the Equation

Using the angle sum property of quadrilaterals:

$$
x + 2x + 90 + 90 = 360
$$

Simplify:

$$
3x + 180 = 360
$$

Subtract 180 from both sides:

$$
3x = 180
$$

Divide both sides by 3:

$$
x = 60
$$

---

### Step 3: Find the Largest Angle

Since the largest angle is $ 2x $, we substitute $ x = 60 $:

$$
2x = 2 \times 60 = 120
$$

---

### Step 4: Verify Convexity

All angles must be **less than 180 degrees** for the quadrilateral to be convex. The angles are:

- $ 60^\circ $
- $ 120^\circ $
- $ 90^\circ $
- $ 90^\circ $

All are less than 180, so the quadrilateral is indeed convex.

---

### Final Answer

$$
\boxed{120}
$$

Solution: The interior angles of a quadrilateral must add up to 360.  (You can solve for this using the formula: $S = (n-2)(180)$, where S is the sum of the interior angles, and $n$ is the number of sides in the polygon.  However, if you want to get this problem quickly, you should have the value memorized.)  Since two of the angles are right, the other two angles must add up to 180.  Name the smaller angle $x$ - since the larger angle is double the smaller, we have $3x = 180 \rightarrow x = 60$, and $2x = 120$.  Thus, there are $\boxed{120}$ degrees in the larger angle.


Explanation:
Okay, so I need to find the maximum value of the function $ f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $ where $ -1 \leq x, y \leq 1 $. Hmm, let me think about how to approach this.

First, I know that both $ x $ and $ y $ are constrained between -1 and 1. The function involves square roots of $ 1 - y^2 $ and $ 1 - x^2 $, which makes me think of trigonometric substitutions. Maybe I can set $ x = \sin \alpha $ and $ y = \sin \beta $ for some angles $ \alpha $ and $ \beta $? Let me try that.

If I let $ x = \sin \alpha $ and $ y = \sin \beta $, then $ \sqrt{1 - y^2} = \cos \beta $ and $ \sqrt{1 - x^2} = \cos \alpha $, assuming that $ \alpha $ and $ \beta $ are in the range where cosine is non-negative, which would be between $ -\pi/2 $ and $ \pi/2 $. But since $ x $ and $ y $ can be negative, maybe I need to consider the absolute values? Wait, but if I take $ \alpha $ and $ \beta $ in the range where cosine is positive, then $ x $ and $ y $ can be positive or negative depending on the angle. Hmm, maybe I need to adjust for that. Alternatively, maybe I can use substitution with cosine directly?

Alternatively, maybe I can consider that $ \sqrt{1 - y^2} $ is the same as $ \cos \theta $ if $ y = \sin \theta $, but I need to make sure that the square roots are real, which they are as long as $ |x| \leq 1 $ and $ |y| \leq 1 $, which is given.

So, proceeding with substitution: Let me set $ x = \sin \alpha $ and $ y = \sin \beta $, where $ \alpha, \beta \in [-\pi/2, \pi/2] $. Then the function becomes:

$ f(\alpha, \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha $.

Wait a second! That simplifies to $ \sin(\alpha + \beta) $, because $ \sin \alpha \cos \beta + \cos \alpha \sin \beta = \sin(\alpha + \beta) $. So, the original function simplifies to $ \sin(\alpha + \beta) $. Therefore, the maximum value of $ f(x, y) $ is the maximum of $ \sin(\alpha + \beta) $, where $ \alpha, \beta \in [-\pi/2, \pi/2] $.

But then, the maximum value of $ \sin(\theta) $ is 1, so if $ \alpha + \beta = \pi/2 $, then $ \sin(\pi/2) = 1 $. However, we need to check if $ \alpha + \beta = \pi/2 $ is achievable given that $ \alpha $ and $ \beta $ are each between $ -\pi/2 $ and $ \pi/2 $.

Let me check. The maximum possible value of $ \alpha + \beta $ would be $ \pi/2 + \pi/2 = \pi $, and the minimum would be $ -\pi/2 - \pi/2 = -\pi $. Therefore, $ \theta = \alpha + \beta $ ranges from $ -\pi $ to $ \pi $. The maximum of $ \sin(\theta) $ over this interval is indeed 1, achieved at $ \theta = \pi/2 $. So, if there exist $ \alpha $ and $ \beta $ such that $ \alpha + \beta = \pi/2 $, then the maximum value is 1.

But we need to check if such $ \alpha $ and $ \beta $ are within the allowed range. For example, if $ \alpha = \pi/2 $, then $ \beta = 0 $, but $ \alpha = \pi/2 $ is allowed since $ x = \sin(\pi/2) = 1 $, which is within the domain. Similarly, $ \beta = 0 $ gives $ y = 0 $, which is also allowed. Therefore, when $ x = 1 $, $ y = 0 $, then $ f(1, 0) = 1 * \sqrt{1 - 0} + 0 * \sqrt{1 - 1} = 1 * 1 + 0 = 1 $. Similarly, if $ x = 0 $, $ y = 1 $, then $ f(0, 1) = 0 * \sqrt{1 - 1} + 1 * \sqrt{1 - 0} = 0 + 1 = 1 $. So, the maximum value is 1? But wait, is there a case where the function can be larger than 1?

Wait, but according to the substitution, the function simplifies to $ \sin(\alpha + \beta) $, which has a maximum of 1. However, maybe there's a mistake here? Let me check with another example. Suppose $ x = y = \frac{\sqrt{2}}{2} $. Then $ f(x, y) = \frac{\sqrt{2}}{2} \sqrt{1 - \frac{1}{2}} + \frac{\sqrt{2}}{2} \sqrt{1 - \frac{1}{2}} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{2} + \frac{1}{2} = 1 $. So that also gives 1. Hmm. What if I take $ x = y = 1 $? Then $ f(1, 1) = 1 * \sqrt{1 - 1} + 1 * \sqrt{1 - 1} = 0 + 0 = 0 $. So that's lower. What if I take $ x = y = \frac{1}{\sqrt{2}} $? Then as above, it's 1. What if I take $ x = \sin(\pi/4) $, $ y = \sin(\pi/4) $, which is same as above. So seems like 1 is achievable. But is there a way to get higher?

Wait, but according to the substitution, the function is equal to $ \sin(\alpha + \beta) $, so maximum is 1. But maybe there's a mistake in substitution? Let me check again.

Original function: $ x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. If I set $ x = \sin \alpha $, $ y = \sin \beta $, then $ \sqrt{1 - y^2} = \cos \beta $, but only if $ \beta $ is in the range where cosine is non-negative. However, if $ \beta $ is in $ [-\pi/2, \pi/2] $, then $ \cos \beta $ is non-negative, so that's okay. Therefore, the substitution is valid. Therefore, the function becomes $ \sin \alpha \cos \beta + \sin \beta \cos \alpha = \sin(\alpha + \beta) $. Therefore, the maximum value is indeed 1. But wait, let me check another case where maybe $ \alpha + \beta $ is something else. For example, suppose $ \alpha = \pi/2 $, $ \beta = \pi/2 $. Then $ x = 1 $, $ y = 1 $, which gives 0 as above. If $ \alpha = \pi/2 $, $ \beta = 0 $, then $ x = 1 $, $ y = 0 $, which gives 1. Similarly, if $ \alpha = 0 $, $ \beta = \pi/2 $, then $ x = 0 $, $ y = 1 $, which gives 1. So seems like 1 is the maximum. But wait, is there a case where $ \alpha + \beta $ is greater than $ \pi/2 $, but $ \sin(\alpha + \beta) $ is still 1? For example, $ \alpha + \beta = \pi/2 + 2\pi n $, but since $ \alpha + \beta $ is between $ -\pi $ and $ \pi $, the only point where sine is 1 is $ \pi/2 $. So that's the only maximum. Therefore, the maximum value is 1.

But wait, let me think again. Suppose I take $ x $ and $ y $ such that they are not in the range of sine. Wait, but $ x $ and $ y $ are between -1 and 1, so they can be represented as sine of some angle. So substitution is valid. Therefore, the maximum is 1. However, I have a feeling that maybe there's a higher value? Let me try some other values. Let me take $ x = \sqrt{3}/2 $, $ y = 1/2 $. Then $ f(x, y) = (\sqrt{3}/2) * \sqrt{1 - (1/2)^2} + (1/2) * \sqrt{1 - (3/4)} $. Calculating:

First term: $ \sqrt{3}/2 * \sqrt{3/4} = \sqrt{3}/2 * (\sqrt{3}/2) = 3/4 $.

Second term: $ 1/2 * \sqrt{1 - 3/4} = 1/2 * \sqrt{1/4} = 1/2 * 1/2 = 1/4 $.

Total: 3/4 + 1/4 = 1. Still 1. Hmm. What if I take $ x = 0.6 $, $ y = 0.8 $. Then $ \sqrt{1 - y^2} = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.6 $, and $ \sqrt{1 - x^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8 $. Then $ f(x, y) = 0.6 * 0.6 + 0.8 * 0.8 = 0.36 + 0.64 = 1 $. Again 1. So seems like regardless of the values, as long as $ x $ and $ y $ are such that $ x = \sin \alpha $, $ y = \sin \beta $, then the function is $ \sin(\alpha + \beta) $, which has maximum 1. Therefore, the maximum value is 1.

But wait, let me check if there's a case where $ \alpha + \beta $ is such that $ \sin(\alpha + \beta) $ is greater than 1? No, because sine function has maximum 1. Therefore, the maximum value of the function is 1. However, let me check if there are other critical points inside the domain where the maximum could be higher. Maybe I need to use calculus here?

Alternatively, maybe I can consider using the Cauchy-Schwarz inequality? Let me think. The function is $ x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. Let me consider vectors. Suppose I think of $ (x, y) $ and $ (\sqrt{1 - y^2}, \sqrt{1 - x^2}) $. Then the function is the dot product of these two vectors. The Cauchy-Schwarz inequality says that the dot product is less than or equal to the product of the magnitudes. Let me compute that.

The magnitude of the first vector is $ \sqrt{x^2 + y^2} $, and the magnitude of the second vector is $ \sqrt{(1 - y^2) + (1 - x^2)} = \sqrt{2 - x^2 - y^2} $. Therefore, by Cauchy-Schwarz:

$ f(x, y) \leq \sqrt{x^2 + y^2} \cdot \sqrt{2 - x^2 - y^2} $.

Let me denote $ t = x^2 + y^2 $. Then the upper bound becomes $ \sqrt{t(2 - t)} $. The maximum of $ \sqrt{t(2 - t)} $ occurs when $ t(2 - t) $ is maximized. The function $ t(2 - t) = -t^2 + 2t $ is a quadratic with maximum at $ t = 1 $, and maximum value 1. Therefore, the upper bound is $ \sqrt{1} = 1 $. Therefore, by Cauchy-Schwarz, $ f(x, y) \leq 1 $. And since we have examples where $ f(x, y) = 1 $, the maximum is indeed 1.

Therefore, the maximum value is 1. But wait, let me check if there are other cases where the maximum could be higher. For example, if we consider negative values? Suppose $ x $ and $ y $ are negative. Let me try $ x = -1 $, $ y = 0 $. Then $ f(-1, 0) = -1 * \sqrt{1 - 0} + 0 * \sqrt{1 - 1} = -1 + 0 = -1 $. Similarly, $ x = 0 $, $ y = -1 $ gives -1. If I take $ x = -\sqrt{3}/2 $, $ y = -1/2 $, then $ f(x, y) = (-\sqrt{3}/2) * \sqrt{1 - 1/4} + (-1/2) * \sqrt{1 - 3/4} = (-\sqrt{3}/2)(\sqrt{3}/2) + (-1/2)(1/2) = (-3/4) + (-1/4) = -1 $. So negative values give negative results. Therefore, the maximum is indeed 1.

