<Question>
Find the maximum value of
\[f(x,y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2},\]where $-1 \le x,$ $y \le 1.$
</Question>

<think>
<Plan_1>
Okay, so I need to find the maximum value of the function f(x, y) = x√(1 - y²) + y√(1 - x²) where both x and y are between -1 and 1. Hmm, let's start by understanding the problem. The function has two variables, x and y, and it's a combination of terms involving each variable multiplied by the square root of the other variable's term. The square roots suggest that the expressions inside must be non-negative, which they are since x and y are between -1 and 1. So, the domains are okay.

First, maybe I should check the boundaries. Since x and y are both in [-1, 1], maybe the maximum occurs at one of the endpoints. Let me test some points.

<prompt_1.1> Evaluate the function at the points (1,1), (1,0), (0,1), and (0,0) to check for maximum values. </prompt_1.1>

But maybe there's a higher value somewhere inside the domain. Let's try some other points.

<prompt_1.2> Evaluate the function at the points (1/2,1/2) and (-1/2,-1/2) to see if the maximum is higher inside the domain. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's try (1,1): f(1,1) = 1*√(1 - 1) + 1*√(1 - 1) = 0 + 0 = 0. That's not good. How about (1,0)? f(1,0) = 1*√(1 - 0) + 0*√(1 - 1) = 1*1 + 0 = 1. Similarly, (0,1) would give the same result: 0*√(1 - 1) + 1*√(1 - 0) = 1. And (0,0) is 0. So, on the axes, the maximum is 1.
</execution_1.1>

<execution_1.2>
What if x = y = 1/2? Then f(1/2, 1/2) would be (1/2)√(1 - (1/2)^2) + (1/2)√(1 - (1/2)^2) = 2*(1/2)*√(3/4) = 2*(1/2)*(√3)/2 = √3/2 ≈ 0.866. That's higher than 0. But what if x and y are both negative? Let's try x = y = -1/2. Then f(-1/2, -1/2) = (-1/2)√(1 - (1/4)) + (-1/2)√(1 - (1/4)) = -√3/2 ≈ -0.866. So negative, so not the maximum.

What if x = 1? Then f(1, y) = 1*√(1 - y²) + y*√(1 - 1) = √(1 - y²) + 0 = √(1 - y²). The maximum of this would be when y = 0, giving √1 = 1. Similarly, if y = 1, then f(x, 1) = x*0 + 1*√(1 - x²) = √(1 - x²), which is maximized at x = 0, giving 1. So on the boundaries where either x or y is 1 or -1, the maximum seems to be 1. But earlier, when both x and y are 1/2, we got √3/2 ≈ 0.866, which is less than 1. Wait, but maybe there's a point inside the domain where the function is higher than 1?
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.2:
Wait, let's check another point. Suppose x = y. Let’s set x = y, then the function becomes f(x, x) = x√(1 - x²) + x√(1 - x²) = 2x√(1 - x²). To find the maximum of this, we can take the derivative with respect to x. Let’s compute that.

<prompt_2.1> Find the maximum of the function when x = y by taking the derivative. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Let’s denote g(x) = 2x√(1 - x²). Then g’(x) = 2√(1 - x²) + 2x*(1/(2√(1 - x²)))*(-2x) = 2√(1 - x²) - (2x²)/√(1 - x²). Setting this equal to zero:

2√(1 - x²) - (2x²)/√(1 - x²) = 0

Multiply both sides by √(1 - x²):

2(1 - x²) - 2x² = 0

2 - 2x² - 2x² = 0

2 - 4x² = 0

4x² = 2

x² = 1/2

x = ±√(1/2) = ±√2/2 ≈ ±0.707

Since x is between -1 and 1, let's take x = √2/2 ≈ 0.707. Then g(x) = 2*(√2/2)*√(1 - (1/2)) = √2 * √(1/2) = √2*(√2/2) = (2)/2 = 1. So the maximum when x = y is 1.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Hmm, so that suggests that the maximum is 1. But wait, is that the case? Let me check another point.

<prompt_3.1> Evaluate the function at the point (√2/2, √2/2) to confirm the maximum value. </prompt_3.1>

But wait, maybe there's a point where x ≠ y that gives a higher value? Let me check. Suppose x is something else. Let's try x = √3/2 ≈ 0.866, y = 1/2.

<prompt_3.2> Evaluate the function at the points (√3/2, 1/2) and (1/2, √3/2) to see if a higher maximum is possible. </prompt_3.2>

But let's check another possibility. Suppose x and y are such that both terms are positive and maybe their sum is larger. For example, if x is larger and y is smaller, but not sure.

Alternatively, maybe use calculus. To find the maximum of f(x, y), we can take partial derivatives with respect to x and y, set them to zero, and solve.

