Explanation:
Okay, so Rick is thinking of a positive factor of 14, and Steve is thinking of a positive factor of 42. They are both thinking of the same number. The question is asking how many possible numbers they could be thinking of. Hmm, let me think about this step by step.

First, I need to recall what factors are. A factor of a number is an integer that divides that number exactly without leaving a remainder. So, for example, the factors of 14 are numbers that can multiply by another integer to get 14. Similarly for 42.

Let me start by listing all the positive factors of 14. Let me think... 14 is a relatively small number, so this shouldn't be too hard. The factors of 14 would be 1, 2, 7, and 14. Let me check that. 1 times 14 is 14, 2 times 7 is 14. Are there any others? Well, 3 doesn't divide 14 because 14 divided by 3 is about 4.666... which isn't an integer. 4? 14 divided by 4 is 3.5, not an integer. 5? 14 divided by 5 is 2.8, nope. 6? 14 divided by 6 is about 2.333... So no. So yeah, the factors of 14 are 1, 2, 7, 14. Got that.

Now, Steve is thinking of a positive factor of 42. Let me list those as well. 42 is a bigger number, so there will be more factors. Let me think. Starting from 1: 1 is a factor. Then 2, because 42 divided by 2 is 21. Then 3, because 42 divided by 3 is 14. Then 6, because 6 times 7 is 42. Then 7, which we already have. Then 14, which is also a factor. Then 21, because 21 times 2 is 42. And finally 42 itself. Let me check if I missed any. Let me list them in order:

1, 2, 3, 6, 7, 14, 21, 42. Let me verify each one:

1 × 42 = 42

2 × 21 = 42

3 × 14 = 42

6 × 7 = 42

Yes, that's all. So the factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42.

Now, the problem says that Rick and Steve are thinking of the same number. So that number has to be a common factor of both 14 and 42. Therefore, we need to find the common factors of 14 and 42. Then, the number of possible numbers they could be thinking of is the number of common factors.

Alternatively, since Rick's number is a factor of 14 and Steve's number is a factor of 42, and they are the same number, so the intersection of the two sets of factors. So let me find the common factors between the two.

First, let me list the factors of 14 again: 1, 2, 7, 14.

Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42.

So the common factors would be numbers that are in both lists. Let me check each factor of 14 against the factors of 42.

Starting with 1: 1 is in both lists. So that's a common factor.

Next, 2: 2 is in both lists. So that's another common factor.

Then 7: 7 is in both lists. So that's another one.

Then 14: 14 is in both lists. So that's another one.

So the common factors are 1, 2, 7, 14. Therefore, there are 4 common factors. Therefore, the number of possible numbers they could be thinking of is 4. So the answer is 4?

Wait, let me make sure I didn't miss anything. Let me check again.

Factors of 14: 1, 2, 7, 14.

Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42.

Common factors: 1, 2, 7, 14. Yep, that's four numbers. So that would mean there are four possible numbers they could be thinking of. Therefore, the answer is 4.

But let me think again if there's another way to approach this problem. Maybe using the greatest common divisor (GCD) concept? Because the common factors of two numbers are exactly the factors of their GCD. So maybe I can compute GCD(14, 42) and then find the number of factors of that GCD.

Let me try that. The GCD of 14 and 42. Well, since 14 divides 42 (because 42 divided by 14 is 3), the GCD of 14 and 42 is 14. Therefore, the common factors of 14 and 42 are the factors of 14. Which are 1, 2, 7, 14. So again, that gives us 4 factors. Therefore, the answer is 4.

So both methods give me the same answer, so that seems to confirm that the answer is 4.

But just to be thorough, let me check if there's any possibility that I made a mistake in listing the factors. For example, maybe I missed a factor of 14 or 42?

Factors of 14: 1, 2, 7, 14. Let me check if there are any others. Let me think of prime factors. 14 is 2 × 7. Therefore, the factors are all combinations of these primes. So 1, 2, 7, and 2×7=14. So that's correct.

Factors of 42: 42 is 2 × 3 × 7. Therefore, the factors are all combinations of these primes. Let me list them:

1 (which is 2^0 × 3^0 × 7^0)

2 (2^1 × 3^0 × 7^0)

3 (2^0 × 3^1 × 7^0)

6 (2^1 × 3^1 × 7^0)

7 (2^0 × 3^0 × 7^1)

14 (2^1 × 3^0 × 7^1)

21 (2^0 × 3^1 × 7^1)

42 (2^1 × 3^1 × 7^1)

So that's all the factors. So that's 8 factors, which matches the list I had before. So that's correct.

Therefore, the common factors between 14 and 42 are the factors of GCD(14, 42) = 14, which are 1, 2, 7, 14. So again, 4 numbers.

Therefore, the answer is 4.

But just to make sure, let me think if there's any other angle. For example, maybe the problem is considering something else? Like, maybe negative factors? But the problem says positive factors, so negative numbers are out of the question.

Also, the problem says "positive factor", so we don't have to consider anything else.

Alternatively, maybe there's a trick here? For example, maybe the number they are thinking of is a common factor, but maybe there's some other constraint? Let me check the problem again:

"Rick is thinking of a positive factor of 14 and Steve is thinking of a positive factor of 42. If Rick and Steve are thinking of the same number, how many possible numbers could they be thinking of?"

No, there's no other constraints. It's just that they are thinking of the same number, which is a factor of both 14 and 42. Therefore, the common factors. Which are 1, 2, 7, 14. So 4 numbers.

Therefore, I think the answer is 4.

But just to be 100% sure, let me check each number:

Is 1 a factor of 14? Yes. Is 1 a factor of 42? Yes. So that's valid.

Is 2 a factor of 14? Yes. 2 is a factor of 42? Yes. Valid.

Is 7 a factor of 14? Yes. 7 is a factor of 42? Yes. Valid.

Is 14 a factor of 14? Yes. 14 is a factor of 42? Yes. Valid.

So all four numbers are valid. Therefore, there are 4 possible numbers.

Therefore, the answer is 4.

I think that's solid. I don't see any mistakes in my reasoning. Both methods of listing the factors and using GCD lead to the same answer, and checking each number confirms that they are all valid. So yeah, I think that's the correct answer.

**Final Answer**
The number of possible numbers they could be thinking of is \boxed{4}.

To solve the problem, we need to determine how many **positive integers** are common to both the set of **factors of 14** and the set of **factors of 42**. These common numbers are the **common factors** of 14 and 42.

---

### Step 1: List the Factors of Each Number

**Factors of 14:**
We find all positive integers that divide 14 without a remainder:
- $1 \times 14 = 14$
- $2 \times 7 = 14$

So, the factors of 14 are:
$$
\{1, 2, 7, 14\}
$$

**Factors of 42:**
Similarly, we find all positive integers that divide 42:
- $1 \times 42 = 42$
- $2 \times 21 = 42$
- $3 \times 14 = 42$
- $6 \times 7 = 42$

So, the factors of 42 are:
$$
\{1, 2, 3, 6, 7, 14, 21, 42\}
$$

---

### Step 2: Find the Common Factors

Now, we find the intersection of the two sets:
- Factors of 14: $\{1, 2, 7, 14\}$
- Factors of 42: $\{1, 2, 3, 6, 7, 14, 21, 42\}$

The common elements are:
$$
\{1, 2, 7, 14\}
$$

So, there are **4** numbers that are common factors of both 14 and 42.

---

### Step 3: Confirm Using GCD

An alternative and more efficient method is to use the **Greatest Common Divisor (GCD)** of 14 and 42.

- $ \text{GCD}(14, 42) = 14 $, since 14 divides 42 exactly.

The **common factors** of 14 and 42 are exactly the **factors of their GCD**, which is 14. As we already listed, the factors of 14 are:
$$
\{1, 2, 7, 14\}
$$

Again, this confirms that there are **4** common factors.

---

### Final Answer

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so Rick is thinking of a positive factor of 14, and Steve is thinking of a positive factor of 42. They are both thinking of the same number. The question is asking how many possible numbers they could be thinking of. Hmm, let me think about this step by step.

First, I need to recall what factors are. A factor of a number is an integer that divides that number exactly without leaving a remainder. So, for example, the factors of 14 are numbers that can multiply by another integer to get 14. Similarly for 42.

Let me start by listing all the positive factors of 14. Let me think... 14 is a relatively small number, so this shouldn't be too hard. The factors of 14 would be 1, 2, 7, and 14. Let me check that. 1 times 14 is 14, 2 times 7 is 14. Are there any others? Well, 3 doesn't divide 14 because 14 divided by 3 is about 4.666... which isn't an integer. 4? 14 divided by 4 is 3.5, not an integer. 5? 14 divided by 5 is 2.8, nope. 6? 14 divided by 6 is about 2.333... So no. So yeah, the factors of 14 are 1, 2, 7, 14. Got that.

Now, Steve is thinking of a positive factor of 42. Let me list those as well. 42 is a bigger number, so there will be more factors. Let me think. Starting from 1: 1 is a factor. Then 2, because 42 divided by 2 is 21. Then 3, because 42 divided by 3 is 14. Then 6, because 6 times 7 is 42. Then 7, which we already have. Then 14, which is also a factor. Then 21, because 21 times 2 is 42. And finally 42 itself. Let me check if I missed any. Let me list them in order:

1, 2, 3, 6, 7, 14, 21, 42. Let me verify each one:

1 × 42 = 42

2 × 21 = 42

3 × 14 = 42

6 × 7 = 42

Yes, that's all. So the factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42.

Now, the problem says that Rick and Steve are thinking of the same number. So that number has to be a common factor of both 14 and 42. Therefore, we need to find the common factors of 14 and 42. Then, the number of possible numbers they could be thinking of is the number of common factors.

Alternatively, since Rick's number is a factor of 14 and Steve's number is a factor of 42, and they are the same number, so the intersection of the two sets of factors. So let me find the common factors between the two.

First, let me list the factors of 14 again: 1, 2, 7, 14.

Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42.

So the common factors would be numbers that are in both lists. Let me check each factor of 14 against the factors of 42.

Starting with 1: 1 is in both lists. So that's a common factor.

Next, 2: 2 is in both lists. So that's another common factor.

Then 7: 7 is in both lists. So that's another one.

Then 14: 14 is in both lists. So that's another one.

So the common factors are 1, 2, 7, 14. Therefore, there are 4 common factors. Therefore, the number of possible numbers they could be thinking of is 4. So the answer is 4?

Wait, let me make sure I didn't miss anything. Let me check again.

Factors of 14: 1, 2, 7, 14.

Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42.

Common factors: 1, 2, 7, 14. Yep, that's four numbers. So that would mean there are four possible numbers they could be thinking of. Therefore, the answer is 4.