But let me think again. Suppose I take $ x $ and $ y $ such that they are not related by the sine addition formula. For example, suppose I take $ x = y $. Let me set $ x = y $, then the function becomes $ x \sqrt{1 - x^2} + x \sqrt{1 - x^2} = 2x \sqrt{1 - x^2} $. Let me find the maximum of this function for $ x \in [-1, 1] $. Let me set $ g(x) = 2x \sqrt{1 - x^2} $. Take derivative: $ g'(x) = 2 \sqrt{1 - x^2} + 2x * (-2x)/(2 \sqrt{1 - x^2}) ) = 2 \sqrt{1 - x^2} - (2x^2)/\sqrt{1 - x^2} $. Combine terms: $ [2(1 - x^2) - 2x^2]/\sqrt{1 - x^2} = [2 - 4x^2]/\sqrt{1 - x^2} $. Setting derivative to zero: $ 2 - 4x^2 = 0 \implies x^2 = 1/2 \implies x = \pm \frac{\sqrt{2}}{2} $. Then $ g(x) = 2 * (\sqrt{2}/2) * \sqrt{1 - 1/2} = \sqrt{2} * \sqrt{1/2} = \sqrt{2} * \frac{1}{\sqrt{2}} = 1 $. So again, maximum is 1. Therefore, even when restricting to $ x = y $, the maximum is 1.

Alternatively, maybe there's a case where $ x $ and $ y $ are not equal, but the function is higher? Let me try some other values. Suppose $ x = 0.8 $, $ y = 0.6 $. Then $ f(x, y) = 0.8 * \sqrt{1 - 0.36} + 0.6 * \sqrt{1 - 0.64} = 0.8 * 0.8 + 0.6 * 0.6 = 0.64 + 0.36 = 1 $. Again 1. So seems like regardless of how I choose $ x $ and $ y $, the maximum is 1.

But let me think again about the substitution. If $ f(x, y) = \sin(\alpha + \beta) $, then the maximum is 1. However, is there a case where $ \alpha + \beta $ is such that $ \sin(\alpha + \beta) $ is greater than 1? No, because sine is bounded by 1. Therefore, the maximum is indeed 1.

But wait, let me check if there's a case where $ \alpha + \beta $ is outside of the range $ [-\pi, \pi] $. But since $ \alpha $ and $ \beta $ are each between $ -\pi/2 $ and $ \pi/2 $, their sum is between $ -\pi $ and $ \pi $, so $ \sin(\alpha + \beta) $ is between $ -1 $ and $ 1 $. Therefore, the maximum is 1.

Alternatively, maybe I can use Lagrange multipliers to find critical points. Let me try that approach.

We need to maximize $ f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $.

First, let me consider the domain $ -1 \leq x, y \leq 1 $. The function is differentiable in the interior of the domain, so critical points can be found by setting the partial derivatives to zero.

Compute partial derivatives:

First, partial derivative with respect to x:

$ \frac{\partial f}{\partial x} = \sqrt{1 - y^2} + y * \frac{-2x}{2 \sqrt{1 - x^2}} = \sqrt{1 - y^2} - \frac{y x}{\sqrt{1 - x^2}} $.

Similarly, partial derivative with respect to y:

$ \frac{\partial f}{\partial y} = \frac{-2y}{2 \sqrt{1 - y^2}} * x + \sqrt{1 - x^2} = \frac{-x y}{\sqrt{1 - y^2}} + \sqrt{1 - x^2} $.

Set both partial derivatives to zero:

1. $ \sqrt{1 - y^2} - \frac{y x}{\sqrt{1 - x^2}} = 0 $
2. $ \frac{-x y}{\sqrt{1 - y^2}} + \sqrt{1 - x^2} = 0 $

Let me write equations:

From equation 1: $ \sqrt{1 - y^2} = \frac{y x}{\sqrt{1 - x^2}} $

From equation 2: $ \sqrt{1 - x^2} = \frac{x y}{\sqrt{1 - y^2}} $

Notice that equation 1 and equation 2 are similar. Let me square both sides of equation 1 to eliminate the square roots:

$ 1 - y^2 = \frac{y^2 x^2}{1 - x^2} $

Multiply both sides by $ 1 - x^2 $:

$ (1 - y^2)(1 - x^2) = y^2 x^2 $

Expand left side:

$ 1 - x^2 - y^2 + x^2 y^2 = y^2 x^2 $

Subtract $ y^2 x^2 $ from both sides:

$ 1 - x^2 - y^2 = 0 $

Therefore, $ x^2 + y^2 = 1 $

So, critical points lie on the circle $ x^2 + y^2 = 1 $. Now, let's check equation 2. From equation 2:

$ \sqrt{1 - x^2} = \frac{x y}{\sqrt{1 - y^2}} $

Square both sides:

$ 1 - x^2 = \frac{x^2 y^2}{1 - y^2} $

Multiply both sides by $ 1 - y^2 $:

$ (1 - x^2)(1 - y^2) = x^2 y^2 $

Which is the same as equation from before, leading to $ x^2 + y^2 = 1 $. Therefore, the critical points lie on the circle $ x^2 + y^2 = 1 $.

So, now, we can parametrize $ x $ and $ y $ on the unit circle. Let me set $ x = \cos \theta $, $ y = \sin \theta $, but wait, but earlier substitution was $ x = \sin \alpha $, $ y = \sin \beta $. However, here, since $ x^2 + y^2 = 1 $, we can parametrize $ x = \cos \theta $, $ y = \sin \theta $, where $ \theta \in [0, 2\pi) $. Let me try that.

So, substituting into the original function:

$ f(x, y) = \cos \theta \sqrt{1 - \sin^2 \theta} + \sin \theta \sqrt{1 - \cos^2 \theta} $

Simplify:

$ \cos \theta \sqrt{\cos^2 \theta} + \sin \theta \sqrt{\sin^2 \theta} $

Which is $ \cos \theta |\cos \theta| + \sin \theta |\sin \theta| $

But since $ \theta $ can be in any range, we need to consider the signs. However, if we assume $ \theta $ is in the first quadrant where $ \cos \theta $ and $ \sin \theta $ are positive, then this simplifies to $ \cos^2 \theta + \sin^2 \theta = 1 $. Therefore, in this case, the function is 1. However, if $ \theta $ is in other quadrants, the absolute values may change the result. For example, if $ \theta $ is in the second quadrant, $ \cos \theta $ is negative, $ \sin \theta $ is positive. Then:

$ f(x, y) = \cos \theta (-\cos \theta) + \sin \theta (\sin \theta) = -\cos^2 \theta + \sin^2 \theta $. Which is $ -\cos 2\theta $. The maximum of this would be 1 when $ \cos 2\theta = -1 $, i.e., $ 2\theta = \pi \implies \theta = \pi/2 $. But at $ \theta = \pi/2 $, $ x = 0 $, $ y = 1 $, which gives $ f(0, 1) = 0 * 0 + 1 * 1 = 1 $. Similarly, if $ \theta $ is in the third quadrant, both $ \cos \theta $ and $ \sin \theta $ are negative, so:

$ f(x, y) = \cos \theta (-\cos \theta) + \sin \theta (-\sin \theta) = -\cos^2 \theta - \sin^2 \theta = -1 $. Similarly, in the fourth quadrant, $ \cos \theta $ positive, $ \sin \theta $ negative:

$ f(x, y) = \cos \theta (\cos \theta) + \sin \theta (-\sin \theta) = \cos^2 \theta - \sin^2 \theta = \cos 2\theta $, which has maximum 1 when $ 2\theta = 0 \implies \theta = 0 $, which gives $ x = 1 $, $ y = 0 $, and $ f(1, 0) = 1 $.

Therefore, regardless of the quadrant, the maximum value of $ f(x, y) $ on the unit circle is 1, achieved at points where $ \theta $ is 0, $ \pi/2 $, $ \pi $, $ 3\pi/2 $, etc. However, at $ \theta = \pi $, $ x = -1 $, $ y = 0 $, which gives $ f(-1, 0) = -1 * 1 + 0 * 0 = -1 $, which is the minimum. Similarly for $ 3\pi/2 $, it's $ 0 * 1 + (-1) * 0 = 0 $. Wait, no, $ \theta = 3\pi/2 $, $ x = \cos(3\pi/2) = 0 $, $ y = \sin(3\pi/2) = -1 $, so $ f(0, -1) = 0 * \sqrt{1 - 1} + (-1) * \sqrt{1 - 0} = 0 - 1 = -1 $.

Therefore, the maximum on the unit circle is indeed 1, achieved at $ (\pm 1, 0) $ and $ (0, \pm 1) $. However, we need to check if there are other critical points outside the unit circle? Wait, earlier we found that critical points must lie on the unit circle $ x^2 + y^2 = 1 $. Therefore, the maximum on the critical points is 1.

But we also need to check the boundaries of the domain. The domain is a square $ [-1, 1] \times [-1, 1] $. The maximum could occur either in the interior (which we checked, and found that critical points on the unit circle give maximum 1) or on the boundary.

So, let's check the boundaries. The boundaries occur when either $ x = \pm 1 $, $ y = \pm 1 $, or when one variable is fixed and the other varies.

First, consider the case when $ x = 1 $. Then $ f(1, y) = 1 * \sqrt{1 - y^2} + y * \sqrt{1 - 1} = \sqrt{1 - y^2} + 0 = \sqrt{1 - y^2} $. The maximum of this occurs when $ y = 0 $, giving $ \sqrt{1} = 1 $. Similarly, when $ x = -1 $, $ f(-1, y) = -\sqrt{1 - y^2} + y * 0 = -\sqrt{1 - y^2} $, which has maximum 0 (at $ y = \pm 1 $) and minimum -1 (at $ y = 0 $).

Similarly, when $ y = 1 $, $ f(x, 1) = x * 0 + 1 * \sqrt{1 - x^2} = \sqrt{1 - x^2} $, maximum 1 at $ x = 0 $. When $ y = -1 $, $ f(x, -1) = x * 0 + (-1) * \sqrt{1 - x^2} = -\sqrt{1 - x^2} $, maximum 0.

Now, check the other boundaries where one variable is fixed. For example, when $ x = 0 $, $ f(0, y) = 0 * \sqrt{1 - y^2} + y * \sqrt{1 - 0} = y * 1 = y $. The maximum of this is 1 (at $ y = 1 $) and minimum -1 (at $ y = -1 $). Similarly, when $ y = 0 $, $ f(x, 0) = x * 1 + 0 * \sqrt{1 - x^2} = x $, maximum 1 (at $ x = 1 $) and minimum -1 (at $ x = -1 $).

Now, check the other edges where variables are not fixed at ±1 or 0. For example, consider the edge where $ x = y $. Wait, but we already considered that earlier. Alternatively, consider the edges where $ x $ or $ y $ is fixed at some value between -1 and 1. However, since we already checked the boundaries where variables are fixed at ±1 or 0, and found that maximum is 1, perhaps there are no higher values on the other edges.

Alternatively, consider the case where one variable is fixed, say $ x $ is fixed, and $ y $ varies. For example, take $ x $ fixed between -1 and 1, and find maximum over $ y $. But since we already considered the boundaries where $ x = \pm 1 $, $ y = \pm 1 $, and $ x = 0 $, $ y = 0 $, maybe there are no higher values.

Alternatively, think about the case where $ x $ and $ y $ are both varying, but on the boundary of the domain. However, since the maximum on the interior is 1, and on the boundaries also 1, then the overall maximum is 1.

Therefore, combining all these results, the maximum value of $ f(x, y) $ is 1.

But just to be thorough, let me check another approach. Suppose I consider the function $ f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. Let me consider squaring the function to see if that helps. However, squaring might complicate things because of cross terms. Let me try:

$ f(x, y)^2 = x^2 (1 - y^2) + y^2 (1 - x^2) + 2xy \sqrt{(1 - y^2)(1 - x^2)} $

Simplify:

$ x^2 - x^2 y^2 + y^2 - x^2 y^2 + 2xy \sqrt{(1 - x^2)(1 - y^2)} $

Which is:

$ x^2 + y^2 - 2x^2 y^2 + 2xy \sqrt{(1 - x^2)(1 - y^2)} $

Not sure if helpful. However, if we suppose that maximum occurs at 1, then $ f(x, y)^2 = 1 $, so:

$ x^2 + y^2 - 2x^2 y^2 + 2xy \sqrt{(1 - x^2)(1 - y^2)} = 1 $

But this seems complicated. However, if we take $ x = 1 $, $ y = 0 $, then:

Left side: $ 1 + 0 - 0 + 0 = 1 $, which matches. Similarly for other points.