<prompt_3.3> Use calculus to find the critical points of the function and determine the maximum value. </prompt_3.3>
</Plan_3>

<Execution_3>
<execution_3.1>
If x = y = √2/2 ≈ 0.707, then f(x, y) = 1, as we found. But when x and y are 1, the function is 0, so 1 is higher. But is there a point where f(x, y) > 1?
</execution_3.1>

<execution_3.2>
Then f(x, y) = (√3/2)*√(1 - (1/4)) + (1/2)*√(1 - (3/4)) = (√3/2)*√(3/4) + (1/2)*√(1/4) = (√3/2)*(√3/2) + (1/2)*(1/2) = (3/4) + (1/4) = 1. So that's also 1.

Wait, so even if x and y are different, as long as one is √3/2 and the other is 1/2, the sum is 1. Hmm. So maybe the maximum is indeed 1. But let me check another combination. Suppose x = √(2)/2 ≈ 0.707 and y = √(2)/2. Then we already saw that gives 1. What if x is something else? Let's say x = 0. Let's see, then f(0, y) = 0 + y*√1 = y. So maximum at y=1, which gives 1. Similarly, if y=0, then f(x,0) = x*0 + 0 = 0. So, on the edges where one variable is 0, the function is either 0 or y, which is at most 1.
</execution_3.2>

<execution_3.3>
Let’s compute the partial derivatives.

First, ∂f/∂x = √(1 - y²) + y*(1/(2√(1 - x²)))*(-2x) = √(1 - y²) - (xy)/√(1 - x²)

Similarly, ∂f/∂y = √(1 - x²) + x*(1/(2√(1 - y²)))*(-2y) = √(1 - x²) - (xy)/√(1 - y²)

Set both partial derivatives to zero:

1. √(1 - y²) - (xy)/√(1 - x²) = 0
2. √(1 - x²) - (xy)/√(1 - y²) = 0

Hmm, so we have a system of equations. Let me write them as:

√(1 - y²) = (xy)/√(1 - x²)  ... (1)

√(1 - x²) = (xy)/√(1 - y²)  ... (2)

Let’s square both sides of equation (1) to eliminate the square roots:

1 - y² = (x² y²)/(1 - x²)

Multiply both sides by (1 - x²):

(1 - y²)(1 - x²) = x² y²

Expand the left side:

1 - x² - y² + x² y² = x² y²

Subtract x² y² from both sides:

1 - x² - y² = 0

So 1 = x² + y²

Similarly, if we square equation (2) and do the same, we would get the same result. So from the critical points, we must have x² + y² = 1.

So the critical points lie on the unit circle. But since x and y are in [-1, 1], this makes sense.
</execution_3.3>
</Execution_3>

<Plan_4>
Based on execution_3.3:
So, substituting x² + y² = 1 into the original function. Let's see.

Given x² + y² = 1, then y² = 1 - x². Let's express f(x, y) in terms of x.

<prompt_4.1> Simplify the function using the constraint x² + y² = 1 and find the maximum value. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
f(x, y) = x√(1 - y²) + y√(1 - x²) = x√(x²) + y√(y²) = x|x| + y|y|.

But since x and y are real numbers between -1 and 1, and given that x² + y² =1, x and y can be positive or negative. However, if x and y are both positive, then |x| = x and |y| = y, so f(x, y) = x² + y² = 1. If one is positive and the other negative, then f(x, y) = x² - y² or -x² + y². If both are negative, then f(x, y) = -x² - y² = -1. So, in the case where x and y are both positive and on the unit circle, f(x, y) =1. If one is positive and the other negative, then depending on which is larger in magnitude, the value could be less than 1 or even negative. For example, if x is positive and y is negative, then f(x, y) = x² - y². Since x² + y² =1, this becomes x² - (1 - x²) = 2x² -1. The maximum of this occurs when x² is maximized. Since x² ≤1, the maximum would be 2*1 -1=1. Similarly, if y is positive and x is negative, it's the same. So in those cases, the maximum is still 1. However, if x and y are both negative, then f(x, y) = -x² - y² = -1, which is the minimum.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Therefore, the maximum value of f(x, y) is 1, achieved when x and y are both positive and satisfy x² + y² =1, or when one is positive and the other is negative such that x² + y²=1. Wait, but when one is positive and the other is negative, we saw that the maximum is still 1. For example, if x is 1 and y is 0, we get 1. If x is √(2)/2 and y is √(2)/2, we get 1. If x is √3/2 and y is 1/2, we also get 1.

But let me check if there's a case where the function exceeds 1. Suppose x and y are both greater than 0 and on the unit circle. Then f(x, y) = x² + y² =1. If x and y are on the unit circle but one is negative, then as we saw, the function can be 1 or less. If both are negative, it's -1. So the maximum is indeed 1.

But wait, earlier when we considered x = y = √2/2 ≈0.707, which satisfies x² + y² = 0.5 +0.5=1, then f(x, y)=1. So that point is on the unit circle and gives f(x, y)=1.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Therefore, the maximum value of $f(x, y)$ under the given constraints is $\boxed{1}$.
</Final_answer>
</think>