But let me think again if there's another way to approach this problem. Maybe using the greatest common divisor (GCD) concept? Because the common factors of two numbers are exactly the factors of their GCD. So maybe I can compute GCD(14, 42) and then find the number of factors of that GCD.

Let me try that. The GCD of 14 and 42. Well, since 14 divides 42 (because 42 divided by 14 is 3), the GCD of 14 and 42 is 14. Therefore, the common factors of 14 and 42 are the factors of 14. Which are 1, 2, 7, 14. So again, that gives us 4 factors. Therefore, the answer is 4.

So both methods give me the same answer, so that seems to confirm that the answer is 4.

But just to be thorough, let me check if there's any possibility that I made a mistake in listing the factors. For example, maybe I missed a factor of 14 or 42?

Factors of 14: 1, 2, 7, 14. Let me check if there are any others. Let me think of prime factors. 14 is 2 × 7. Therefore, the factors are all combinations of these primes. So 1, 2, 7, and 2×7=14. So that's correct.

Factors of 42: 42 is 2 × 3 × 7. Therefore, the factors are all combinations of these primes. Let me list them:

1 (which is 2^0 × 3^0 × 7^0)

2 (2^1 × 3^0 × 7^0)

3 (2^0 × 3^1 × 7^0)

6 (2^1 × 3^1 × 7^0)

7 (2^0 × 3^0 × 7^1)

14 (2^1 × 3^0 × 7^1)

21 (2^0 × 3^1 × 7^1)

42 (2^1 × 3^1 × 7^1)

So that's all the factors. So that's 8 factors, which matches the list I had before. So that's correct.

Therefore, the common factors between 14 and 42 are the factors of GCD(14, 42) = 14, which are 1, 2, 7, 14. So again, 4 numbers.

Therefore, the answer is 4.

But just to make sure, let me think if there's any other angle. For example, maybe the problem is considering something else? Like, maybe negative factors? But the problem says positive factors, so negative numbers are out of the question.

Also, the problem says "positive factor", so we don't have to consider anything else.

Alternatively, maybe there's a trick here? For example, maybe the number they are thinking of is a common factor, but maybe there's some other constraint? Let me check the problem again:

"Rick is thinking of a positive factor of 14 and Steve is thinking of a positive factor of 42. If Rick and Steve are thinking of the same number, how many possible numbers could they be thinking of?"

No, there's no other constraints. It's just that they are thinking of the same number, which is a factor of both 14 and 42. Therefore, the common factors. Which are 1, 2, 7, 14. So 4 numbers.

Therefore, I think the answer is 4.

But just to be 100% sure, let me check each number:

Is 1 a factor of 14? Yes. Is 1 a factor of 42? Yes. So that's valid.

Is 2 a factor of 14? Yes. 2 is a factor of 42? Yes. Valid.

Is 7 a factor of 14? Yes. 7 is a factor of 42? Yes. Valid.

Is 14 a factor of 14? Yes. 14 is a factor of 42? Yes. Valid.

So all four numbers are valid. Therefore, there are 4 possible numbers.

Therefore, the answer is 4.

I think that's solid. I don't see any mistakes in my reasoning. Both methods of listing the factors and using GCD lead to the same answer, and checking each number confirms that they are all valid. So yeah, I think that's the correct answer.

**Final Answer**
The number of possible numbers they could be thinking of is \boxed{4}.

To solve the problem, we need to determine how many **positive integers** are common to both the set of **factors of 14** and the set of **factors of 42**. These common numbers are the **common factors** of 14 and 42.

---

### Step 1: List the Factors of Each Number

**Factors of 14:**
We find all positive integers that divide 14 without a remainder:
- $1 \times 14 = 14$
- $2 \times 7 = 14$

So, the factors of 14 are:
$$
\{1, 2, 7, 14\}
$$

**Factors of 42:**
Similarly, we find all positive integers that divide 42:
- $1 \times 42 = 42$
- $2 \times 21 = 42$
- $3 \times 14 = 42$
- $6 \times 7 = 42$

So, the factors of 42 are:
$$
\{1, 2, 3, 6, 7, 14, 21, 42\}
$$

---

### Step 2: Find the Common Factors

Now, we find the intersection of the two sets:
- Factors of 14: $\{1, 2, 7, 14\}$
- Factors of 42: $\{1, 2, 3, 6, 7, 14, 21, 42\}$

The common elements are:
$$
\{1, 2, 7, 14\}
$$

So, there are **4** numbers that are common factors of both 14 and 42.

---

### Step 3: Confirm Using GCD

An alternative and more efficient method is to use the **Greatest Common Divisor (GCD)** of 14 and 42.

- $ \text{GCD}(14, 42) = 14 $, since 14 divides 42 exactly.

The **common factors** of 14 and 42 are exactly the **factors of their GCD**, which is 14. As we already listed, the factors of 14 are:
$$
\{1, 2, 7, 14\}
$$

Again, this confirms that there are **4** common factors.

---

### Final Answer

$$
\boxed{4}
$$

Solution: We will find the positive divisors of 14 by finding pairs that multiply to 14.   We begin our list as follows, $$1 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 14.$$  Checking $2$, we find that $2\cdot 7=14$, so our list becomes $$1 \quad 2 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 7 \quad 14.$$  Checking $3$, $4$, $5$, and $6$, we find that none of these are divisors of $14$, so our final list is $$1 \quad 2 \quad 7 \quad 14.$$  Next, we use the buddy method to determine the factors of $42$.  We begin our list as follows, $$1\quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 42.$$  Checking $2$, we find that $2\cdot 21=42$, so our list becomes $$1\quad 2 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 21 \quad 42.$$  Checking $3$, we find that $3\cdot 14=42$, so our list becomes $$1\quad 2 \quad 3 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 14 \quad 21 \quad 42.$$  Checking $4$ and $5$ we find that $4$ and $5$ are not divisors of $42$.  Checking $6$, we find that $6\cdot 7=42$, so our list becomes $$1\quad 2 \quad 3 \quad 6 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 7 \quad 14 \quad 21 \quad 42.$$  Since $7$ is already on our list, our final list is $$1\quad 2 \quad 3 \quad 6 \quad 7 \quad 14 \quad 21 \quad 42.$$  We compare our lists for the factors of $14$ and the factors of $42$ to see that the factors that $14$ and $42$ share are $1$, $2$, $7$, and $14$.  Therefore, Rick and Steve could be thinking of $\boxed{4}$ possible numbers.  Note that since $14$ is a factor of $42$, all of the factors of $14$ are also factors of $42$.


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. So none of the angles are reflex angles. That might be important later if I need to check my answers.

Now, the problem states 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. So let me note that down:

- Let the smallest angle be x degrees.
- Then the largest angle is 2x degrees.
- The other two angles are each 90 degrees.

Since there are four angles in a quadrilateral, adding them all up should give 360 degrees. So I can set up an equation:

x (smallest angle) + 2x (largest angle) + 90 (right angle) + 90 (another right angle) = 360 degrees.

Let me write that out:

x + 2x + 90 + 90 = 360

Simplifying the left side:

x + 2x is 3x, and 90 + 90 is 180. So:

3x + 180 = 360

Now, subtract 180 from both sides:

3x = 360 - 180
3x = 180

Then divide both sides by 3:

x = 60

So the smallest angle is 60 degrees, and the largest angle is twice that, so 120 degrees. Wait, but hold on. Let me check if that makes sense.

So the four angles would be 60°, 120°, 90°, and 90°. Adding them up: 60 + 120 + 90 + 90 = 360. Yep, that adds up.

But wait, the problem says it's a convex quadrilateral. So all angles must be less than 180°. The largest angle here is 120°, which is less than 180°, so that's okay. The other angles are 90°, which is fine. The smallest angle is 60°, which is also fine.

But let me think again. Is there a possibility that the largest angle could be something else? For example, maybe the problem is referring to the largest angle being twice the smallest, but maybe there's another angle that's larger? Wait, no. The problem says "the measure of the largest angle is twice the measure of the smallest angle," so that implies that the other two angles are not larger than that. Since the other two angles are right angles, which are 90°, so if the largest angle is 120°, that's bigger than 90°, so that's okay.

But let me check if there's another way to interpret the problem. Maybe the other two angles are not necessarily right angles? Wait, no, the problem says "the other two angles are both right angles." So that's clear. So the four angles are: smallest angle (x), largest angle (2x), and two right angles (90° each).

So the calculation seems straightforward. But let me just verify once more.

Total angles: 360.

x + 2x + 90 + 90 = 360.

3x + 180 = 360.

3x = 180.

x = 60.

So largest angle is 120.

But wait, is there a case where the largest angle could be more than 120? For example, if the smallest angle was smaller? But the problem says that the largest angle is twice the smallest. So if the smallest angle was smaller, then the largest would be even smaller? Wait, no. If the smallest angle is smaller, then the largest angle would be twice that, but then the other angles are fixed at 90. Wait, but if the smallest angle is smaller, then the largest angle would be smaller? Wait, no. If the smallest angle is smaller, then twice that would be smaller? Wait, no. If x is smaller, then 2x is smaller? Wait, no. If x is smaller, then 2x is smaller? Wait, no. If x is smaller, then 2x is smaller? Wait, for example, if x is 30, then 2x is 60. If x is 60, then 2x is 120. So if x is smaller, then 2x is smaller. But in that case, the other angles are 90. So if the largest angle is 2x, then if x is smaller, 2x would be smaller than 120. But then, if x is smaller, then the other angles are 90. So if 2x is smaller than 90, then the largest angle would be 90. But the problem says that the largest angle is twice the smallest angle. So if 2x is the largest angle, then 2x must be larger than the other angles. The other angles are 90. So 2x must be larger than 90. Therefore, x must be larger than 45.

In our previous calculation, x was 60, so 2x is 120, which is larger than 90. So that works. If x was 45, then 2x would be 90, which is equal to the other angles. But the problem says the largest angle is twice the smallest. If 2x is equal to the other angles, then there are multiple angles with the same measure. But the problem says "the measure of the largest angle is twice the measure of the smallest angle." So if 2x is equal to 90, then the largest angle is 90, which is not larger than the other angles. So that would contradict the idea that it's the largest. Therefore, x must be greater than 45.

So in our case, x is 60, which is greater than 45, so 2x is 120, which is larger than 90. So that works.

But let me check if there's another possibility. Suppose that the two right angles are not the ones that are not the largest or smallest. Wait, but the problem says "the other two angles are both right angles." So the two angles that are not the largest and not the smallest are right angles. So that would mean that the largest angle is 2x, the smallest is x, and the other two are 90.

Alternatively, could the two right angles be the largest and smallest? But no, because the largest angle is twice the smallest. If one of the right angles was the largest angle, then the smallest angle would be half of 90, which is 45. Then the other angles would be 90, 45, and 90. But then the largest angle is 90, which is not twice the smallest angle (which would be 45). Wait, 90 is twice 45. Wait, that actually works. Wait, hold on.