Alternatively, maybe consider using substitution variables. Let me set $ a = x $, $ b = y $, then the function is $ a \sqrt{1 - b^2} + b \sqrt{1 - a^2} $. Let me consider variables $ a $ and $ b $ such that $ a, b \in [-1, 1] $.

Alternatively, think of this as a function of two variables and use symmetry. Notice that the function is symmetric in a way if we swap $ x $ and $ y $, but not entirely symmetric. However, if we consider $ x = y $, we saw that maximum is 1.

Alternatively, think of the function as $ x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. Let me consider variables $ x $ and $ y $ such that $ x = \sin \alpha $, $ y = \sin \beta $, then as before, function becomes $ \sin(\alpha + \beta) $, which has maximum 1.

Therefore, all approaches lead to the conclusion that the maximum value is 1.

But just to be 100% sure, let me check if there's a case where $ \alpha + \beta $ is something else. For example, suppose $ \alpha = \pi/2 $, $ \beta = \pi/2 $, but then $ x = 1 $, $ y = 1 $, which gives 0. If $ \alpha = \pi/2 $, $ \beta = \pi/2 - \epsilon $, then $ x = 1 $, $ y = \sin(\pi/2 - \epsilon) = \cos \epsilon \approx 1 - \epsilon^2/2 $. Then $ f(x, y) = 1 * \sqrt{1 - y^2} + y * \sqrt{1 - 1} = \sqrt{1 - y^2} + 0 $. Since $ y \approx 1 $, $ \sqrt{1 - y^2} \approx \sqrt{1 - (1 - \epsilon^2)} = \sqrt{\epsilon^2} = \epsilon $, which approaches 0 as $ \epsilon \to 0 $. Therefore, the maximum is not achieved here.

Alternatively, take $ \alpha = \pi/2 - \theta $, $ \beta = \pi/2 - \theta $, then $ x = \sin(\pi/2 - \theta) = \cos \theta $, $ y = \cos \theta $. Then $ f(x, y) = \cos \theta \sqrt{1 - \cos^2 \theta} + \cos \theta \sqrt{1 - \cos^2 \theta} = 2 \cos \theta \sin \theta = \sin 2\theta $. The maximum of this is 1 when $ 2\theta = \pi/2 \implies \theta = \pi/4 $. Then $ x = \cos(\pi/4) = \sqrt{2}/2 $, $ y = \sqrt{2}/2 $, which gives $ f(x, y) = 1 $, as before.

Therefore, all these cases confirm that the maximum is indeed 1.

Therefore, after multiple approaches—trigonometric substitution, Cauchy-Schwarz inequality, Lagrange multipliers, checking boundaries, and testing specific values—it seems that the maximum value of the function is 1.

**Final Answer**
\boxed{1}

To find the maximum value of the function
$$
f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2}
$$
with the constraint $ -1 \leq x, y \leq 1 $, we proceed by analyzing the structure of the function and applying appropriate mathematical tools.

---

### Step 1: Trigonometric Substitution

We observe that the square root terms $ \sqrt{1 - y^2} $ and $ \sqrt{1 - x^2} $ suggest a trigonometric substitution. Let us define:
$$
x = \sin \alpha, \quad y = \sin \beta
$$
for some angles $ \alpha, \beta \in [-\pi/2, \pi/2] $, where the square roots are non-negative. Then:
$$
\sqrt{1 - y^2} = \cos \beta, \quad \sqrt{1 - x^2} = \cos \alpha
$$
Substituting into the function:
$$
f(x, y) = \sin \alpha \cos \beta + \sin \beta \cos \alpha = \sin(\alpha + \beta)
$$
This is a well-known trigonometric identity.

---

### Step 2: Analyze the Range of the Function

The function now becomes:
$$
f(x, y) = \sin(\alpha + \beta)
$$
with $ \alpha, \beta \in [-\pi/2, \pi/2] $, so the sum $ \alpha + \beta \in [-\pi, \pi] $. The sine function achieves its maximum value of 1 at $ \alpha + \beta = \pi/2 $, which is within the range.

Thus, the maximum value of $ f(x, y) $ is:
$$
\max f(x, y) = \sin(\pi/2) = 1
$$

---

### Step 3: Verify Achievability of the Maximum

We need to check whether the maximum value of 1 is actually attainable. For this, we find values of $ x $ and $ y $ such that $ f(x, y) = 1 $.

Let’s consider:
- $ x = 1 $, $ y = 0 $:
  $$
  f(1, 0) = 1 \cdot \sqrt{1 - 0^2} + 0 \cdot \sqrt{1 - 1^2} = 1
  $$
- $ x = 0 $, $ y = 1 $:
  $$
  f(0, 1) = 0 \cdot \sqrt{1 - 1^2} + 1 \cdot \sqrt{1 - 0^2} = 1
  $$

Both cases yield $ f(x, y) = 1 $, confirming that the maximum is indeed attainable.

---

### Step 4: Confirm via Alternative Methods

We can also use the **Cauchy-Schwarz inequality** to bound the function:
$$
f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2}
$$
Let’s consider the vectors:
$$
\vec{u} = (x, y), \quad \vec{v} = (\sqrt{1 - y^2}, \sqrt{1 - x^2})
$$
Then:
$$
f(x, y) = \vec{u} \cdot \vec{v} \leq \|\vec{u}\| \cdot \|\vec{v}\|
$$
Compute the magnitudes:
$$
\|\vec{u}\| = \sqrt{x^2 + y^2}, \quad \|\vec{v}\| = \sqrt{(1 - y^2) + (1 - x^2)} = \sqrt{2 - x^2 - y^2}
$$
So:
$$
f(x, y) \leq \sqrt{x^2 + y^2} \cdot \sqrt{2 - x^2 - y^2}
$$
Let $ t = x^2 + y^2 $, then the upper bound becomes:
$$
\sqrt{t(2 - t)}
$$
This is maximized when $ t = 1 $, giving:
$$
\sqrt{1 \cdot (2 - 1)} = \sqrt{1} = 1
$$
Thus, the maximum value of $ f(x, y) $ is again confirmed to be 1.

---

### Step 5: Boundary and Critical Point Analysis

We also check the boundaries of the domain $ [-1, 1] \times [-1, 1] $, and find that the maximum value of 1 is achieved at points like $ (1, 0) $, $ (0, 1) $, and similar symmetric points. Using Lagrange multipliers, we find that the critical points lie on the unit circle $ x^2 + y^2 = 1 $, and again, the maximum value of the function on this circle is 1.

---

### Final Conclusion

All approaches—trigonometric substitution, Cauchy-Schwarz inequality, boundary analysis, and critical point analysis—confirm that the **maximum value** of the function is:

$$
\boxed{1}
$$

[Assistant]
Explanation:
Okay, so I need to find the maximum value of the function $ f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $ where $ -1 \leq x, y \leq 1 $. Hmm, let me think about how to approach this.

First, I know that both $ x $ and $ y $ are constrained between -1 and 1. The function involves square roots of $ 1 - y^2 $ and $ 1 - x^2 $, which makes me think of trigonometric substitutions. Maybe I can set $ x = \sin \alpha $ and $ y = \sin \beta $ for some angles $ \alpha $ and $ \beta $? Let me try that.

If I let $ x = \sin \alpha $ and $ y = \sin \beta $, then $ \sqrt{1 - y^2} = \cos \beta $ and $ \sqrt{1 - x^2} = \cos \alpha $, assuming that $ \alpha $ and $ \beta $ are in the range where cosine is non-negative, which would be between $ -\pi/2 $ and $ \pi/2 $. But since $ x $ and $ y $ can be negative, maybe I need to consider the absolute values? Wait, but if I take $ \alpha $ and $ \beta $ in the range where cosine is positive, then $ x $ and $ y $ can be positive or negative depending on the angle. Hmm, maybe I need to adjust for that. Alternatively, maybe I can use substitution with cosine directly?

Alternatively, maybe I can consider that $ \sqrt{1 - y^2} $ is the same as $ \cos \theta $ if $ y = \sin \theta $, but I need to make sure that the square roots are real, which they are as long as $ |x| \leq 1 $ and $ |y| \leq 1 $, which is given.

So, proceeding with substitution: Let me set $ x = \sin \alpha $ and $ y = \sin \beta $, where $ \alpha, \beta \in [-\pi/2, \pi/2] $. Then the function becomes:

$ f(\alpha, \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha $.

Wait a second! That simplifies to $ \sin(\alpha + \beta) $, because $ \sin \alpha \cos \beta + \cos \alpha \sin \beta = \sin(\alpha + \beta) $. So, the original function simplifies to $ \sin(\alpha + \beta) $. Therefore, the maximum value of $ f(x, y) $ is the maximum of $ \sin(\alpha + \beta) $, where $ \alpha, \beta \in [-\pi/2, \pi/2] $.

But then, the maximum value of $ \sin(\theta) $ is 1, so if $ \alpha + \beta = \pi/2 $, then $ \sin(\pi/2) = 1 $. However, we need to check if $ \alpha + \beta = \pi/2 $ is achievable given that $ \alpha $ and $ \beta $ are each between $ -\pi/2 $ and $ \pi/2 $.

Let me check. The maximum possible value of $ \alpha + \beta $ would be $ \pi/2 + \pi/2 = \pi $, and the minimum would be $ -\pi/2 - \pi/2 = -\pi $. Therefore, $ \theta = \alpha + \beta $ ranges from $ -\pi $ to $ \pi $. The maximum of $ \sin(\theta) $ over this interval is indeed 1, achieved at $ \theta = \pi/2 $. So, if there exist $ \alpha $ and $ \beta $ such that $ \alpha + \beta = \pi/2 $, then the maximum value is 1.

But we need to check if such $ \alpha $ and $ \beta $ are within the allowed range. For example, if $ \alpha = \pi/2 $, then $ \beta = 0 $, but $ \alpha = \pi/2 $ is allowed since $ x = \sin(\pi/2) = 1 $, which is within the domain. Similarly, $ \beta = 0 $ gives $ y = 0 $, which is also allowed. Therefore, when $ x = 1 $, $ y = 0 $, then $ f(1, 0) = 1 * \sqrt{1 - 0} + 0 * \sqrt{1 - 1} = 1 * 1 + 0 = 1 $. Similarly, if $ x = 0 $, $ y = 1 $, then $ f(0, 1) = 0 * \sqrt{1 - 1} + 1 * \sqrt{1 - 0} = 0 + 1 = 1 $. So, the maximum value is 1? But wait, is there a case where the function can be larger than 1?

Wait, but according to the substitution, the function simplifies to $ \sin(\alpha + \beta) $, which has a maximum of 1. However, maybe there's a mistake here? Let me check with another example. Suppose $ x = y = \frac{\sqrt{2}}{2} $. Then $ f(x, y) = \frac{\sqrt{2}}{2} \sqrt{1 - \frac{1}{2}} + \frac{\sqrt{2}}{2} \sqrt{1 - \frac{1}{2}} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{2} + \frac{1}{2} = 1 $. So that also gives 1. Hmm. What if I take $ x = y = 1 $? Then $ f(1, 1) = 1 * \sqrt{1 - 1} + 1 * \sqrt{1 - 1} = 0 + 0 = 0 $. So that's lower. What if I take $ x = y = \frac{1}{\sqrt{2}} $? Then as above, it's 1. What if I take $ x = \sin(\pi/4) $, $ y = \sin(\pi/4) $, which is same as above. So seems like 1 is achievable. But is there a way to get higher?

Wait, but according to the substitution, the function is equal to $ \sin(\alpha + \beta) $, so maximum is 1. But maybe there's a mistake in substitution? Let me check again.