Wait, if the largest angle is 90, and the smallest angle is 45, then the other two angles are 90. But then the largest angle is 90, which is twice the smallest angle (45). But then the problem says "the other two angles are both right angles." So if the largest angle is 90, then the other angles would be 45, 90, 90, and 90. Wait, but then there are three angles of 90 and one of 45. But the problem says "the measure of the largest angle is twice the measure of the smallest angle," which would be true here. But then the problem says "the other two angles are both right angles." Wait, if there are three angles of 90, then the other two angles would be... Wait, the problem says "the other two angles are both right angles." So if the largest angle is 90, then the smallest angle is 45, and the other two angles are right angles. So that would be four angles: 45, 90, 90, 90. So the other two angles (apart from the largest and smallest) are both right angles. So that would satisfy the problem's conditions.

But then in this case, the largest angle is 90, which is twice the smallest angle (45). But then the problem says "the measure of the largest angle is twice the measure of the smallest angle," so that would be okay. But then why did we get 120 before?

Wait, so there seems to be two possibilities? But that can't be. Wait, maybe I need to check which one is valid.

Wait, in the first case, we had angles of 60, 120, 90, 90. In the second case, angles of 45, 90, 90, 90. Both satisfy the conditions?

But the problem says "the measure of the largest angle is twice the measure of the smallest angle." In the second case, the largest angle is 90, which is twice the smallest angle (45). So that works. But in that case, the largest angle is 90, which is a right angle. But the problem says "the other two angles are both right angles." So in this case, there are three right angles. But the problem says "the other two angles are both right angles." So if the largest angle is 90, then the other two angles (apart from the largest and smallest) are both right angles. So that would be okay.

But then why did we get 120 before? Because depending on which angle is considered the largest. Wait, but in the second case, the largest angle is 90, which is a right angle. But in the first case, the largest angle is 120. So which one is correct?

Wait, the problem says "the measure of the largest angle is twice the measure of the smallest angle." So if there are two different possibilities, how do we know which one is correct?

Wait, maybe there's a misunderstanding here. Let me re-examine the problem statement:

"In a convex quadrilateral, the measure of the largest angle is twice the measure of the smallest angle, and the other two angles are both right angles. How many degrees are in the largest angle?"

So the problem says that the largest angle is twice the smallest angle. So if there are three angles of 90°, then the largest angle is 90°, which is twice the smallest angle (45°). But then the problem says "the other two angles are both right angles." So if the largest angle is 90°, then the other two angles (apart from the largest and smallest) are both right angles. So that would be two angles of 90°, and the smallest angle is 45°, and the largest angle is 90°. Wait, but then the largest angle is 90°, which is the same as the other angles. So in that case, there are three angles of 90°, and one of 45°.

But the problem says "the measure of the largest angle is twice the measure of the smallest angle." So if the largest angle is 90°, then the smallest angle is 45°, which is half of 90°, so that's okay.

But then why did we get 120° earlier? Because depending on how we set up the equation.

Wait, maybe the problem is that in the first case, when we set up the equation, we assumed that the largest angle is 2x, and the smallest angle is x, and the other two angles are 90°. But if the largest angle is 90°, then that would mean that 2x = 90°, so x = 45°, and then the other two angles are 90°, so the four angles would be 45°, 90°, 90°, 90°, which adds up to 45 + 90 + 90 + 90 = 315°, which is not 360°. Wait, that's a problem!

Wait, hold on. If the largest angle is 90°, then the other angles are 90°, 90°, and x. Wait, no. 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 if the largest angle is 90°, then the smallest angle is 45°, and the other two angles are right angles. So that would be four angles: 45°, 90°, 90°, and 90°. But adding them up: 45 + 90 + 90 + 90 = 315°, which is not 360°. Therefore, that's impossible.

Ah! There's the mistake. Earlier, when I thought that maybe the largest angle could be 90°, I forgot that the sum of the angles must be 360°. So if the largest angle is 90°, then the other angles would have to add up to 360 - 90 = 270°. But if the other two angles are right angles (90° each), then that's 90 + 90 = 180°, and the smallest angle would be 270 - 180 = 90°, which would make the smallest angle also 90°, but then the largest angle is also 90°, so they are equal. But the problem says the largest angle is twice the smallest angle. Therefore, that case is invalid.

Therefore, my initial calculation was correct. The only valid case is when the largest angle is 120°, the smallest is 60°, and the other two angles are 90°, adding up to 60 + 120 + 90 + 90 = 360°.

So where did the confusion come from? When I considered the case where the largest angle is 90°, I incorrectly assumed that the other two angles are right angles, but then the smallest angle would have to be 45°, but that would make the total sum less than 360°, which is impossible. Therefore, that case is invalid.

Therefore, the correct answer is 120°.

But let me check once more. Let me write down the equation again.

Let me define:

Let the smallest angle be x.

The largest angle is 2x.

The other two angles are 90° each.

Total sum: x + 2x + 90 + 90 = 360.

So 3x + 180 = 360.

3x = 180.

x = 60.

Therefore, largest angle is 2x = 120°.

Yes, that seems correct.

Another way to think about it: since two angles are right angles, that's 180°, so the remaining two angles must add up to 180°.

Let me denote the smallest angle as x and the largest as 2x. Then x + 2x = 3x = 180°, so x = 60°, and 2x = 120°.

Yes, that's another way to get the same result.

Therefore, the largest angle is 120 degrees.

I think that's solid. I don't see any mistakes in this reasoning. The confusion earlier was due to an incorrect assumption about the angles, but upon checking the total sum, that case was invalid. Therefore, the correct answer is 120 degrees.

**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

We are told:

- The **largest angle** is **twice** the **smallest angle**.
- The **other two angles** are **right angles**, i.e., **90 degrees** each.

Let the **smallest angle** be $ x $ degrees.
Then the **largest angle** is $ 2x $ degrees.
The other two angles are each $ 90^\circ $.

---

### Step 2: Set Up the Equation

Using the total sum of the interior angles:

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

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:

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

---

### Step 4: Verify the Solution

Let’s check that the sum of all angles is indeed 360°:

$$
60^\circ + 120^\circ + 90^\circ + 90^\circ = 360^\circ
$$

This confirms the solution is correct.

Also, since the quadrilateral is **convex**, all angles must be **less than 180°**, which is true for all angles in this case.

---

### 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. So none of the angles are reflex angles. That might be important later if I need to check my answers.

Now, the problem states 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. So let me note that down:

- Let the smallest angle be x degrees.
- Then the largest angle is 2x degrees.
- The other two angles are each 90 degrees.

Since there are four angles in a quadrilateral, adding them all up should give 360 degrees. So I can set up an equation:

x (smallest angle) + 2x (largest angle) + 90 (right angle) + 90 (another right angle) = 360 degrees.

Let me write that out:

x + 2x + 90 + 90 = 360

Simplifying the left side:

x + 2x is 3x, and 90 + 90 is 180. So:

3x + 180 = 360

Now, subtract 180 from both sides:

3x = 360 - 180
3x = 180

Then divide both sides by 3:

x = 60

So the smallest angle is 60 degrees, and the largest angle is twice that, so 120 degrees. Wait, but hold on. Let me check if that makes sense.

So the four angles would be 60°, 120°, 90°, and 90°. Adding them up: 60 + 120 + 90 + 90 = 360. Yep, that adds up.

But wait, the problem says it's a convex quadrilateral. So all angles must be less than 180°. The largest angle here is 120°, which is less than 180°, so that's okay. The other angles are 90°, which is fine. The smallest angle is 60°, which is also fine.

But let me think again. Is there a possibility that the largest angle could be something else? For example, maybe the problem is referring to the largest angle being twice the smallest, but maybe there's another angle that's larger? Wait, no. The problem says "the measure of the largest angle is twice the measure of the smallest angle," so that implies that the other two angles are not larger than that. Since the other two angles are right angles, which are 90°, so if the largest angle is 120°, that's bigger than 90°, so that's okay.

But let me check if there's another way to interpret the problem. Maybe the other two angles are not necessarily right angles? Wait, no, the problem says "the other two angles are both right angles." So that's clear. So the four angles are: smallest angle (x), largest angle (2x), and two right angles (90° each).

So the calculation seems straightforward. But let me just verify once more.

Total angles: 360.

x + 2x + 90 + 90 = 360.

3x + 180 = 360.

3x = 180.

x = 60.

So largest angle is 120.

But wait, is there a case where the largest angle could be more than 120? For example, if the smallest angle was smaller? But the problem says that the largest angle is twice the smallest. So if the smallest angle was smaller, then the largest would be even smaller? Wait, no. If the smallest angle is smaller, then the largest angle would be twice that, but then the other angles are fixed at 90. Wait, but if the smallest angle is smaller, then the largest angle would be smaller? Wait, no. If the smallest angle is smaller, then twice that would be smaller? Wait, no. If x is smaller, then 2x is smaller? Wait, no. If x is smaller, then 2x is smaller? Wait, no. If x is smaller, then 2x is smaller? Wait, for example, if x is 30, then 2x is 60. If x is 60, then 2x is 120. So if x is smaller, then 2x is smaller. But in that case, the other angles are 90. So if the largest angle is 2x, then if x is smaller, 2x would be smaller than 120. But then, if x is smaller, then the other angles are 90. So if 2x is smaller than 90, then the largest angle would be 90. But the problem says that the largest angle is twice the smallest angle. So if 2x is the largest angle, then 2x must be larger than the other angles. The other angles are 90. So 2x must be larger than 90. Therefore, x must be larger than 45.

In our previous calculation, x was 60, so 2x is 120, which is larger than 90. So that works. If x was 45, then 2x would be 90, which is equal to the other angles. But the problem says the largest angle is twice the smallest. If 2x is equal to the other angles, then there are multiple angles with the same measure. But the problem says "the measure of the largest angle is twice the measure of the smallest angle." So if 2x is equal to 90, then the largest angle is 90, which is not larger than the other angles. So that would contradict the idea that it's the largest. Therefore, x must be greater than 45.

So in our case, x is 60, which is greater than 45, so 2x is 120, which is larger than 90. So that works.

But let me check if there's another possibility. Suppose that the two right angles are not the ones that are not the largest or smallest. Wait, but the problem says "the other two angles are both right angles." So the two angles that are not the largest and not the smallest are right angles. So that would mean that the largest angle is 2x, the smallest is x, and the other two are 90.

Alternatively, could the two right angles be the largest and smallest? But no, because the largest angle is twice the smallest. If one of the right angles was the largest angle, then the smallest angle would be half of 90, which is 45. Then the other angles would be 90, 45, and 90. But then the largest angle is 90, which is not twice the smallest angle (which would be 45). Wait, 90 is twice 45. Wait, that actually works. Wait, hold on.

Wait, if the largest angle is 90, and the smallest angle is 45, then the other two angles are 90. But then the largest angle is 90, which is twice the smallest angle (45). But then the problem says "the other two angles are both right angles." So if the largest angle is 90, then the other angles would be 45, 90, 90, and 90. Wait, but then there are three angles of 90 and one of 45. But the problem says "the measure of the largest angle is twice the measure of the smallest angle," which would be true here. But then the problem says "the other two angles are both right angles." Wait, if there are three angles of 90, then the other two angles would be... Wait, the problem says "the other two angles are both right angles." So if the largest angle is 90, then the smallest angle is 45, and the other two angles are right angles. So that would be four angles: 45, 90, 90, 90. So the other two angles (apart from the largest and smallest) are both right angles. So that would satisfy the problem's conditions.

But then in this case, the largest angle is 90, which is twice the smallest angle (45). But then the problem says "the measure of the largest angle is twice the measure of the smallest angle," so that would be okay. But then why did we get 120 before?

Wait, so there seems to be two possibilities? But that can't be. Wait, maybe I need to check which one is valid.

Wait, in the first case, we had angles of 60, 120, 90, 90. In the second case, angles of 45, 90, 90, 90. Both satisfy the conditions?

But the problem says "the measure of the largest angle is twice the measure of the smallest angle." In the second case, the largest angle is 90, which is twice the smallest angle (45). So that works. But in that case, the largest angle is 90, which is a right angle. But the problem says "the other two angles are both right angles." So in this case, there are three right angles. But the problem says "the other two angles are both right angles." So if the largest angle is 90, then the other two angles (apart from the largest and smallest) are both right angles. So that would be okay.

But then why did we get 120 before? Because depending on which angle is considered the largest. Wait, but in the second case, the largest angle is 90, which is a right angle. But in the first case, the largest angle is 120. So which one is correct?

Wait, the problem says "the measure of the largest angle is twice the measure of the smallest angle." So if there are two different possibilities, how do we know which one is correct?

Wait, maybe there's a misunderstanding here. Let me re-examine the problem statement:

"In a convex quadrilateral, the measure of the largest angle is twice the measure of the smallest angle, and the other two angles are both right angles. How many degrees are in the largest angle?"

So the problem says that the largest angle is twice the smallest angle. So if there are three angles of 90°, then the largest angle is 90°, which is twice the smallest angle (45°). But then the problem says "the other two angles are both right angles." So if the largest angle is 90°, then the other two angles (apart from the largest and smallest) are both right angles. So that would be two angles of 90°, and the smallest angle is 45°, and the largest angle is 90°. Wait, but then the largest angle is 90°, which is the same as the other angles. So in that case, there are three angles of 90°, and one of 45°.

But the problem says "the measure of the largest angle is twice the measure of the smallest angle." So if the largest angle is 90°, then the smallest angle is 45°, which is half of 90°, so that's okay.

But then why did we get 120° earlier? Because depending on how we set up the equation.

Wait, maybe the problem is that in the first case, when we set up the equation, we assumed that the largest angle is 2x, and the smallest angle is x, and the other two angles are 90°. But if the largest angle is 90°, then that would mean that 2x = 90°, so x = 45°, and then the other two angles are 90°, so the four angles would be 45°, 90°, 90°, 90°, which adds up to 45 + 90 + 90 + 90 = 315°, which is not 360°. Wait, that's a problem!

Wait, hold on. If the largest angle is 90°, then the other angles are 90°, 90°, and x. Wait, no. 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 if the largest angle is 90°, then the smallest angle is 45°, and the other two angles are right angles. So that would be four angles: 45°, 90°, 90°, and 90°. But adding them up: 45 + 90 + 90 + 90 = 315°, which is not 360°. Therefore, that's impossible.

Ah! There's the mistake. Earlier, when I thought that maybe the largest angle could be 90°, I forgot that the sum of the angles must be 360°. So if the largest angle is 90°, then the other angles would have to add up to 360 - 90 = 270°. But if the other two angles are right angles (90° each), then that's 90 + 90 = 180°, and the smallest angle would be 270 - 180 = 90°, which would make the smallest angle also 90°, but then the largest angle is also 90°, so they are equal. But the problem says the largest angle is twice the smallest angle. Therefore, that case is invalid.

Therefore, my initial calculation was correct. The only valid case is when the largest angle is 120°, the smallest is 60°, and the other two angles are 90°, adding up to 60 + 120 + 90 + 90 = 360°.

So where did the confusion come from? When I considered the case where the largest angle is 90°, I incorrectly assumed that the other two angles are right angles, but then the smallest angle would have to be 45°, but that would make the total sum less than 360°, which is impossible. Therefore, that case is invalid.

Therefore, the correct answer is 120°.

But let me check once more. Let me write down the equation again.

Let me define:

Let the smallest angle be x.

The largest angle is 2x.

The other two angles are 90° each.

Total sum: x + 2x + 90 + 90 = 360.

So 3x + 180 = 360.

3x = 180.

x = 60.

Therefore, largest angle is 2x = 120°.

Yes, that seems correct.

Another way to think about it: since two angles are right angles, that's 180°, so the remaining two angles must add up to 180°.

Let me denote the smallest angle as x and the largest as 2x. Then x + 2x = 3x = 180°, so x = 60°, and 2x = 120°.

Yes, that's another way to get the same result.

Therefore, the largest angle is 120 degrees.

I think that's solid. I don't see any mistakes in this reasoning. The confusion earlier was due to an incorrect assumption about the angles, but upon checking the total sum, that case was invalid. Therefore, the correct answer is 120 degrees.

**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

We are told:

- The **largest angle** is **twice** the **smallest angle**.
- The **other two angles** are **right angles**, i.e., **90 degrees** each.

Let the **smallest angle** be $ x $ degrees.
Then the **largest angle** is $ 2x $ degrees.
The other two angles are each $ 90^\circ $.

---

### Step 2: Set Up the Equation

Using the total sum of the interior angles:

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

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:

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

---

### Step 4: Verify the Solution

Let’s check that the sum of all angles is indeed 360°:

$$
60^\circ + 120^\circ + 90^\circ + 90^\circ = 360^\circ
$$

This confirms the solution is correct.

Also, since the quadrilateral is **convex**, all angles must be **less than 180°**, which is true for all angles in this case.

---

### 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 $ -\frac{\pi}{2} $ and $ \frac{\pi}{2} $. But since $ x $ and $ y $ can be negative, maybe I need to consider the absolute values? Wait, but if $ x $ is negative, then $ \sqrt{1 - x^2} $ is still positive, so maybe the substitution is okay as long as I take care of the signs. Hmm, maybe I should consider $ x = \sin \alpha $ and $ y = \sin \beta $ with $ \alpha, \beta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $, so that cosine is non-negative.

So substituting, the function becomes:
$$
f(\alpha, \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha
$$
Wait a second, that looks familiar. Isn't that the sine addition formula? Because $ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $. So, $ f(\alpha, \beta) = \sin(\alpha + \beta) $.

Oh! So the original function simplifies to $ \sin(\alpha + \beta) $, where $ x = \sin \alpha $, $ y = \sin \beta $, and $ \alpha, \beta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Therefore, the maximum value of $ f(x, y) $ is the maximum of $ \sin(\alpha + \beta) $, given that $ \alpha $ and $ \beta $ are in that interval.

But what is the range of $ \alpha + \beta $? Since each of $ \alpha $ and $ \beta $ can be between $ -\frac{\pi}{2} $ and $ \frac{\pi}{2} $, their sum can range from $ -\pi $ to $ \pi $. The sine function reaches its maximum value of 1 at $ \frac{\pi}{2} $, so if $ \alpha + \beta = \frac{\pi}{2} $, then $ f(x, y) = 1 $. But we need to check if there exist $ \alpha $ and $ \beta $ in the interval $ [-\frac{\pi}{2}, \frac{\pi}{2}] $ such that $ \alpha + \beta = \frac{\pi}{2} $.

For example, if $ \alpha = \frac{\pi}{2} $, then $ \beta = 0 $, but $ \alpha = \frac{\pi}{2} $ is allowed since $ x = \sin(\frac{\pi}{2}) = 1 $, which is within the domain. Similarly, $ \beta = 0 $ gives $ y = 0 $. So that's valid. Therefore, the maximum value of $ f(x, y) $ is 1? Wait, but let me check if there are other possibilities.

Alternatively, maybe there's a case where $ \alpha + \beta $ is larger? But since the maximum of $ \alpha + \beta $ is $ \pi $, and $ \sin(\pi) = 0 $, which is less than 1. Similarly, the minimum is $ -\pi $, and $ \sin(-\pi) = 0 $. So the maximum of $ \sin(\alpha + \beta) $ is indeed 1.

But wait, let me verify with specific values. Let me take $ 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 that checks out.

But is there a case where the function is larger than 1? Let me try some other values. 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{2}{4} + \frac{2}{4} = 1 $. So again, 1.

What if I take $ x = y = \sin \theta $, then $ f(x, y) = \sin \theta \sqrt{1 - \sin^2 \theta} + \sin \theta \sqrt{1 - \sin^2 \theta} = 2 \sin \theta \cos \theta = \sin 2\theta $. The maximum of $ \sin 2\theta $ is 1, so again, same result.

But maybe there's a case where the function is larger? Let me try with different values. Suppose $ x = \frac{1}{2} $, $ y = \frac{\sqrt{3}}{2} $. Then $ f(x, y) = \frac{1}{2} \sqrt{1 - \frac{3}{4}} + \frac{\sqrt{3}}{2} \sqrt{1 - \frac{1}{4}} = \frac{1}{2} \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1 $. Still 1.

Hmm, so maybe 1 is indeed the maximum? But let me think again. Suppose I take $ x $ and $ y $ such that $ \alpha + \beta = \frac{\pi}{2} $. For example, $ \alpha = \frac{\pi}{2} - \beta $. Then $ x = \sin(\frac{\pi}{2} - \beta) = \cos \beta $, and $ y = \sin \beta $. Then $ f(x, y) = \cos \beta \sqrt{1 - \sin^2 \beta} + \sin \beta \sqrt{1 - \cos^2 \beta} = \cos \beta \cdot \cos \beta + \sin \beta \cdot \sin \beta = \cos^2 \beta + \sin^2 \beta = 1 $. So regardless of $ \beta $, this gives 1. Therefore, the maximum is indeed 1.

But wait, is there a case where $ \alpha + \beta $ is such that $ \sin(\alpha + \beta) $ is greater than 1? No, because the sine function has a maximum of 1. Therefore, the maximum value of $ f(x, y) $ is 1.