Original function: $ x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. If I set $ x = \sin \alpha $, $ y = \sin \beta $, then $ \sqrt{1 - y^2} = \cos \beta $, but only if $ \beta $ is in the range where cosine is non-negative. However, if $ \beta $ is in $ [-\pi/2, \pi/2] $, then $ \cos \beta $ is non-negative, so that's okay. Therefore, the substitution is valid. Therefore, the function becomes $ \sin \alpha \cos \beta + \sin \beta \cos \alpha = \sin(\alpha + \beta) $. Therefore, the maximum value is indeed 1. But wait, let me check another case where maybe $ \alpha + \beta $ is something else. For example, suppose $ \alpha = \pi/2 $, $ \beta = \pi/2 $. Then $ x = 1 $, $ y = 1 $, which gives 0 as above. If $ \alpha = \pi/2 $, $ \beta = 0 $, then $ x = 1 $, $ y = 0 $, which gives 1. Similarly, if $ \alpha = 0 $, $ \beta = \pi/2 $, then $ x = 0 $, $ y = 1 $, which gives 1. So seems like 1 is the maximum. But wait, is there a case where $ \alpha + \beta $ is greater than $ \pi/2 $, but $ \sin(\alpha + \beta) $ is still 1? For example, $ \alpha + \beta = \pi/2 + 2\pi n $, but since $ \alpha + \beta $ is between $ -\pi $ and $ \pi $, the only point where sine is 1 is $ \pi/2 $. So that's the only maximum. Therefore, the maximum value is 1.

But wait, let me think again. Suppose I take $ x $ and $ y $ such that they are not in the range of sine. Wait, but $ x $ and $ y $ are between -1 and 1, so they can be represented as sine of some angle. So substitution is valid. Therefore, the maximum is 1. However, I have a feeling that maybe there's a higher value? Let me try some other values. Let me take $ x = \sqrt{3}/2 $, $ y = 1/2 $. Then $ f(x, y) = (\sqrt{3}/2) * \sqrt{1 - (1/2)^2} + (1/2) * \sqrt{1 - (3/4)} $. Calculating:

First term: $ \sqrt{3}/2 * \sqrt{3/4} = \sqrt{3}/2 * (\sqrt{3}/2) = 3/4 $.

Second term: $ 1/2 * \sqrt{1 - 3/4} = 1/2 * \sqrt{1/4} = 1/2 * 1/2 = 1/4 $.

Total: 3/4 + 1/4 = 1. Still 1. Hmm. What if I take $ x = 0.6 $, $ y = 0.8 $. Then $ \sqrt{1 - y^2} = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.6 $, and $ \sqrt{1 - x^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8 $. Then $ f(x, y) = 0.6 * 0.6 + 0.8 * 0.8 = 0.36 + 0.64 = 1 $. Again 1. So seems like regardless of the values, as long as $ x $ and $ y $ are such that $ x = \sin \alpha $, $ y = \sin \beta $, then the function is $ \sin(\alpha + \beta) $, which has maximum 1. Therefore, the maximum value is 1.

But wait, let me check if there's a case where $ \alpha + \beta $ is such that $ \sin(\alpha + \beta) $ is greater than 1? No, because sine function has maximum 1. Therefore, the maximum value of the function is 1. However, let me check if there are other critical points inside the domain where the maximum could be higher. Maybe I need to use calculus here?

Alternatively, maybe I can consider using the Cauchy-Schwarz inequality? Let me think. The function is $ x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. Let me consider vectors. Suppose I think of $ (x, y) $ and $ (\sqrt{1 - y^2}, \sqrt{1 - x^2}) $. Then the function is the dot product of these two vectors. The Cauchy-Schwarz inequality says that the dot product is less than or equal to the product of the magnitudes. Let me compute that.

The magnitude of the first vector is $ \sqrt{x^2 + y^2} $, and the magnitude of the second vector is $ \sqrt{(1 - y^2) + (1 - x^2)} = \sqrt{2 - x^2 - y^2} $. Therefore, by Cauchy-Schwarz:

$ f(x, y) \leq \sqrt{x^2 + y^2} \cdot \sqrt{2 - x^2 - y^2} $.

Let me denote $ t = x^2 + y^2 $. Then the upper bound becomes $ \sqrt{t(2 - t)} $. The maximum of $ \sqrt{t(2 - t)} $ occurs when $ t(2 - t) $ is maximized. The function $ t(2 - t) = -t^2 + 2t $ is a quadratic with maximum at $ t = 1 $, and maximum value 1. Therefore, the upper bound is $ \sqrt{1} = 1 $. Therefore, by Cauchy-Schwarz, $ f(x, y) \leq 1 $. And since we have examples where $ f(x, y) = 1 $, the maximum is indeed 1.

Therefore, the maximum value is 1. But wait, let me check if there are other cases where the maximum could be higher. For example, if we consider negative values? Suppose $ x $ and $ y $ are negative. Let me try $ x = -1 $, $ y = 0 $. Then $ f(-1, 0) = -1 * \sqrt{1 - 0} + 0 * \sqrt{1 - 1} = -1 + 0 = -1 $. Similarly, $ x = 0 $, $ y = -1 $ gives -1. If I take $ x = -\sqrt{3}/2 $, $ y = -1/2 $, then $ f(x, y) = (-\sqrt{3}/2) * \sqrt{1 - 1/4} + (-1/2) * \sqrt{1 - 3/4} = (-\sqrt{3}/2)(\sqrt{3}/2) + (-1/2)(1/2) = (-3/4) + (-1/4) = -1 $. So negative values give negative results. Therefore, the maximum is indeed 1.

But let me think again. Suppose I take $ x $ and $ y $ such that they are not related by the sine addition formula. For example, suppose I take $ x = y $. Let me set $ x = y $, then the function becomes $ x \sqrt{1 - x^2} + x \sqrt{1 - x^2} = 2x \sqrt{1 - x^2} $. Let me find the maximum of this function for $ x \in [-1, 1] $. Let me set $ g(x) = 2x \sqrt{1 - x^2} $. Take derivative: $ g'(x) = 2 \sqrt{1 - x^2} + 2x * (-2x)/(2 \sqrt{1 - x^2}) ) = 2 \sqrt{1 - x^2} - (2x^2)/\sqrt{1 - x^2} $. Combine terms: $ [2(1 - x^2) - 2x^2]/\sqrt{1 - x^2} = [2 - 4x^2]/\sqrt{1 - x^2} $. Setting derivative to zero: $ 2 - 4x^2 = 0 \implies x^2 = 1/2 \implies x = \pm \frac{\sqrt{2}}{2} $. Then $ g(x) = 2 * (\sqrt{2}/2) * \sqrt{1 - 1/2} = \sqrt{2} * \sqrt{1/2} = \sqrt{2} * \frac{1}{\sqrt{2}} = 1 $. So again, maximum is 1. Therefore, even when restricting to $ x = y $, the maximum is 1.

Alternatively, maybe there's a case where $ x $ and $ y $ are not equal, but the function is higher? Let me try some other values. Suppose $ x = 0.8 $, $ y = 0.6 $. Then $ f(x, y) = 0.8 * \sqrt{1 - 0.36} + 0.6 * \sqrt{1 - 0.64} = 0.8 * 0.8 + 0.6 * 0.6 = 0.64 + 0.36 = 1 $. Again 1. So seems like regardless of how I choose $ x $ and $ y $, the maximum is 1.

But let me think again about the substitution. If $ f(x, y) = \sin(\alpha + \beta) $, then the maximum is 1. However, is there a case where $ \alpha + \beta $ is such that $ \sin(\alpha + \beta) $ is greater than 1? No, because sine is bounded by 1. Therefore, the maximum is indeed 1.

But wait, let me check if there's a case where $ \alpha + \beta $ is outside of the range $ [-\pi, \pi] $. But since $ \alpha $ and $ \beta $ are each between $ -\pi/2 $ and $ \pi/2 $, their sum is between $ -\pi $ and $ \pi $, so $ \sin(\alpha + \beta) $ is between $ -1 $ and $ 1 $. Therefore, the maximum is 1.

Alternatively, maybe I can use Lagrange multipliers to find critical points. Let me try that approach.

We need to maximize $ f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $.

First, let me consider the domain $ -1 \leq x, y \leq 1 $. The function is differentiable in the interior of the domain, so critical points can be found by setting the partial derivatives to zero.

Compute partial derivatives:

First, partial derivative with respect to x:

$ \frac{\partial f}{\partial x} = \sqrt{1 - y^2} + y * \frac{-2x}{2 \sqrt{1 - x^2}} = \sqrt{1 - y^2} - \frac{y x}{\sqrt{1 - x^2}} $.

Similarly, partial derivative with respect to y:

$ \frac{\partial f}{\partial y} = \frac{-2y}{2 \sqrt{1 - y^2}} * x + \sqrt{1 - x^2} = \frac{-x y}{\sqrt{1 - y^2}} + \sqrt{1 - x^2} $.

Set both partial derivatives to zero:

1. $ \sqrt{1 - y^2} - \frac{y x}{\sqrt{1 - x^2}} = 0 $
2. $ \frac{-x y}{\sqrt{1 - y^2}} + \sqrt{1 - x^2} = 0 $

Let me write equations:

From equation 1: $ \sqrt{1 - y^2} = \frac{y x}{\sqrt{1 - x^2}} $

From equation 2: $ \sqrt{1 - x^2} = \frac{x y}{\sqrt{1 - y^2}} $

Notice that equation 1 and equation 2 are similar. Let me square both sides of equation 1 to eliminate the square roots:

$ 1 - y^2 = \frac{y^2 x^2}{1 - x^2} $

Multiply both sides by $ 1 - x^2 $:

$ (1 - y^2)(1 - x^2) = y^2 x^2 $

Expand left side:

$ 1 - x^2 - y^2 + x^2 y^2 = y^2 x^2 $

Subtract $ y^2 x^2 $ from both sides:

$ 1 - x^2 - y^2 = 0 $

Therefore, $ x^2 + y^2 = 1 $

So, critical points lie on the circle $ x^2 + y^2 = 1 $. Now, let's check equation 2. From equation 2:

$ \sqrt{1 - x^2} = \frac{x y}{\sqrt{1 - y^2}} $

Square both sides:

$ 1 - x^2 = \frac{x^2 y^2}{1 - y^2} $

Multiply both sides by $ 1 - y^2 $:

$ (1 - x^2)(1 - y^2) = x^2 y^2 $

Which is the same as equation from before, leading to $ x^2 + y^2 = 1 $. Therefore, the critical points lie on the circle $ x^2 + y^2 = 1 $.

So, now, we can parametrize $ x $ and $ y $ on the unit circle. Let me set $ x = \cos \theta $, $ y = \sin \theta $, but wait, but earlier substitution was $ x = \sin \alpha $, $ y = \sin \beta $. However, here, since $ x^2 + y^2 = 1 $, we can parametrize $ x = \cos \theta $, $ y = \sin \theta $, where $ \theta \in [0, 2\pi) $. Let me try that.

So, substituting into the original function:

$ f(x, y) = \cos \theta \sqrt{1 - \sin^2 \theta} + \sin \theta \sqrt{1 - \cos^2 \theta} $

Simplify:

$ \cos \theta \sqrt{\cos^2 \theta} + \sin \theta \sqrt{\sin^2 \theta} $

Which is $ \cos \theta |\cos \theta| + \sin \theta |\sin \theta| $

But since $ \theta $ can be in any range, we need to consider the signs. However, if we assume $ \theta $ is in the first quadrant where $ \cos \theta $ and $ \sin \theta $ are positive, then this simplifies to $ \cos^2 \theta + \sin^2 \theta = 1 $. Therefore, in this case, the function is 1. However, if $ \theta $ is in other quadrants, the absolute values may change the result. For example, if $ \theta $ is in the second quadrant, $ \cos \theta $ is negative, $ \sin \theta $ is positive. Then:

$ f(x, y) = \cos \theta (-\cos \theta) + \sin \theta (\sin \theta) = -\cos^2 \theta + \sin^2 \theta $. Which is $ -\cos 2\theta $. The maximum of this would be 1 when $ \cos 2\theta = -1 $, i.e., $ 2\theta = \pi \implies \theta = \pi/2 $. But at $ \theta = \pi/2 $, $ x = 0 $, $ y = 1 $, which gives $ f(0, 1) = 0 * 0 + 1 * 1 = 1 $. Similarly, if $ \theta $ is in the third quadrant, both $ \cos \theta $ and $ \sin \theta $ are negative, so:

$ f(x, y) = \cos \theta (-\cos \theta) + \sin \theta (-\sin \theta) = -\cos^2 \theta - \sin^2 \theta = -1 $. Similarly, in the fourth quadrant, $ \cos \theta $ positive, $ \sin \theta $ negative:

$ f(x, y) = \cos \theta (\cos \theta) + \sin \theta (-\sin \theta) = \cos^2 \theta - \sin^2 \theta = \cos 2\theta $, which has maximum 1 when $ 2\theta = 0 \implies \theta = 0 $, which gives $ x = 1 $, $ y = 0 $, and $ f(1, 0) = 1 $.