But let me check if there are any constraints I might have missed. For example, when I did the substitution $ x = \sin \alpha $, $ y = \sin \beta $, I assumed that $ \alpha $ and $ \beta $ are in the range where cosine is non-negative, which is true because $ \alpha, \beta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Therefore, the square roots are valid.

Alternatively, maybe there's another approach without substitution. Let me try using calculus. Let's consider the function $ f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. To find its maximum, we can take partial derivatives with respect to x and y, set them equal to zero, and solve.

First, compute the partial derivative with respect to x:

$ \frac{\partial f}{\partial x} = \sqrt{1 - y^2} + x \cdot \frac{d}{dx} \sqrt{1 - y^2} + y \cdot \frac{d}{dx} \sqrt{1 - x^2} $. Wait, no. Wait, actually, the function is $ x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. So when taking the partial derivative with respect to x, the first term is $ x \sqrt{1 - y^2} $, so derivative is $ \sqrt{1 - y^2} + x \cdot 0 $, since $ \sqrt{1 - y^2} $ is treated as a constant with respect to x. Then the second term is $ y \sqrt{1 - x^2} $, derivative is $ y \cdot \frac{-x}{\sqrt{1 - x^2}} $. Therefore:

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

Similarly, partial derivative with respect to y:

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

So setting both partial derivatives to zero:

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

Let me write these equations:

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

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

So if I square both sides of equation 1:

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

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

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

Expanding the left side:

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

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

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

Therefore:

$ x^2 + y^2 = 1 $

So the critical points lie on the unit circle. Interesting. Now, let's check equation 2. If we square equation 2:

$ 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 1. Therefore, both equations reduce to $ x^2 + y^2 = 1 $. Therefore, the critical points lie on the unit circle.

So, we can parametrize $ x $ and $ y $ such that $ x^2 + y^2 = 1 $. Let me use the same substitution as before. Let me set $ x = \cos \theta $, $ y = \sin \theta $, but wait, earlier I used sine for x and y. Wait, but if $ x^2 + y^2 = 1 $, then we can parametrize as $ x = \cos \theta $, $ y = \sin \theta $, but then $ \sqrt{1 - y^2} = \cos \theta $, and $ \sqrt{1 - x^2} = \sin \theta $. Wait, but then the function becomes:

$ f(x, y) = \cos \theta \cdot \cos \theta + \sin \theta \cdot \sin \theta = \cos^2 \theta + \sin^2 \theta = 1 $. So again, the function is equal to 1. Therefore, all critical points on the unit circle give the value 1.

But wait, but earlier when I considered the substitution $ x = \sin \alpha $, $ y = \sin \beta $, I found that $ f(x, y) = \sin(\alpha + \beta) $, which can be 1. So, in this case, the critical points on the unit circle give the value 1. However, we also need to check the boundaries of the domain.

Wait, the original domain is $ -1 \leq x, y \leq 1 $. So, the critical points are inside the domain, but we also need to check the boundaries. However, when we found that the critical points lie on the unit circle, but the boundaries of the domain would be when either x or y is ±1.

Wait, but when x or y is ±1, then the other variable can be anything? Wait, no. The boundaries of the domain are when x or y is at their extremes, i.e., x = 1, x = -1, y = 1, y = -1. But we need to check the function on these boundaries.

So, for example, let's check when x = 1. Then, the function becomes $ f(1, y) = 1 \cdot \sqrt{1 - y^2} + y \cdot \sqrt{1 - 1} = \sqrt{1 - y^2} + 0 = \sqrt{1 - y^2} $. The maximum of this expression occurs when $ y = 0 $, giving $ \sqrt{1 - 0} = 1 $. Similarly, when y = 1, the function becomes $ x \cdot 0 + 1 \cdot \sqrt{1 - x^2} = \sqrt{1 - x^2} $, which is maximized at x = 0, giving 1.

Similarly, when x = -1, the function becomes $ -1 \cdot \sqrt{1 - y^2} + y \cdot \sqrt{1 - 1} = -\sqrt{1 - y^2} $, which has maximum value -1, but since we are looking for maximum, this is not relevant. Similarly for y = -1, the function becomes $ x \cdot \sqrt{1 - 1} + (-1) \cdot \sqrt{1 - x^2} = -\sqrt{1 - x^2} $, which again has maximum -1.

Therefore, on the boundaries, the maximum value is 1.

But also, we need to check other boundaries? For example, when both x and y are at their extremes? But if both x and y are 1, then f(1,1) = 1*0 + 1*0 = 0. Similarly, other combinations would give lower values.

Therefore, the maximum value of the function is 1, achieved when either (x, y) = (1, 0), (0, 1), or any point on the unit circle where $ \alpha + \beta = \frac{\pi}{2} $, which corresponds to points like $ x = \cos \theta $, $ y = \sin \theta $, but wait, earlier substitution showed that when $ x = \sin \alpha $, $ y = \sin \beta $, then $ f(x, y) = \sin(\alpha + \beta) $, so maximum 1.

But also, when we parametrized with $ x = \cos \theta $, $ y = \sin \theta $, then f(x, y) = 1. So, in both cases, the maximum is 1.

But let me check another case. Suppose x and y are both positive, but not on the unit circle. For example, take x = y = t, where t is between 0 and 1. Then f(t, t) = t * sqrt(1 - t^2) + t * sqrt(1 - t^2) = 2t sqrt(1 - t^2). Let's find the maximum of this function. Let me set g(t) = 2t sqrt(1 - t^2). Take derivative: g’(t) = 2 sqrt(1 - t^2) + 2t * ( -2t ) / (2 sqrt(1 - t^2)) ) = 2 sqrt(1 - t^2) - 2t^2 / sqrt(1 - t^2) = [2(1 - t^2) - 2t^2]/sqrt(1 - t^2) = [2 - 4t^2]/sqrt(1 - t^2). Setting derivative to zero: 2 - 4t^2 = 0 => t^2 = 1/2 => t = 1/√2. Then, g(1/√2) = 2*(1/√2)*sqrt(1 - 1/2) = 2*(1/√2)*(1/√2) = 2*(1/2) = 1. So again, maximum is 1.

Therefore, even in this case, the maximum is 1.

So, all approaches seem to lead to the conclusion that the maximum value is 1. However, let me check if there's a case where the function could be greater than 1. Suppose I take x and y such that they are not on the unit circle. For example, take x = 0.8, y = 0.6. Then, sqrt(1 - y^2) = sqrt(1 - 0.36) = sqrt(0.64) = 0.8, and sqrt(1 - x^2) = sqrt(1 - 0.64) = sqrt(0.36) = 0.6. Then f(x, y) = 0.8*0.8 + 0.6*0.6 = 0.64 + 0.36 = 1. So again, 1.

Wait, but x^2 + y^2 = 0.64 + 0.36 = 1, so that's on the unit circle. So again, same result.

Another example: take x = 0.6, y = 0.8. Then same thing.

But suppose I take x = 0.9, y = 0.435. Let me check if x^2 + y^2 is something. Let me compute 0.9^2 + 0.435^2 = 0.81 + 0.189225 = 0.999225, which is close to 1. So, if I take x = 0.9, y ≈ sqrt(1 - 0.81) = sqrt(0.19) ≈ 0.435. Then f(x, y) = 0.9 * sqrt(1 - y^2) + y * sqrt(1 - x^2) = 0.9 * sqrt(0.19) + 0.435 * sqrt(0.19). Wait, sqrt(1 - y^2) is sqrt(0.81) = 0.9, and sqrt(1 - x^2) is sqrt(0.19) ≈ 0.435. Therefore, f(x, y) = 0.9 * 0.9 + 0.435 * 0.435 ≈ 0.81 + 0.189 ≈ 0.999, which is approximately 1. So again, close to 1.

Therefore, it seems that whenever x and y are on the unit circle, the function equals 1, and when they are not, it's less than or equal to 1.

But let me think again. Suppose I take x and y such that they are not on the unit circle. For example, take x = 0.5, y = 0.5. Then f(x, y) = 0.5 * sqrt(1 - 0.25) + 0.5 * sqrt(1 - 0.25) = 0.5 * sqrt(0.75) + 0.5 * sqrt(0.75) = sqrt(0.75) ≈ 0.866, which is less than 1.

Another example: take x = 1, y = 0. Then f(x, y) = 1 * 1 + 0 * 0 = 1.

So, all these examples confirm that the maximum is 1.

But let me think again about the substitution approach. If we set $ x = \sin \alpha $, $ y = \sin \beta $, then $ f(x, y) = \sin(\alpha + \beta) $, which has maximum 1. However, is there a case where $ \alpha + \beta $ could be more than $ \frac{\pi}{2} $, but still $ \sin(\alpha + \beta) $ is greater than 1? No, because sine function's maximum is 1.

Alternatively, if we consider $ \alpha + \beta = \frac{\pi}{2} $, then $ \sin(\alpha + \beta) = 1 $. If $ \alpha + \beta $ is greater than $ \frac{\pi}{2} $, then $ \sin(\alpha + \beta) $ would start decreasing. For example, $ \alpha + \beta = \frac{3\pi}{4} $, then $ \sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2} \approx 0.707 $, which is less than 1. Similarly, if $ \alpha + \beta = \frac{\pi}{2} + \epsilon $, then $ \sin(\alpha + \beta) = \cos \epsilon $, which is less than 1 for $ \epsilon > 0 $.

Therefore, the maximum occurs when $ \alpha + \beta = \frac{\pi}{2} $, which gives the maximum value of 1.

But let me check if there are any constraints on $ \alpha $ and $ \beta $. Since $ x = \sin \alpha $ and $ y = \sin \beta $, and $ x, y \in [-1, 1] $, then $ \alpha, \beta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Therefore, $ \alpha + \beta \in [-\pi, \pi] $. However, the maximum of $ \sin(\alpha + \beta) $ in this interval is indeed 1, achieved when $ \alpha + \beta = \frac{\pi}{2} $.

Therefore, combining all these approaches, it seems that the maximum value of the function is 1.

But just to be thorough, let me check if there are any other critical points. Earlier, we found that critical points lie on the unit circle $ x^2 + y^2 = 1 $. However, when we set the partial derivatives to zero, we arrived at that condition. But maybe there are other critical points? For example, when the partial derivatives don't exist?

But the function $ f(x, y) $ is differentiable everywhere in the interior of the domain $ (-1, 1) \times (-1, 1) $, since the square roots are smooth there. At the boundaries, the function may not be differentiable, but we already checked the boundaries and found that the maximum is 1.

Therefore, the maximum value is indeed 1.