Therefore, regardless of the quadrant, the maximum value of $ f(x, y) $ on the unit circle is 1, achieved at points where $ \theta $ is 0, $ \pi/2 $, $ \pi $, $ 3\pi/2 $, etc. However, at $ \theta = \pi $, $ x = -1 $, $ y = 0 $, which gives $ f(-1, 0) = -1 * 1 + 0 * 0 = -1 $, which is the minimum. Similarly for $ 3\pi/2 $, it's $ 0 * 1 + (-1) * 0 = 0 $. Wait, no, $ \theta = 3\pi/2 $, $ x = \cos(3\pi/2) = 0 $, $ y = \sin(3\pi/2) = -1 $, so $ f(0, -1) = 0 * \sqrt{1 - 1} + (-1) * \sqrt{1 - 0} = 0 - 1 = -1 $.

Therefore, the maximum on the unit circle is indeed 1, achieved at $ (\pm 1, 0) $ and $ (0, \pm 1) $. However, we need to check if there are other critical points outside the unit circle? Wait, earlier we found that critical points must lie on the unit circle $ x^2 + y^2 = 1 $. Therefore, the maximum on the critical points is 1.

But we also need to check the boundaries of the domain. The domain is a square $ [-1, 1] \times [-1, 1] $. The maximum could occur either in the interior (which we checked, and found that critical points on the unit circle give maximum 1) or on the boundary.

So, let's check the boundaries. The boundaries occur when either $ x = \pm 1 $, $ y = \pm 1 $, or when one variable is fixed and the other varies.

First, consider the case when $ x = 1 $. Then $ f(1, y) = 1 * \sqrt{1 - y^2} + y * \sqrt{1 - 1} = \sqrt{1 - y^2} + 0 = \sqrt{1 - y^2} $. The maximum of this occurs when $ y = 0 $, giving $ \sqrt{1} = 1 $. Similarly, when $ x = -1 $, $ f(-1, y) = -\sqrt{1 - y^2} + y * 0 = -\sqrt{1 - y^2} $, which has maximum 0 (at $ y = \pm 1 $) and minimum -1 (at $ y = 0 $).

Similarly, when $ y = 1 $, $ f(x, 1) = x * 0 + 1 * \sqrt{1 - x^2} = \sqrt{1 - x^2} $, maximum 1 at $ x = 0 $. When $ y = -1 $, $ f(x, -1) = x * 0 + (-1) * \sqrt{1 - x^2} = -\sqrt{1 - x^2} $, maximum 0.

Now, check the other boundaries where one variable is fixed. For example, when $ x = 0 $, $ f(0, y) = 0 * \sqrt{1 - y^2} + y * \sqrt{1 - 0} = y * 1 = y $. The maximum of this is 1 (at $ y = 1 $) and minimum -1 (at $ y = -1 $). Similarly, when $ y = 0 $, $ f(x, 0) = x * 1 + 0 * \sqrt{1 - x^2} = x $, maximum 1 (at $ x = 1 $) and minimum -1 (at $ x = -1 $).

Now, check the other edges where variables are not fixed at ±1 or 0. For example, consider the edge where $ x = y $. Wait, but we already considered that earlier. Alternatively, consider the edges where $ x $ or $ y $ is fixed at some value between -1 and 1. However, since we already checked the boundaries where variables are fixed at ±1 or 0, and found that maximum is 1, perhaps there are no higher values on the other edges.

Alternatively, consider the case where one variable is fixed, say $ x $ is fixed, and $ y $ varies. For example, take $ x $ fixed between -1 and 1, and find maximum over $ y $. But since we already considered the boundaries where $ x = \pm 1 $, $ y = \pm 1 $, and $ x = 0 $, $ y = 0 $, maybe there are no higher values.

Alternatively, think about the case where $ x $ and $ y $ are both varying, but on the boundary of the domain. However, since the maximum on the interior is 1, and on the boundaries also 1, then the overall maximum is 1.

Therefore, combining all these results, the maximum value of $ f(x, y) $ is 1.

But just to be thorough, let me check another approach. Suppose I consider the function $ f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. Let me consider squaring the function to see if that helps. However, squaring might complicate things because of cross terms. Let me try:

$ f(x, y)^2 = x^2 (1 - y^2) + y^2 (1 - x^2) + 2xy \sqrt{(1 - y^2)(1 - x^2)} $

Simplify:

$ x^2 - x^2 y^2 + y^2 - x^2 y^2 + 2xy \sqrt{(1 - x^2)(1 - y^2)} $

Which is:

$ x^2 + y^2 - 2x^2 y^2 + 2xy \sqrt{(1 - x^2)(1 - y^2)} $

Not sure if helpful. However, if we suppose that maximum occurs at 1, then $ f(x, y)^2 = 1 $, so:

$ x^2 + y^2 - 2x^2 y^2 + 2xy \sqrt{(1 - x^2)(1 - y^2)} = 1 $

But this seems complicated. However, if we take $ x = 1 $, $ y = 0 $, then:

Left side: $ 1 + 0 - 0 + 0 = 1 $, which matches. Similarly for other points.

Alternatively, maybe consider using substitution variables. Let me set $ a = x $, $ b = y $, then the function is $ a \sqrt{1 - b^2} + b \sqrt{1 - a^2} $. Let me consider variables $ a $ and $ b $ such that $ a, b \in [-1, 1] $.

Alternatively, think of this as a function of two variables and use symmetry. Notice that the function is symmetric in a way if we swap $ x $ and $ y $, but not entirely symmetric. However, if we consider $ x = y $, we saw that maximum is 1.

Alternatively, think of the function as $ x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. Let me consider variables $ x $ and $ y $ such that $ x = \sin \alpha $, $ y = \sin \beta $, then as before, function becomes $ \sin(\alpha + \beta) $, which has maximum 1.

Therefore, all approaches lead to the conclusion that the maximum value is 1.

But just to be 100% sure, let me check if there's a case where $ \alpha + \beta $ is something else. For example, suppose $ \alpha = \pi/2 $, $ \beta = \pi/2 $, but then $ x = 1 $, $ y = 1 $, which gives 0. If $ \alpha = \pi/2 $, $ \beta = \pi/2 - \epsilon $, then $ x = 1 $, $ y = \sin(\pi/2 - \epsilon) = \cos \epsilon \approx 1 - \epsilon^2/2 $. Then $ f(x, y) = 1 * \sqrt{1 - y^2} + y * \sqrt{1 - 1} = \sqrt{1 - y^2} + 0 $. Since $ y \approx 1 $, $ \sqrt{1 - y^2} \approx \sqrt{1 - (1 - \epsilon^2)} = \sqrt{\epsilon^2} = \epsilon $, which approaches 0 as $ \epsilon \to 0 $. Therefore, the maximum is not achieved here.

Alternatively, take $ \alpha = \pi/2 - \theta $, $ \beta = \pi/2 - \theta $, then $ x = \sin(\pi/2 - \theta) = \cos \theta $, $ y = \cos \theta $. Then $ f(x, y) = \cos \theta \sqrt{1 - \cos^2 \theta} + \cos \theta \sqrt{1 - \cos^2 \theta} = 2 \cos \theta \sin \theta = \sin 2\theta $. The maximum of this is 1 when $ 2\theta = \pi/2 \implies \theta = \pi/4 $. Then $ x = \cos(\pi/4) = \sqrt{2}/2 $, $ y = \sqrt{2}/2 $, which gives $ f(x, y) = 1 $, as before.

Therefore, all these cases confirm that the maximum is indeed 1.

Therefore, after multiple approaches—trigonometric substitution, Cauchy-Schwarz inequality, Lagrange multipliers, checking boundaries, and testing specific values—it seems that the maximum value of the function is 1.

**Final Answer**
\boxed{1}

To find the maximum value of the function
$$
f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2}
$$
with the constraint $ -1 \leq x, y \leq 1 $, we proceed by analyzing the structure of the function and applying appropriate mathematical tools.

---

### Step 1: Trigonometric Substitution

We observe that the square root terms $ \sqrt{1 - y^2} $ and $ \sqrt{1 - x^2} $ suggest a trigonometric substitution. Let us define:
$$
x = \sin \alpha, \quad y = \sin \beta
$$
for some angles $ \alpha, \beta \in [-\pi/2, \pi/2] $, where the square roots are non-negative. Then:
$$
\sqrt{1 - y^2} = \cos \beta, \quad \sqrt{1 - x^2} = \cos \alpha
$$
Substituting into the function:
$$
f(x, y) = \sin \alpha \cos \beta + \sin \beta \cos \alpha = \sin(\alpha + \beta)
$$
This is a well-known trigonometric identity.

---

### Step 2: Analyze the Range of the Function

The function now becomes:
$$
f(x, y) = \sin(\alpha + \beta)
$$
with $ \alpha, \beta \in [-\pi/2, \pi/2] $, so the sum $ \alpha + \beta \in [-\pi, \pi] $. The sine function achieves its maximum value of 1 at $ \alpha + \beta = \pi/2 $, which is within the range.

Thus, the maximum value of $ f(x, y) $ is:
$$
\max f(x, y) = \sin(\pi/2) = 1
$$

---

### Step 3: Verify Achievability of the Maximum

We need to check whether the maximum value of 1 is actually attainable. For this, we find values of $ x $ and $ y $ such that $ f(x, y) = 1 $.

Let’s consider:
- $ x = 1 $, $ y = 0 $:
  $$
  f(1, 0) = 1 \cdot \sqrt{1 - 0^2} + 0 \cdot \sqrt{1 - 1^2} = 1
  $$
- $ x = 0 $, $ y = 1 $:
  $$
  f(0, 1) = 0 \cdot \sqrt{1 - 1^2} + 1 \cdot \sqrt{1 - 0^2} = 1
  $$

Both cases yield $ f(x, y) = 1 $, confirming that the maximum is indeed attainable.

---

### Step 4: Confirm via Alternative Methods

We can also use the **Cauchy-Schwarz inequality** to bound the function:
$$
f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2}
$$
Let’s consider the vectors:
$$
\vec{u} = (x, y), \quad \vec{v} = (\sqrt{1 - y^2}, \sqrt{1 - x^2})
$$
Then:
$$
f(x, y) = \vec{u} \cdot \vec{v} \leq \|\vec{u}\| \cdot \|\vec{v}\|
$$
Compute the magnitudes:
$$
\|\vec{u}\| = \sqrt{x^2 + y^2}, \quad \|\vec{v}\| = \sqrt{(1 - y^2) + (1 - x^2)} = \sqrt{2 - x^2 - y^2}
$$
So:
$$
f(x, y) \leq \sqrt{x^2 + y^2} \cdot \sqrt{2 - x^2 - y^2}
$$
Let $ t = x^2 + y^2 $, then the upper bound becomes:
$$
\sqrt{t(2 - t)}
$$
This is maximized when $ t = 1 $, giving:
$$
\sqrt{1 \cdot (2 - 1)} = \sqrt{1} = 1
$$
Thus, the maximum value of $ f(x, y) $ is again confirmed to be 1.

---

### Step 5: Boundary and Critical Point Analysis

We also check the boundaries of the domain $ [-1, 1] \times [-1, 1] $, and find that the maximum value of 1 is achieved at points like $ (1, 0) $, $ (0, 1) $, and similar symmetric points. Using Lagrange multipliers, we find that the critical points lie on the unit circle $ x^2 + y^2 = 1 $, and again, the maximum value of the function on this circle is 1.

---

### Final Conclusion

All approaches—trigonometric substitution, Cauchy-Schwarz inequality, boundary analysis, and critical point analysis—confirm that the **maximum value** of the function is:

$$
\boxed{1}
$$

Solution:
Let $a$ and $b$ be real numbers.  Then $(a - b)^2 \ge 0,$ which is equivalent to
\[ab \le \frac{a^2 + b^2}{2}.\](This looks like AM-GM, but here, we want to show that it holds for all real numbers, not just nonnegative real numbers.)

Setting $a = x$ and $b = \sqrt{1 - y^2},$ we get
\[x \sqrt{1 - y^2} \le \frac{x^2 + 1 - y^2}{2}.\]Setting $a = y$ and $b = \sqrt{1 - x^2},$ we get
\[y \sqrt{1 - x^2} \le \frac{y^2 + 1 - x^2}{2}.\]Therefore,
\[x \sqrt{1 - y^2} +y \sqrt{1 - x^2} \le \frac{x^2 + 1 - y^2}{2} + \frac{y^2 + 1 - x^2}{2} = 1.\]Since $f(1,0) = 1,$ the maximum value is $\boxed{1}.$


Explanation:
Okay, so I need to find the area of triangle OAB, given the vectors OA and OB. Let me recall how to do this. I think the area of a triangle formed by two vectors can be found using the cross product. The formula is something like half the magnitude of the cross product of the two vectors. Let me check if that's right.

Yes, I remember that if you have two vectors from the same point, say OA and OB, then the area of the triangle OAB is (1/2) times the magnitude of OA cross OB. So, first, I need to compute the cross product of vectors OA and OB, then find its magnitude, and then divide by 2. That should give me the area.

Alright, let me write down the vectors again. Vector OA is given as (1, 2, 3) and vector OB is (-3, -2, 1). So, OA = [1, 2, 3] and OB = [-3, -2, 1]. Now, I need to compute OA × OB.

The cross product of two vectors in three dimensions can be calculated using the determinant of a matrix with the standard unit vectors i, j, k in the first row, the components of the first vector in the second row, and the components of the second vector in the third row. Let me set that up.

So, OA × OB = |i   j   k|
             |1   2   3|
             |-3 -2   1|

Calculating this determinant, I can expand it along the first row. So, it would be:

i * (2*1 - 3*(-2)) - j * (1*1 - 3*(-3)) + k * (1*(-2) - 2*(-3))

Let me compute each component step by step.

First, the i component: 2*1 is 2, and 3*(-2) is -6, but since it's minus 3*(-2), it becomes +6. So, 2 + 6 = 8. Therefore, the i component is 8i.

Wait, hold on. Let me make sure. The formula for the determinant is:

i*( (2)(1) - (3)(-2) ) - j*( (1)(1) - (3)(-3) ) + k*( (1)(-2) - (2)(-3) )

Yes, that's right. So for the i component: 2*1 - 3*(-2) = 2 + 6 = 8. So that's 8i.

Then the j component: it's subtracted, so - [ (1)(1) - (3)(-3) ] = - [1 + 9] = -10. Wait, hold on. Let me check again. The j component is - [ (1)(1) - (3)(-3) ]? Wait, no. Wait, the formula is:

For the cross product, the j component is subtracted, so it's - [ (OA_x * OB_z - OA_z * OB_x) ]? Wait, maybe I should recall the formula properly.

Alternatively, maybe I should use the standard formula for cross product components. Let me recall:

If vector OA = (a1, a2, a3) and vector OB = (b1, b2, b3), then OA × OB = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1). Wait, is that right? Wait, no, the cross product components are:

First component: a2b3 - a3b2

Second component: a3b1 - a1b3

Third component: a1b2 - a2b1

But I need to be careful with the signs. Alternatively, using the determinant method:

i*(a2b3 - a3b2) - j*(a1b3 - a3b1) + k*(a1b2 - a2b1)

Which is equivalent to (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1). So, yes, that's correct.

So, applying this to our vectors OA = (1, 2, 3) and OB = (-3, -2, 1):

First component (i): a2b3 - a3b2 = (2)(1) - (3)(-2) = 2 + 6 = 8

Second component (j): a3b1 - a1b3 = (3)(-3) - (1)(1) = -9 - 1 = -10

Third component (k): a1b2 - a2b1 = (1)(-2) - (2)(-3) = -2 + 6 = 4

Therefore, OA × OB = (8, -10, 4). Wait, but in the determinant method, the j component is subtracted, so when I calculated earlier, I had - [ (1)(1) - (3)(-3) ] which would be - [1 + 9] = -10. So that matches. So the cross product vector is (8, -10, 4). Let me confirm once again:

First component: 2*1 - 3*(-2) = 2 + 6 = 8. Correct.

Second component: 3*(-3) - 1*1 = -9 -1 = -10. Correct.

Third component: 1*(-2) - 2*(-3) = -2 + 6 = 4. Correct. So OA × OB = (8, -10, 4).

Now, the magnitude of this cross product vector is sqrt(8^2 + (-10)^2 + 4^2). Let me compute that.

First, 8 squared is 64, (-10) squared is 100, and 4 squared is 16. Adding them up: 64 + 100 + 16 = 180. Therefore, the magnitude is sqrt(180). Let me simplify sqrt(180).

180 factors into 36 * 5, since 36*5=180. Therefore sqrt(36*5) = 6*sqrt(5). So the magnitude of OA × OB is 6√5.

Therefore, the area of triangle OAB is half of that, so (1/2)*6√5 = 3√5.

Wait, let me check again. So cross product magnitude is sqrt(8² + (-10)² + 4²) = sqrt(64 + 100 + 16) = sqrt(180). Yes, sqrt(180) is indeed 6√5. Then half of that is 3√5. So the area is 3√5.

But let me just make sure I didn't make any mistakes in the cross product calculation. Let me recompute OA × OB.

OA is (1, 2, 3), OB is (-3, -2, 1). So cross product components:

i: (2)(1) - (3)(-2) = 2 + 6 = 8. Correct.

j: -( (1)(1) - (3)(-3) ) = - (1 + 9) = -10. Correct.

k: (1)(-2) - (2)(-3) = -2 + 6 = 4. Correct. So cross product is (8, -10, 4).

Yes, that seems right. Then magnitude squared is 8² + (-10)² + 4² = 64 + 100 + 16 = 180. So magnitude is sqrt(180) = 6*sqrt(5). Then area is half that, so 3*sqrt(5).

Therefore, the area of triangle OAB is 3√5.

But just to be thorough, let me think if there's another way to approach this problem. For example, using coordinates, maybe compute the lengths of OA, OB, and AB, then use Heron's formula? Although that might be more complicated, but let me try it as a check.

First, compute the lengths of OA, OB, and AB.

Length of OA: sqrt(1² + 2² + 3²) = sqrt(1 + 4 + 9) = sqrt(14).

Length of OB: sqrt((-3)^2 + (-2)^2 + 1^2) = sqrt(9 + 4 + 1) = sqrt(14). Interesting, both OA and OB have the same length, sqrt(14).

Now, length of AB. Since AB = OB - OA. Wait, vector AB is OB - OA? Wait, no. Wait, vector OA is from O to A, vector OB is from O to B. So vector AB is OB - OA. So coordinates of AB would be (-3 - 1, -2 - 2, 1 - 3) = (-4, -4, -2). Therefore, the length of AB is sqrt((-4)^2 + (-4)^2 + (-2)^2) = sqrt(16 + 16 + 4) = sqrt(36) = 6.

So sides of triangle OAB are OA = sqrt(14), OB = sqrt(14), AB = 6.

Now, using Heron's formula, the semi-perimeter s = (sqrt(14) + sqrt(14) + 6)/2 = (2*sqrt(14) + 6)/2 = sqrt(14) + 3.

Then area = sqrt( s(s - OA)(s - OB)(s - AB) )

So plugging in:

s = sqrt(14) + 3

s - OA = sqrt(14) + 3 - sqrt(14) = 3

Similarly, s - OB = 3

s - AB = sqrt(14) + 3 - 6 = sqrt(14) - 3

Therefore, area = sqrt( [sqrt(14) + 3] * 3 * 3 * [sqrt(14) - 3] )

Let me compute that. First, note that [sqrt(14) + 3][sqrt(14) - 3] is a difference of squares: (sqrt(14))² - 3² = 14 - 9 = 5.

Therefore, the expression under the square root becomes 5 * 3 * 3 = 45. Therefore, area = sqrt(45) = 3*sqrt(5).

Which matches the previous result! So that's reassuring. Therefore, the area is indeed 3√5.

So both methods give the same answer, so I think that's correct.

Alternatively, another way to think about it is that the area of the parallelogram formed by OA and OB is |OA × OB|, so the area of the triangle is half that. Which is exactly what I did first.

Therefore, I can be confident that the area of triangle OAB is 3√5.

Just to recap, steps were:

1. Find vectors OA and OB.

2. Compute their cross product.

3. Find the magnitude of the cross product.

4. Divide by 2 to get the area of the triangle.

Alternatively, using Heron's formula with the lengths of the sides also gives the same answer, which is a good check.

Therefore, the answer is 3√5.

**Final Answer**
The area of triangle $ OAB $ is \boxed{3\sqrt{5}}.

To find the area of triangle $ OAB $, we are given the position vectors of points $ A $ and $ B $ with respect to the origin $ O $:

$$
\vec{OA} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \quad \vec{OB} = \begin{pmatrix} -3 \\ -2 \\ 1 \end{pmatrix}
$$

---

### Step 1: Use the Cross Product to Find the Area

The area of triangle $ OAB $ can be calculated using the formula:

$$
\text{Area} = \frac{1}{2} \left\| \vec{OA} \times \vec{OB} \right\|
$$

We compute the cross product $ \vec{OA} \times \vec{OB} $ using the determinant method:

$$
\vec{OA} \times \vec{OB} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & 3 \\
-3 & -2 & 1
\end{vmatrix}
= \mathbf{i}(2 \cdot 1 - 3 \cdot (-2)) - \mathbf{j}(1 \cdot 1 - 3 \cdot (-3)) + \mathbf{k}(1 \cdot (-2) - 2 \cdot (-3))
$$

$$
= \mathbf{i}(2 + 6) - \mathbf{j}(1 + 9) + \mathbf{k}(-2 + 6)
= 8\mathbf{i} - 10\mathbf{j} + 4\mathbf{k}
$$

So,

$$
\vec{OA} \times \vec{OB} = \begin{pmatrix} 8 \\ -10 \\ 4 \end{pmatrix}
$$

---

### Step 2: Compute the Magnitude of the Cross Product

$$
\left\| \vec{OA} \times \vec{OB} \right\| = \sqrt{8^2 + (-10)^2 + 4^2} = \sqrt{64 + 100 + 16} = \sqrt{180}
$$

Simplify:

$$
\sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5}
$$

---

### Step 3: Compute the Area of the Triangle

$$
\text{Area} = \frac{1}{2} \cdot 6\sqrt{5} = 3\sqrt{5}
$$

---

### Final Answer

$$
\boxed{3\sqrt{5}}
$$

[Assistant]
Explanation:
Okay, so I need to find the area of triangle OAB, given the vectors OA and OB. Let me recall how to do this. I think the area of a triangle formed by two vectors can be found using the cross product. The formula is something like half the magnitude of the cross product of the two vectors. Let me check if that's right.

Yes, I remember that if you have two vectors from the same point, say OA and OB, then the area of the triangle OAB is (1/2) times the magnitude of OA cross OB. So, first, I need to compute the cross product of vectors OA and OB, then find its magnitude, and then divide by 2. That should give me the area.

Alright, let me write down the vectors again. Vector OA is given as (1, 2, 3) and vector OB is (-3, -2, 1). So, OA = [1, 2, 3] and OB = [-3, -2, 1]. Now, I need to compute OA × OB.