Wait, but let me think again. Suppose I take $ x = \sin \alpha $, $ y = \sin \beta $, and $ \alpha + \beta = \frac{\pi}{2} $. Then $ \beta = \frac{\pi}{2} - \alpha $. Therefore, $ y = \sin(\frac{\pi}{2} - \alpha) = \cos \alpha $. Therefore, $ x = \sin \alpha $, $ y = \cos \alpha $. Therefore, $ x^2 + y^2 = \sin^2 \alpha + \cos^2 \alpha = 1 $, which is consistent with the previous result. Therefore, all these points lie on the unit circle.

Therefore, the maximum value of the function is 1, achieved when $ x = \sin \alpha $, $ y = \cos \alpha $ for some $ \alpha $, or equivalently, when $ x $ and $ y $ are such that $ x^2 + y^2 = 1 $ and $ \alpha + \beta = \frac{\pi}{2} $.

But let me check if there's a case where the function could be greater than 1. Suppose I take $ x = 1 $, $ y = 0 $, which gives 1. If I take $ x = \cos \theta $, $ y = \sin \theta $, then $ f(x, y) = \cos \theta \cdot \sqrt{1 - \sin^2 \theta} + \sin \theta \cdot \sqrt{1 - \cos^2 \theta} = \cos \theta \cdot |\cos \theta| + \sin \theta \cdot |\sin \theta| $. But since $ \theta $ is in the range where $ \cos \theta $ and $ \sin \theta $ can be positive or negative. However, if we take $ \theta $ in the first quadrant, then $ \cos \theta $ and $ \sin \theta $ are positive, so $ f(x, y) = \cos^2 \theta + \sin^2 \theta = 1 $. If $ \theta $ is in other quadrants, then the absolute values would make it still 1? Wait, for example, if $ \theta $ is in the second quadrant, $ \cos \theta $ is negative, $ \sin \theta $ is positive. Then $ f(x, y) = \cos \theta \cdot |\cos \theta| + \sin \theta \cdot |\sin \theta| = \cos \theta (-\cos \theta) + \sin \theta \cdot \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 $, so $ \theta = \frac{\pi}{2} $. But at $ \theta = \frac{\pi}{2} $, $ x = 0 $, $ y = 1 $, and $ f(x, y) = 0 * 0 + 1 * 0 = 0 $? Wait, no. Wait, if $ \theta = \frac{\pi}{2} $, then $ x = \cos \frac{\pi}{2} = 0 $, $ y = \sin \frac{\pi}{2} = 1 $. Then $ f(x, y) = 0 * \sqrt{1 - 1} + 1 * \sqrt{1 - 0} = 0 + 1 = 1 $. Wait, but according to the previous expression, $ -\cos^2 \theta + \sin^2 \theta = -0 + 1 = 1 $. So that's correct.

Similarly, if $ \theta $ is in the third quadrant, both $ \cos \theta $ and $ \sin \theta $ are negative. Then $ f(x, y) = \cos \theta \cdot |\cos \theta| + \sin \theta \cdot |\sin \theta| = \cos \theta (-\cos \theta) + \sin \theta (-\sin \theta) = -\cos^2 \theta - \sin^2 \theta = -1 $, which is the minimum.

Similarly, in the fourth quadrant, $ \cos \theta $ is positive, $ \sin \theta $ is negative. Then $ f(x, y) = \cos \theta \cdot \cos \theta + \sin \theta \cdot (-\sin \theta) = \cos^2 \theta - \sin^2 \theta = \cos 2\theta $, which has maximum 1 when $ 2\theta = 0 $, i.e., $ \theta = 0 $, which gives $ x = 1 $, $ y = 0 $, and $ f(x, y) = 1 $.

Therefore, regardless of the quadrant, the maximum value of the function is 1.

Therefore, after all these checks, I can be confident that the maximum value of $ f(x, y) $ is indeed 1.

But just to make sure, let me consider another approach. Let me use the Cauchy-Schwarz inequality.

The function is $ f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. Let me think of this as the dot product of two vectors: $ (x, y) $ and $ (\sqrt{1 - y^2}, \sqrt{1 - x^2}) $. Wait, but the second vector depends on both x and y, so it's not straightforward.

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

Alternatively, think of this as $ a \sqrt{1 - b^2} + b \sqrt{1 - a^2} \leq \sqrt{(a^2 + b^2)( (1 - b^2) + (1 - a^2) )} $ by Cauchy-Schwarz. Let me check:

By Cauchy-Schwarz, $ (a \sqrt{1 - b^2} + b \sqrt{1 - a^2})^2 \leq (a^2 + b^2)[(1 - b^2) + (1 - a^2)] $.

Calculating the right-hand side:

$ (a^2 + b^2)(2 - a^2 - b^2) $. Let me denote $ S = a^2 + b^2 $. Then the expression becomes $ S(2 - S) $. This is a quadratic in S: $ -S^2 + 2S $, which has maximum at $ S = 1 $, with maximum value $ -1 + 2 = 1 $. Therefore, the square of the function is less than or equal to 1, so the function is less than or equal to 1.

Therefore, the maximum value is 1, achieved when equality holds in Cauchy-Schwarz. Equality holds when $ \frac{a}{\sqrt{1 - b^2}} = \frac{b}{\sqrt{1 - a^2}} $. Let me solve this equation.

Cross-multiplying: $ a \sqrt{1 - a^2} = b \sqrt{1 - b^2} $. Squaring both sides: $ a^2 (1 - a^2) = b^2 (1 - b^2) $.

Which implies $ a^2 - a^4 = b^2 - b^4 $. Rearranged: $ a^2 - b^2 = a^4 - b^4 $. Factor both sides:

Left side: $ (a - b)(a + b) $

Right side: $ (a^2 - b^2)(a^2 + b^2) = (a - b)(a + b)(a^2 + b^2) $

Therefore, equation becomes:

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

If $ a \neq b $, we can divide both sides by $ (a - b)(a + b) $, assuming $ a + b \neq 0 $, leading to $ 1 = a^2 + b^2 $.

If $ a = b $, then the original equation $ a \sqrt{1 - a^2} = a \sqrt{1 - a^2} $ holds for any a.

Therefore, equality holds when either $ a = b $ or $ a^2 + b^2 = 1 $.

So, if $ a = b $, then the function becomes $ a \sqrt{1 - a^2} + a \sqrt{1 - a^2} = 2a \sqrt{1 - a^2} $, which we already saw has maximum 1 when $ a = \frac{1}{\sqrt{2}} $.

Alternatively, if $ a^2 + b^2 = 1 $, then the function is equal to 1, as shown earlier.

Therefore, the maximum value of the function is indeed 1, achieved in multiple cases.

Therefore, after multiple approaches—trigonometric substitution, calculus, Cauchy-Schwarz inequality—it seems that the maximum value of the function is 1.

Just to recap, all methods lead to the conclusion that the maximum is 1, achieved when (x, y) is on the unit circle with specific relations between x and y, or at the boundary points (1, 0) and (0, 1).

Therefore, I can confidently say that the maximum value of $ f(x, y) $ 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 [-\frac{\pi}{2}, \frac{\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 identity from trigonometry.

---

### Step 2: Maximizing the Sine Function

The function now becomes:
$$
f(x, y) = \sin(\alpha + \beta)
$$
with the constraint that $ \alpha, \beta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Therefore, the sum $ \alpha + \beta \in [-\pi, \pi] $.

The sine function achieves its maximum value of 1 at $ \alpha + \beta = \frac{\pi}{2} $, which is within the allowed range. Thus, the maximum value of $ f(x, y) $ is:
$$
\sin\left(\frac{\pi}{2}\right) = 1
$$

---

### Step 3: Verifying the Maximum

We now verify that this maximum is achievable. For instance, take:
$$
\alpha = \frac{\pi}{2}, \quad \beta = 0
\Rightarrow x = \sin\left(\frac{\pi}{2}\right) = 1, \quad y = \sin(0) = 0
$$
Then:
$$
f(1, 0) = 1 \cdot \sqrt{1 - 0^2} + 0 \cdot \sqrt{1 - 1^2} = 1
$$
Similarly, taking $ \alpha = 0, \beta = \frac{\pi}{2} $, we get $ f(0, 1) = 1 $.

Also, if $ x = \cos \theta $, $ y = \sin \theta $, then:
$$
f(x, y) = \cos \theta \cdot \sqrt{1 - \sin^2 \theta} + \sin \theta \cdot \sqrt{1 - \cos^2 \theta} = \cos^2 \theta + \sin^2 \theta = 1
$$
So, the function achieves the value 1 on the unit circle $ x^2 + y^2 = 1 $, and this is the maximum possible value.

---

### Step 4: Boundary and Critical Point Analysis

We also consider the boundaries of the domain $ -1 \leq x, y \leq 1 $. For example:
- If $ x = 1 $, then $ f(1, y) = \sqrt{1 - y^2} $, which is maximized at $ y = 0 $, giving $ f(1, 0) = 1 $
- If $ y = 1 $, then $ f(x, 1) = \sqrt{1 - x^2} $, which is maximized at $ x = 0 $, giving $ f(0, 1) = 1 $

Using calculus, we find that the critical points of $ f(x, y) $ occur when $ x^2 + y^2 = 1 $, and on this curve, the function again evaluates to 1.

---

### Step 5: Inequality Approach (Cauchy-Schwarz)

We can also apply the Cauchy-Schwarz inequality:
$$
f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2} \leq \sqrt{(x^2 + y^2)\left((1 - y^2) + (1 - x^2)\right)} = \sqrt{(x^2 + y^2)(2 - x^2 - y^2)}
$$
Let $ S = x^2 + y^2 $. Then the expression becomes:
$$
\sqrt{S(2 - S)}
$$
This is maximized when $ S = 1 $, giving:
$$
\sqrt{1 \cdot (2 - 1)} = \sqrt{1} = 1
$$
Thus, the maximum value of $ f(x, y) $ is again confirmed to be 1.

---

### Final Conclusion

All approaches—trigonometric substitution, calculus, and inequality—lead to the same conclusion. The function achieves its maximum value of 1 when $ x = 1, y = 0 $, $ x = 0, y = 1 $, or when $ x^2 + y^2 = 1 $ with appropriate angle relations.

---

### Final Answer

$$
\boxed{1}
$$

[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 $ -\frac{\pi}{2} $ and $ \frac{\pi}{2} $. But since $ x $ and $ y $ can be negative, maybe I need to consider the absolute values? Wait, but if $ x $ is negative, then $ \sqrt{1 - x^2} $ is still positive, so maybe the substitution is okay as long as I take care of the signs. Hmm, maybe I should consider $ x = \sin \alpha $ and $ y = \sin \beta $ with $ \alpha, \beta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $, so that cosine is non-negative.

So substituting, the function becomes:
$$
f(\alpha, \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha
$$
Wait a second, that looks familiar. Isn't that the sine addition formula? Because $ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $. So, $ f(\alpha, \beta) = \sin(\alpha + \beta) $.