The cross product of two vectors in three dimensions can be calculated using the determinant of a matrix with the standard unit vectors i, j, k in the first row, the components of the first vector in the second row, and the components of the second vector in the third row. Let me set that up.

So, OA × OB = |i   j   k|
             |1   2   3|
             |-3 -2   1|

Calculating this determinant, I can expand it along the first row. So, it would be:

i * (2*1 - 3*(-2)) - j * (1*1 - 3*(-3)) + k * (1*(-2) - 2*(-3))

Let me compute each component step by step.

First, the i component: 2*1 is 2, and 3*(-2) is -6, but since it's minus 3*(-2), it becomes +6. So, 2 + 6 = 8. Therefore, the i component is 8i.

Wait, hold on. Let me make sure. The formula for the determinant is:

i*( (2)(1) - (3)(-2) ) - j*( (1)(1) - (3)(-3) ) + k*( (1)(-2) - (2)(-3) )

Yes, that's right. So for the i component: 2*1 - 3*(-2) = 2 + 6 = 8. So that's 8i.

Then the j component: it's subtracted, so - [ (1)(1) - (3)(-3) ] = - [1 + 9] = -10. Wait, hold on. Let me check again. The j component is - [ (1)(1) - (3)(-3) ]? Wait, no. Wait, the formula is:

For the cross product, the j component is subtracted, so it's - [ (OA_x * OB_z - OA_z * OB_x) ]? Wait, maybe I should recall the formula properly.

Alternatively, maybe I should use the standard formula for cross product components. Let me recall:

If vector OA = (a1, a2, a3) and vector OB = (b1, b2, b3), then OA × OB = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1). Wait, is that right? Wait, no, the cross product components are:

First component: a2b3 - a3b2

Second component: a3b1 - a1b3

Third component: a1b2 - a2b1

But I need to be careful with the signs. Alternatively, using the determinant method:

i*(a2b3 - a3b2) - j*(a1b3 - a3b1) + k*(a1b2 - a2b1)

Which is equivalent to (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1). So, yes, that's correct.

So, applying this to our vectors OA = (1, 2, 3) and OB = (-3, -2, 1):

First component (i): a2b3 - a3b2 = (2)(1) - (3)(-2) = 2 + 6 = 8

Second component (j): a3b1 - a1b3 = (3)(-3) - (1)(1) = -9 - 1 = -10

Third component (k): a1b2 - a2b1 = (1)(-2) - (2)(-3) = -2 + 6 = 4

Therefore, OA × OB = (8, -10, 4). Wait, but in the determinant method, the j component is subtracted, so when I calculated earlier, I had - [ (1)(1) - (3)(-3) ] which would be - [1 + 9] = -10. So that matches. So the cross product vector is (8, -10, 4). Let me confirm once again:

First component: 2*1 - 3*(-2) = 2 + 6 = 8. Correct.

Second component: 3*(-3) - 1*1 = -9 -1 = -10. Correct.

Third component: 1*(-2) - 2*(-3) = -2 + 6 = 4. Correct. So OA × OB = (8, -10, 4).

Now, the magnitude of this cross product vector is sqrt(8^2 + (-10)^2 + 4^2). Let me compute that.

First, 8 squared is 64, (-10) squared is 100, and 4 squared is 16. Adding them up: 64 + 100 + 16 = 180. Therefore, the magnitude is sqrt(180). Let me simplify sqrt(180).

180 factors into 36 * 5, since 36*5=180. Therefore sqrt(36*5) = 6*sqrt(5). So the magnitude of OA × OB is 6√5.

Therefore, the area of triangle OAB is half of that, so (1/2)*6√5 = 3√5.

Wait, let me check again. So cross product magnitude is sqrt(8² + (-10)² + 4²) = sqrt(64 + 100 + 16) = sqrt(180). Yes, sqrt(180) is indeed 6√5. Then half of that is 3√5. So the area is 3√5.

But let me just make sure I didn't make any mistakes in the cross product calculation. Let me recompute OA × OB.

OA is (1, 2, 3), OB is (-3, -2, 1). So cross product components:

i: (2)(1) - (3)(-2) = 2 + 6 = 8. Correct.

j: -( (1)(1) - (3)(-3) ) = - (1 + 9) = -10. Correct.

k: (1)(-2) - (2)(-3) = -2 + 6 = 4. Correct. So cross product is (8, -10, 4).

Yes, that seems right. Then magnitude squared is 8² + (-10)² + 4² = 64 + 100 + 16 = 180. So magnitude is sqrt(180) = 6*sqrt(5). Then area is half that, so 3*sqrt(5).

Therefore, the area of triangle OAB is 3√5.

But just to be thorough, let me think if there's another way to approach this problem. For example, using coordinates, maybe compute the lengths of OA, OB, and AB, then use Heron's formula? Although that might be more complicated, but let me try it as a check.

First, compute the lengths of OA, OB, and AB.

Length of OA: sqrt(1² + 2² + 3²) = sqrt(1 + 4 + 9) = sqrt(14).

Length of OB: sqrt((-3)^2 + (-2)^2 + 1^2) = sqrt(9 + 4 + 1) = sqrt(14). Interesting, both OA and OB have the same length, sqrt(14).

Now, length of AB. Since AB = OB - OA. Wait, vector AB is OB - OA? Wait, no. Wait, vector OA is from O to A, vector OB is from O to B. So vector AB is OB - OA. So coordinates of AB would be (-3 - 1, -2 - 2, 1 - 3) = (-4, -4, -2). Therefore, the length of AB is sqrt((-4)^2 + (-4)^2 + (-2)^2) = sqrt(16 + 16 + 4) = sqrt(36) = 6.

So sides of triangle OAB are OA = sqrt(14), OB = sqrt(14), AB = 6.

Now, using Heron's formula, the semi-perimeter s = (sqrt(14) + sqrt(14) + 6)/2 = (2*sqrt(14) + 6)/2 = sqrt(14) + 3.

Then area = sqrt( s(s - OA)(s - OB)(s - AB) )

So plugging in:

s = sqrt(14) + 3

s - OA = sqrt(14) + 3 - sqrt(14) = 3

Similarly, s - OB = 3

s - AB = sqrt(14) + 3 - 6 = sqrt(14) - 3

Therefore, area = sqrt( [sqrt(14) + 3] * 3 * 3 * [sqrt(14) - 3] )

Let me compute that. First, note that [sqrt(14) + 3][sqrt(14) - 3] is a difference of squares: (sqrt(14))² - 3² = 14 - 9 = 5.

Therefore, the expression under the square root becomes 5 * 3 * 3 = 45. Therefore, area = sqrt(45) = 3*sqrt(5).

Which matches the previous result! So that's reassuring. Therefore, the area is indeed 3√5.

So both methods give the same answer, so I think that's correct.

Alternatively, another way to think about it is that the area of the parallelogram formed by OA and OB is |OA × OB|, so the area of the triangle is half that. Which is exactly what I did first.

Therefore, I can be confident that the area of triangle OAB is 3√5.

Just to recap, steps were:

1. Find vectors OA and OB.

2. Compute their cross product.

3. Find the magnitude of the cross product.

4. Divide by 2 to get the area of the triangle.

Alternatively, using Heron's formula with the lengths of the sides also gives the same answer, which is a good check.

Therefore, the answer is 3√5.

**Final Answer**
The area of triangle $ OAB $ is \boxed{3\sqrt{5}}.

To find the area of triangle $ OAB $, we are given the position vectors of points $ A $ and $ B $ with respect to the origin $ O $:

$$
\vec{OA} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \quad \vec{OB} = \begin{pmatrix} -3 \\ -2 \\ 1 \end{pmatrix}
$$

---

### Step 1: Use the Cross Product to Find the Area

The area of triangle $ OAB $ can be calculated using the formula:

$$
\text{Area} = \frac{1}{2} \left\| \vec{OA} \times \vec{OB} \right\|
$$

We compute the cross product $ \vec{OA} \times \vec{OB} $ using the determinant method:

$$
\vec{OA} \times \vec{OB} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & 3 \\
-3 & -2 & 1
\end{vmatrix}
= \mathbf{i}(2 \cdot 1 - 3 \cdot (-2)) - \mathbf{j}(1 \cdot 1 - 3 \cdot (-3)) + \mathbf{k}(1 \cdot (-2) - 2 \cdot (-3))
$$

$$
= \mathbf{i}(2 + 6) - \mathbf{j}(1 + 9) + \mathbf{k}(-2 + 6)
= 8\mathbf{i} - 10\mathbf{j} + 4\mathbf{k}
$$

So,

$$
\vec{OA} \times \vec{OB} = \begin{pmatrix} 8 \\ -10 \\ 4 \end{pmatrix}
$$

---

### Step 2: Compute the Magnitude of the Cross Product

$$
\left\| \vec{OA} \times \vec{OB} \right\| = \sqrt{8^2 + (-10)^2 + 4^2} = \sqrt{64 + 100 + 16} = \sqrt{180}
$$

Simplify:

$$
\sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5}
$$

---

### Step 3: Compute the Area of the Triangle

$$
\text{Area} = \frac{1}{2} \cdot 6\sqrt{5} = 3\sqrt{5}
$$

---

### Final Answer

$$
\boxed{3\sqrt{5}}
$$

Solution:
The area of the parallelogram generated by $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is given by
\[\|\overrightarrow{OA} \times \overrightarrow{OB}\| = \left\| \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \times \begin{pmatrix} -3 \\ -2 \\ 1 \end{pmatrix} \right\| = \left\| \begin{pmatrix} 8 \\ -10 \\ 4 \end{pmatrix} \right\| = 6 \sqrt{5}.\][asy]
unitsize(0.4 cm);

pair A, B, C, D;

A = (0,0);
B = (7,2);
C = (1,3);
D = B + C;

draw(A--B,Arrow(6));
draw(A--C,Arrow(6));
draw(B--C);
draw(B--D--C,dashed);

label("$O$", A, SW);
label("$A$", B, SE);
label("$B$", C, W);
[/asy]

Therefore, the area of triangle $OAB$ is $\boxed{3 \sqrt{5}}.$


Explanation:
Okay, so I need to simplify the fraction 3 over the square root of 27. Let me think about how to approach this. I remember that simplifying expressions with square roots often involves rationalizing the denominator, which means getting rid of the square root in the denominator. But first, maybe I should simplify the square root of 27 as much as possible?

Alright, let me recall that 27 is a multiple of 9, which is a perfect square. So, maybe I can break down the square root of 27 into the square root of 9 times the square root of 3. Let me check that. Yes, because 9 times 3 is 27. So, √27 = √(9*3). And since the square root of a product is the product of the square roots, that would be √9 * √3. And √9 is 3, right? So that simplifies to 3√3.

So, substituting back into the original expression, the denominator becomes 3√3. Therefore, the original fraction 3 over √27 simplifies to 3 over 3√3. Wait, so that would be 3 divided by 3√3. Hmm, the 3 in the numerator and the 3 in the denominator can cancel each other out. So that leaves me with 1 over √3.

But now I have a square root in the denominator again. I think the standard form is to have no radicals in the denominator. So, I need to rationalize the denominator here. How do I do that? I remember that you can multiply the numerator and the denominator by the same radical to eliminate the square root in the denominator.

In this case, the denominator is √3, so if I multiply both the numerator and the denominator by √3, that should rationalize it. Let me try that.

So, multiplying numerator and denominator by √3 gives me (1 * √3) over (√3 * √3). The numerator becomes √3, and the denominator becomes √3 squared, which is 3. So, that simplifies to √3 over 3.

Wait, so putting it all together, starting from 3/√27, simplifying √27 to 3√3, then simplifying the fraction to 1/√3, and then rationalizing to get √3/3. Is that the simplest form? Let me check if there's another way or if I made a mistake somewhere.

Alternatively, maybe I could have simplified the original expression differently. Let me try another approach. Let's take the original expression: 3 divided by √27. Instead of simplifying √27 first, maybe I can express 3 as a multiple of √27? Wait, that might not be helpful. Let me think.

Alternatively, maybe I can write both numerator and denominator in terms of exponents. Let me recall that √27 is 27^(1/2). And 27 is 3^3. So, √27 is (3^3)^(1/2) which is 3^(3/2). Therefore, the original expression is 3 divided by 3^(3/2).

Using exponent rules, dividing exponents with the same base means subtracting exponents. So, 3 is 3^1, so 3^1 divided by 3^(3/2) is 3^(1 - 3/2) = 3^(-1/2). Which is the same as 1 over 3^(1/2), which is 1 over √3. Then, rationalizing gives √3 over 3. So, same result as before.

Therefore, regardless of the method, I end up with √3 over 3. Let me check if that's correct by approximating the original expression and the simplified one numerically.

Original expression: 3 divided by √27. Let me compute √27 first. √25 is 5, √27 is a bit more than 5. Let me calculate it. 5.1 squared is 26.01, 5.2 squared is 27.04. So √27 is approximately 5.196. Therefore, 3 divided by 5.196 is approximately 0.577.

Now, √3 is approximately 1.732, so √3 divided by 3 is approximately 1.732 / 3 ≈ 0.577. So, same value. Therefore, that seems correct.

Alternatively, if I didn't rationalize, 1 over √3 is approximately 1 / 1.732 ≈ 0.577, which is the same. So, both forms are equivalent, but the rationalized form is considered more simplified.

Therefore, the simplified form of 3 over √27 is √3 over 3.

Wait, but let me check if there's a way to simplify it further. For example, can √3 over 3 be simplified more? I don't think so. √3 is already in its simplest radical form, and 3 is a rational number. So, that's as simplified as it gets.

Alternatively, maybe I can write it as (1/3)√3, but that's essentially the same thing. So, both forms are acceptable, but typically, it's written with the radical in the numerator.

Therefore, I think the answer is √3 over 3.

But let me go back through the steps again to make sure I didn't skip anything.

Original problem: 3 / √27.

Step 1: Simplify √27. Since 27 = 9 * 3, √27 = √(9*3) = √9 * √3 = 3√3.

So, substituting back, 3 / (3√3).

Step 2: Simplify numerator and denominator. 3 divided by 3 is 1, so we have 1 / √3.

Step 3: Rationalize the denominator. Multiply numerator and denominator by √3.

(1 * √3) / (√3 * √3) = √3 / 3.

Yes, that's correct.

Alternatively, if I didn't simplify √27 first, but instead rationalized the original expression. Let me try that approach.

Original expression: 3 / √27.

Multiply numerator and denominator by √27 to rationalize.

So, (3 * √27) / (√27 * √27) = (3√27) / 27.

Then, simplify numerator and denominator.

First, simplify √27. As before, √27 = 3√3. Therefore, numerator becomes 3 * 3√3 = 9√3.

Denominator is 27.

So, 9√3 / 27. Simplify by dividing numerator and denominator by 9: √3 / 3.

Same result.

Therefore, regardless of the method, the simplified form is √3 over 3.

I think that's solid. Let me check if there's any alternative interpretation of the original problem. For example, is there a possibility that the original expression was meant to be something else? Like, maybe 3 divided by sqrt(27), which is what I assumed.

Alternatively, if someone misreads it as sqrt(3)/27, but that's not the case here. The original problem is written as 3 over sqrt(27), so the fraction is 3 divided by the square root of 27.

Another check: Let me compute both the original expression and the simplified expression using exact values.

Original expression: 3 / √27.

Simplified expression: √3 / 3.

Let me square both to see if they are equal.

Original squared: (3 / √27)^2 = 9 / 27 = 1/3.

Simplified squared: (√3 / 3)^2 = 3 / 9 = 1/3.

Same result. Therefore, they are equal.

Therefore, the simplification is correct.

So, after all these steps and checks, I can confidently say that the simplified form of 3 over the square root of 27 is √3 over 3.

**Final Answer**
The simplified form is \boxed{\dfrac{\sqrt{3}}{3}}.

To simplify the expression:

$$
\frac{3}{\sqrt{27}}
$$

we follow a systematic approach, focusing on simplifying the square root in the denominator and rationalizing it if necessary.

---

### Step 1: Simplify the Square Root in the Denominator

We begin by simplifying the square root of 27. Since 27 is not a perfect square, we factor it into a product of a perfect square and another number:

$$
\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}
$$

So, the original expression becomes:

$$
\frac{3}{\sqrt{27}} = \frac{3}{3\sqrt{3}}
$$

---

### Step 2: Simplify the Fraction

Now, we simplify the fraction by canceling the common factor of 3 in the numerator and denominator:

$$
\frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}}
$$

---

### Step 3: Rationalize the Denominator

To express the fraction in a standard form with no radicals in the denominator, we rationalize the denominator. Multiply both the numerator and the denominator by $\sqrt{3}$:

$$
\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}
$$

---

### Final Answer

The simplified form of the expression is:

$$
\boxed{\dfrac{\sqrt{3}}{3}}
$$

[Assistant]
Explanation:
Okay, so I need to simplify the fraction 3 over the square root of 27. Let me think about how to approach this. I remember that simplifying expressions with square roots often involves rationalizing the denominator, which means getting rid of the square root in the denominator. But first, maybe I should simplify the square root of 27 as much as possible?

Alright, let me recall that 27 is a multiple of 9, which is a perfect square. So, maybe I can break down the square root of 27 into the square root of 9 times the square root of 3. Let me check that. Yes, because 9 times 3 is 27. So, √27 = √(9*3). And since the square root of a product is the product of the square roots, that would be √9 * √3. And √9 is 3, right? So that simplifies to 3√3.

So, substituting back into the original expression, the denominator becomes 3√3. Therefore, the original fraction 3 over √27 simplifies to 3 over 3√3. Wait, so that would be 3 divided by 3√3. Hmm, the 3 in the numerator and the 3 in the denominator can cancel each other out. So that leaves me with 1 over √3.

But now I have a square root in the denominator again. I think the standard form is to have no radicals in the denominator. So, I need to rationalize the denominator here. How do I do that? I remember that you can multiply the numerator and the denominator by the same radical to eliminate the square root in the denominator.

In this case, the denominator is √3, so if I multiply both the numerator and the denominator by √3, that should rationalize it. Let me try that.

So, multiplying numerator and denominator by √3 gives me (1 * √3) over (√3 * √3). The numerator becomes √3, and the denominator becomes √3 squared, which is 3. So, that simplifies to √3 over 3.

Wait, so putting it all together, starting from 3/√27, simplifying √27 to 3√3, then simplifying the fraction to 1/√3, and then rationalizing to get √3/3. Is that the simplest form? Let me check if there's another way or if I made a mistake somewhere.

Alternatively, maybe I could have simplified the original expression differently. Let me try another approach. Let's take the original expression: 3 divided by √27. Instead of simplifying √27 first, maybe I can express 3 as a multiple of √27? Wait, that might not be helpful. Let me think.

Alternatively, maybe I can write both numerator and denominator in terms of exponents. Let me recall that √27 is 27^(1/2). And 27 is 3^3. So, √27 is (3^3)^(1/2) which is 3^(3/2). Therefore, the original expression is 3 divided by 3^(3/2).

Using exponent rules, dividing exponents with the same base means subtracting exponents. So, 3 is 3^1, so 3^1 divided by 3^(3/2) is 3^(1 - 3/2) = 3^(-1/2). Which is the same as 1 over 3^(1/2), which is 1 over √3. Then, rationalizing gives √3 over 3. So, same result as before.

Therefore, regardless of the method, I end up with √3 over 3. Let me check if that's correct by approximating the original expression and the simplified one numerically.

Original expression: 3 divided by √27. Let me compute √27 first. √25 is 5, √27 is a bit more than 5. Let me calculate it. 5.1 squared is 26.01, 5.2 squared is 27.04. So √27 is approximately 5.196. Therefore, 3 divided by 5.196 is approximately 0.577.

Now, √3 is approximately 1.732, so √3 divided by 3 is approximately 1.732 / 3 ≈ 0.577. So, same value. Therefore, that seems correct.

Alternatively, if I didn't rationalize, 1 over √3 is approximately 1 / 1.732 ≈ 0.577, which is the same. So, both forms are equivalent, but the rationalized form is considered more simplified.

Therefore, the simplified form of 3 over √27 is √3 over 3.

Wait, but let me check if there's a way to simplify it further. For example, can √3 over 3 be simplified more? I don't think so. √3 is already in its simplest radical form, and 3 is a rational number. So, that's as simplified as it gets.

Alternatively, maybe I can write it as (1/3)√3, but that's essentially the same thing. So, both forms are acceptable, but typically, it's written with the radical in the numerator.

Therefore, I think the answer is √3 over 3.

But let me go back through the steps again to make sure I didn't skip anything.

Original problem: 3 / √27.

Step 1: Simplify √27. Since 27 = 9 * 3, √27 = √(9*3) = √9 * √3 = 3√3.

So, substituting back, 3 / (3√3).

Step 2: Simplify numerator and denominator. 3 divided by 3 is 1, so we have 1 / √3.

Step 3: Rationalize the denominator. Multiply numerator and denominator by √3.

(1 * √3) / (√3 * √3) = √3 / 3.

Yes, that's correct.

Alternatively, if I didn't simplify √27 first, but instead rationalized the original expression. Let me try that approach.

Original expression: 3 / √27.

Multiply numerator and denominator by √27 to rationalize.

So, (3 * √27) / (√27 * √27) = (3√27) / 27.

Then, simplify numerator and denominator.

First, simplify √27. As before, √27 = 3√3. Therefore, numerator becomes 3 * 3√3 = 9√3.

Denominator is 27.

So, 9√3 / 27. Simplify by dividing numerator and denominator by 9: √3 / 3.

Same result.

Therefore, regardless of the method, the simplified form is √3 over 3.

I think that's solid. Let me check if there's any alternative interpretation of the original problem. For example, is there a possibility that the original expression was meant to be something else? Like, maybe 3 divided by sqrt(27), which is what I assumed.

Alternatively, if someone misreads it as sqrt(3)/27, but that's not the case here. The original problem is written as 3 over sqrt(27), so the fraction is 3 divided by the square root of 27.

Another check: Let me compute both the original expression and the simplified expression using exact values.

Original expression: 3 / √27.

Simplified expression: √3 / 3.

Let me square both to see if they are equal.

Original squared: (3 / √27)^2 = 9 / 27 = 1/3.

Simplified squared: (√3 / 3)^2 = 3 / 9 = 1/3.

Same result. Therefore, they are equal.

Therefore, the simplification is correct.

So, after all these steps and checks, I can confidently say that the simplified form of 3 over the square root of 27 is √3 over 3.

**Final Answer**
The simplified form is \boxed{\dfrac{\sqrt{3}}{3}}.

To simplify the expression:

$$
\frac{3}{\sqrt{27}}
$$

we follow a systematic approach, focusing on simplifying the square root in the denominator and rationalizing it if necessary.

---

### Step 1: Simplify the Square Root in the Denominator

We begin by simplifying the square root of 27. Since 27 is not a perfect square, we factor it into a product of a perfect square and another number:

$$
\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}
$$

So, the original expression becomes:

$$
\frac{3}{\sqrt{27}} = \frac{3}{3\sqrt{3}}
$$

---

### Step 2: Simplify the Fraction

Now, we simplify the fraction by canceling the common factor of 3 in the numerator and denominator:

$$
\frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}}
$$

---

### Step 3: Rationalize the Denominator

To express the fraction in a standard form with no radicals in the denominator, we rationalize the denominator. Multiply both the numerator and the denominator by $\sqrt{3}$:

$$
\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}
$$

---

### Final Answer

The simplified form of the expression is:

$$
\boxed{\dfrac{\sqrt{3}}{3}}
$$

Solution:
We have:

$\frac{3}{\sqrt{27}}=\frac{3\sqrt{3}}{\sqrt{81}}=\frac{3\sqrt{3}}{9}=\boxed{\frac{\sqrt{3}}{3}}$.