Oh! So the original function simplifies to $ \sin(\alpha + \beta) $, where $ x = \sin \alpha $, $ y = \sin \beta $, and $ \alpha, \beta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Therefore, the maximum value of $ f(x, y) $ is the maximum of $ \sin(\alpha + \beta) $, given that $ \alpha $ and $ \beta $ are in that interval.

But what is the range of $ \alpha + \beta $? Since each of $ \alpha $ and $ \beta $ can be between $ -\frac{\pi}{2} $ and $ \frac{\pi}{2} $, their sum can range from $ -\pi $ to $ \pi $. The sine function reaches its maximum value of 1 at $ \frac{\pi}{2} $, so if $ \alpha + \beta = \frac{\pi}{2} $, then $ f(x, y) = 1 $. But we need to check if there exist $ \alpha $ and $ \beta $ in the interval $ [-\frac{\pi}{2}, \frac{\pi}{2}] $ such that $ \alpha + \beta = \frac{\pi}{2} $.

For example, if $ \alpha = \frac{\pi}{2} $, then $ \beta = 0 $, but $ \alpha = \frac{\pi}{2} $ is allowed since $ x = \sin(\frac{\pi}{2}) = 1 $, which is within the domain. Similarly, $ \beta = 0 $ gives $ y = 0 $. So that's valid. Therefore, the maximum value of $ f(x, y) $ is 1? Wait, but let me check if there are other possibilities.

Alternatively, maybe there's a case where $ \alpha + \beta $ is larger? But since the maximum of $ \alpha + \beta $ is $ \pi $, and $ \sin(\pi) = 0 $, which is less than 1. Similarly, the minimum is $ -\pi $, and $ \sin(-\pi) = 0 $. So the maximum of $ \sin(\alpha + \beta) $ is indeed 1.

But wait, let me verify with specific values. Let me take $ 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 that checks out.

But is there a case where the function is larger than 1? Let me try some other values. 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{2}{4} + \frac{2}{4} = 1 $. So again, 1.

What if I take $ x = y = \sin \theta $, then $ f(x, y) = \sin \theta \sqrt{1 - \sin^2 \theta} + \sin \theta \sqrt{1 - \sin^2 \theta} = 2 \sin \theta \cos \theta = \sin 2\theta $. The maximum of $ \sin 2\theta $ is 1, so again, same result.

But maybe there's a case where the function is larger? Let me try with different values. Suppose $ x = \frac{1}{2} $, $ y = \frac{\sqrt{3}}{2} $. Then $ f(x, y) = \frac{1}{2} \sqrt{1 - \frac{3}{4}} + \frac{\sqrt{3}}{2} \sqrt{1 - \frac{1}{4}} = \frac{1}{2} \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1 $. Still 1.

Hmm, so maybe 1 is indeed the maximum? But let me think again. Suppose I take $ x $ and $ y $ such that $ \alpha + \beta = \frac{\pi}{2} $. For example, $ \alpha = \frac{\pi}{2} - \beta $. Then $ x = \sin(\frac{\pi}{2} - \beta) = \cos \beta $, and $ y = \sin \beta $. Then $ f(x, y) = \cos \beta \sqrt{1 - \sin^2 \beta} + \sin \beta \sqrt{1 - \cos^2 \beta} = \cos \beta \cdot \cos \beta + \sin \beta \cdot \sin \beta = \cos^2 \beta + \sin^2 \beta = 1 $. So regardless of $ \beta $, this gives 1. Therefore, the maximum is indeed 1.

But wait, is there a case where $ \alpha + \beta $ is such that $ \sin(\alpha + \beta) $ is greater than 1? No, because the sine function has a maximum of 1. Therefore, the maximum value of $ f(x, y) $ is 1.

But let me check if there are any constraints I might have missed. For example, when I did the substitution $ x = \sin \alpha $, $ y = \sin \beta $, I assumed that $ \alpha $ and $ \beta $ are in the range where cosine is non-negative, which is true because $ \alpha, \beta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Therefore, the square roots are valid.

Alternatively, maybe there's another approach without substitution. Let me try using calculus. Let's consider the function $ f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. To find its maximum, we can take partial derivatives with respect to x and y, set them equal to zero, and solve.

First, compute the partial derivative with respect to x:

$ \frac{\partial f}{\partial x} = \sqrt{1 - y^2} + x \cdot \frac{d}{dx} \sqrt{1 - y^2} + y \cdot \frac{d}{dx} \sqrt{1 - x^2} $. Wait, no. Wait, actually, the function is $ x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. So when taking the partial derivative with respect to x, the first term is $ x \sqrt{1 - y^2} $, so derivative is $ \sqrt{1 - y^2} + x \cdot 0 $, since $ \sqrt{1 - y^2} $ is treated as a constant with respect to x. Then the second term is $ y \sqrt{1 - x^2} $, derivative is $ y \cdot \frac{-x}{\sqrt{1 - x^2}} $. Therefore:

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

Similarly, partial derivative with respect to y:

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

So setting both partial derivatives to zero:

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

Let me write these equations:

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

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

So if I square both sides of equation 1:

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

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

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

Expanding the left side:

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

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

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

Therefore:

$ x^2 + y^2 = 1 $

So the critical points lie on the unit circle. Interesting. Now, let's check equation 2. If we square equation 2:

$ 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 1. Therefore, both equations reduce to $ x^2 + y^2 = 1 $. Therefore, the critical points lie on the unit circle.

So, we can parametrize $ x $ and $ y $ such that $ x^2 + y^2 = 1 $. Let me use the same substitution as before. Let me set $ x = \cos \theta $, $ y = \sin \theta $, but wait, earlier I used sine for x and y. Wait, but if $ x^2 + y^2 = 1 $, then we can parametrize as $ x = \cos \theta $, $ y = \sin \theta $, but then $ \sqrt{1 - y^2} = \cos \theta $, and $ \sqrt{1 - x^2} = \sin \theta $. Wait, but then the function becomes:

$ f(x, y) = \cos \theta \cdot \cos \theta + \sin \theta \cdot \sin \theta = \cos^2 \theta + \sin^2 \theta = 1 $. So again, the function is equal to 1. Therefore, all critical points on the unit circle give the value 1.

But wait, but earlier when I considered the substitution $ x = \sin \alpha $, $ y = \sin \beta $, I found that $ f(x, y) = \sin(\alpha + \beta) $, which can be 1. So, in this case, the critical points on the unit circle give the value 1. However, we also need to check the boundaries of the domain.

Wait, the original domain is $ -1 \leq x, y \leq 1 $. So, the critical points are inside the domain, but we also need to check the boundaries. However, when we found that the critical points lie on the unit circle, but the boundaries of the domain would be when either x or y is ±1.

Wait, but when x or y is ±1, then the other variable can be anything? Wait, no. The boundaries of the domain are when x or y is at their extremes, i.e., x = 1, x = -1, y = 1, y = -1. But we need to check the function on these boundaries.

So, for example, let's check when x = 1. Then, the function becomes $ f(1, y) = 1 \cdot \sqrt{1 - y^2} + y \cdot \sqrt{1 - 1} = \sqrt{1 - y^2} + 0 = \sqrt{1 - y^2} $. The maximum of this expression occurs when $ y = 0 $, giving $ \sqrt{1 - 0} = 1 $. Similarly, when y = 1, the function becomes $ x \cdot 0 + 1 \cdot \sqrt{1 - x^2} = \sqrt{1 - x^2} $, which is maximized at x = 0, giving 1.

Similarly, when x = -1, the function becomes $ -1 \cdot \sqrt{1 - y^2} + y \cdot \sqrt{1 - 1} = -\sqrt{1 - y^2} $, which has maximum value -1, but since we are looking for maximum, this is not relevant. Similarly for y = -1, the function becomes $ x \cdot \sqrt{1 - 1} + (-1) \cdot \sqrt{1 - x^2} = -\sqrt{1 - x^2} $, which again has maximum -1.

Therefore, on the boundaries, the maximum value is 1.

But also, we need to check other boundaries? For example, when both x and y are at their extremes? But if both x and y are 1, then f(1,1) = 1*0 + 1*0 = 0. Similarly, other combinations would give lower values.

Therefore, the maximum value of the function is 1, achieved when either (x, y) = (1, 0), (0, 1), or any point on the unit circle where $ \alpha + \beta = \frac{\pi}{2} $, which corresponds to points like $ x = \cos \theta $, $ y = \sin \theta $, but wait, earlier substitution showed that when $ x = \sin \alpha $, $ y = \sin \beta $, then $ f(x, y) = \sin(\alpha + \beta) $, so maximum 1.

But also, when we parametrized with $ x = \cos \theta $, $ y = \sin \theta $, then f(x, y) = 1. So, in both cases, the maximum is 1.

But let me check another case. Suppose x and y are both positive, but not on the unit circle. For example, take x = y = t, where t is between 0 and 1. Then f(t, t) = t * sqrt(1 - t^2) + t * sqrt(1 - t^2) = 2t sqrt(1 - t^2). Let's find the maximum of this function. Let me set g(t) = 2t sqrt(1 - t^2). Take derivative: g’(t) = 2 sqrt(1 - t^2) + 2t * ( -2t ) / (2 sqrt(1 - t^2)) ) = 2 sqrt(1 - t^2) - 2t^2 / sqrt(1 - t^2) = [2(1 - t^2) - 2t^2]/sqrt(1 - t^2) = [2 - 4t^2]/sqrt(1 - t^2). Setting derivative to zero: 2 - 4t^2 = 0 => t^2 = 1/2 => t = 1/√2. Then, g(1/√2) = 2*(1/√2)*sqrt(1 - 1/2) = 2*(1/√2)*(1/√2) = 2*(1/2) = 1. So again, maximum is 1.

Therefore, even in this case, the maximum is 1.

So, all approaches seem to lead to the conclusion that the maximum value is 1. However, let me check if there's a case where the function could be greater than 1. Suppose I take x and y such that they are not on the unit circle. For example, take x = 0.8, y = 0.6. Then, sqrt(1 - y^2) = sqrt(1 - 0.36) = sqrt(0.64) = 0.8, and sqrt(1 - x^2) = sqrt(1 - 0.64) = sqrt(0.36) = 0.6. Then f(x, y) = 0.8*0.8 + 0.6*0.6 = 0.64 + 0.36 = 1. So again, 1.

Wait, but x^2 + y^2 = 0.64 + 0.36 = 1, so that's on the unit circle. So again, same result.

Another example: take x = 0.6, y = 0.8. Then same thing.

But suppose I take x = 0.9, y = 0.435. Let me check if x^2 + y^2 is something. Let me compute 0.9^2 + 0.435^2 = 0.81 + 0.189225 = 0.999225, which is close to 1. So, if I take x = 0.9, y ≈ sqrt(1 - 0.81) = sqrt(0.19) ≈ 0.435. Then f(x, y) = 0.9 * sqrt(1 - y^2) + y * sqrt(1 - x^2) = 0.9 * sqrt(0.19) + 0.435 * sqrt(0.19). Wait, sqrt(1 - y^2) is sqrt(0.81) = 0.9, and sqrt(1 - x^2) is sqrt(0.19) ≈ 0.435. Therefore, f(x, y) = 0.9 * 0.9 + 0.435 * 0.435 ≈ 0.81 + 0.189 ≈ 0.999, which is approximately 1. So again, close to 1.

Therefore, it seems that whenever x and y are on the unit circle, the function equals 1, and when they are not, it's less than or equal to 1.

But let me think again. Suppose I take x and y such that they are not on the unit circle. For example, take x = 0.5, y = 0.5. Then f(x, y) = 0.5 * sqrt(1 - 0.25) + 0.5 * sqrt(1 - 0.25) = 0.5 * sqrt(0.75) + 0.5 * sqrt(0.75) = sqrt(0.75) ≈ 0.866, which is less than 1.

Another example: take x = 1, y = 0. Then f(x, y) = 1 * 1 + 0 * 0 = 1.

So, all these examples confirm that the maximum is 1.

But let me think again about the substitution approach. If we set $ x = \sin \alpha $, $ y = \sin \beta $, then $ f(x, y) = \sin(\alpha + \beta) $, which has maximum 1. However, is there a case where $ \alpha + \beta $ could be more than $ \frac{\pi}{2} $, but still $ \sin(\alpha + \beta) $ is greater than 1? No, because sine function's maximum is 1.

Alternatively, if we consider $ \alpha + \beta = \frac{\pi}{2} $, then $ \sin(\alpha + \beta) = 1 $. If $ \alpha + \beta $ is greater than $ \frac{\pi}{2} $, then $ \sin(\alpha + \beta) $ would start decreasing. For example, $ \alpha + \beta = \frac{3\pi}{4} $, then $ \sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2} \approx 0.707 $, which is less than 1. Similarly, if $ \alpha + \beta = \frac{\pi}{2} + \epsilon $, then $ \sin(\alpha + \beta) = \cos \epsilon $, which is less than 1 for $ \epsilon > 0 $.

Therefore, the maximum occurs when $ \alpha + \beta = \frac{\pi}{2} $, which gives the maximum value of 1.

But let me check if there are any constraints on $ \alpha $ and $ \beta $. Since $ x = \sin \alpha $ and $ y = \sin \beta $, and $ x, y \in [-1, 1] $, then $ \alpha, \beta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Therefore, $ \alpha + \beta \in [-\pi, \pi] $. However, the maximum of $ \sin(\alpha + \beta) $ in this interval is indeed 1, achieved when $ \alpha + \beta = \frac{\pi}{2} $.

Therefore, combining all these approaches, it seems that the maximum value of the function is 1.

But just to be thorough, let me check if there are any other critical points. Earlier, we found that critical points lie on the unit circle $ x^2 + y^2 = 1 $. However, when we set the partial derivatives to zero, we arrived at that condition. But maybe there are other critical points? For example, when the partial derivatives don't exist?

But the function $ f(x, y) $ is differentiable everywhere in the interior of the domain $ (-1, 1) \times (-1, 1) $, since the square roots are smooth there. At the boundaries, the function may not be differentiable, but we already checked the boundaries and found that the maximum is 1.

Therefore, the maximum value is indeed 1.

Wait, but let me think again. Suppose I take $ x = \sin \alpha $, $ y = \sin \beta $, and $ \alpha + \beta = \frac{\pi}{2} $. Then $ \beta = \frac{\pi}{2} - \alpha $. Therefore, $ y = \sin(\frac{\pi}{2} - \alpha) = \cos \alpha $. Therefore, $ x = \sin \alpha $, $ y = \cos \alpha $. Therefore, $ x^2 + y^2 = \sin^2 \alpha + \cos^2 \alpha = 1 $, which is consistent with the previous result. Therefore, all these points lie on the unit circle.

Therefore, the maximum value of the function is 1, achieved when $ x = \sin \alpha $, $ y = \cos \alpha $ for some $ \alpha $, or equivalently, when $ x $ and $ y $ are such that $ x^2 + y^2 = 1 $ and $ \alpha + \beta = \frac{\pi}{2} $.

But let me check if there's a case where the function could be greater than 1. Suppose I take $ x = 1 $, $ y = 0 $, which gives 1. If I take $ x = \cos \theta $, $ y = \sin \theta $, then $ f(x, y) = \cos \theta \cdot \sqrt{1 - \sin^2 \theta} + \sin \theta \cdot \sqrt{1 - \cos^2 \theta} = \cos \theta \cdot |\cos \theta| + \sin \theta \cdot |\sin \theta| $. But since $ \theta $ is in the range where $ \cos \theta $ and $ \sin \theta $ can be positive or negative. However, if we take $ \theta $ in the first quadrant, then $ \cos \theta $ and $ \sin \theta $ are positive, so $ f(x, y) = \cos^2 \theta + \sin^2 \theta = 1 $. If $ \theta $ is in other quadrants, then the absolute values would make it still 1? Wait, for example, if $ \theta $ is in the second quadrant, $ \cos \theta $ is negative, $ \sin \theta $ is positive. Then $ f(x, y) = \cos \theta \cdot |\cos \theta| + \sin \theta \cdot |\sin \theta| = \cos \theta (-\cos \theta) + \sin \theta \cdot \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 $, so $ \theta = \frac{\pi}{2} $. But at $ \theta = \frac{\pi}{2} $, $ x = 0 $, $ y = 1 $, and $ f(x, y) = 0 * 0 + 1 * 0 = 0 $? Wait, no. Wait, if $ \theta = \frac{\pi}{2} $, then $ x = \cos \frac{\pi}{2} = 0 $, $ y = \sin \frac{\pi}{2} = 1 $. Then $ f(x, y) = 0 * \sqrt{1 - 1} + 1 * \sqrt{1 - 0} = 0 + 1 = 1 $. Wait, but according to the previous expression, $ -\cos^2 \theta + \sin^2 \theta = -0 + 1 = 1 $. So that's correct.

Similarly, if $ \theta $ is in the third quadrant, both $ \cos \theta $ and $ \sin \theta $ are negative. Then $ f(x, y) = \cos \theta \cdot |\cos \theta| + \sin \theta \cdot |\sin \theta| = \cos \theta (-\cos \theta) + \sin \theta (-\sin \theta) = -\cos^2 \theta - \sin^2 \theta = -1 $, which is the minimum.

Similarly, in the fourth quadrant, $ \cos \theta $ is positive, $ \sin \theta $ is negative. Then $ f(x, y) = \cos \theta \cdot \cos \theta + \sin \theta \cdot (-\sin \theta) = \cos^2 \theta - \sin^2 \theta = \cos 2\theta $, which has maximum 1 when $ 2\theta = 0 $, i.e., $ \theta = 0 $, which gives $ x = 1 $, $ y = 0 $, and $ f(x, y) = 1 $.

Therefore, regardless of the quadrant, the maximum value of the function is 1.

Therefore, after all these checks, I can be confident that the maximum value of $ f(x, y) $ is indeed 1.

But just to make sure, let me consider another approach. Let me use the Cauchy-Schwarz inequality.

The function is $ f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2} $. Let me think of this as the dot product of two vectors: $ (x, y) $ and $ (\sqrt{1 - y^2}, \sqrt{1 - x^2}) $. Wait, but the second vector depends on both x and y, so it's not straightforward.

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

Alternatively, think of this as $ a \sqrt{1 - b^2} + b \sqrt{1 - a^2} \leq \sqrt{(a^2 + b^2)( (1 - b^2) + (1 - a^2) )} $ by Cauchy-Schwarz. Let me check:

By Cauchy-Schwarz, $ (a \sqrt{1 - b^2} + b \sqrt{1 - a^2})^2 \leq (a^2 + b^2)[(1 - b^2) + (1 - a^2)] $.

Calculating the right-hand side:

$ (a^2 + b^2)(2 - a^2 - b^2) $. Let me denote $ S = a^2 + b^2 $. Then the expression becomes $ S(2 - S) $. This is a quadratic in S: $ -S^2 + 2S $, which has maximum at $ S = 1 $, with maximum value $ -1 + 2 = 1 $. Therefore, the square of the function is less than or equal to 1, so the function is less than or equal to 1.

Therefore, the maximum value is 1, achieved when equality holds in Cauchy-Schwarz. Equality holds when $ \frac{a}{\sqrt{1 - b^2}} = \frac{b}{\sqrt{1 - a^2}} $. Let me solve this equation.

Cross-multiplying: $ a \sqrt{1 - a^2} = b \sqrt{1 - b^2} $. Squaring both sides: $ a^2 (1 - a^2) = b^2 (1 - b^2) $.

Which implies $ a^2 - a^4 = b^2 - b^4 $. Rearranged: $ a^2 - b^2 = a^4 - b^4 $. Factor both sides:

Left side: $ (a - b)(a + b) $

Right side: $ (a^2 - b^2)(a^2 + b^2) = (a - b)(a + b)(a^2 + b^2) $

Therefore, equation becomes:

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

If $ a \neq b $, we can divide both sides by $ (a - b)(a + b) $, assuming $ a + b \neq 0 $, leading to $ 1 = a^2 + b^2 $.

If $ a = b $, then the original equation $ a \sqrt{1 - a^2} = a \sqrt{1 - a^2} $ holds for any a.

Therefore, equality holds when either $ a = b $ or $ a^2 + b^2 = 1 $.

So, if $ a = b $, then the function becomes $ a \sqrt{1 - a^2} + a \sqrt{1 - a^2} = 2a \sqrt{1 - a^2} $, which we already saw has maximum 1 when $ a = \frac{1}{\sqrt{2}} $.

Alternatively, if $ a^2 + b^2 = 1 $, then the function is equal to 1, as shown earlier.

Therefore, the maximum value of the function is indeed 1, achieved in multiple cases.

Therefore, after multiple approaches—trigonometric substitution, calculus, Cauchy-Schwarz inequality—it seems that the maximum value of the function is 1.

Just to recap, all methods lead to the conclusion that the maximum is 1, achieved when (x, y) is on the unit circle with specific relations between x and y, or at the boundary points (1, 0) and (0, 1).

Therefore, I can confidently say that the maximum value of $ f(x, y) $ 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 [-\frac{\pi}{2}, \frac{\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 identity from trigonometry.

---

### Step 2: Maximizing the Sine Function

The function now becomes:
$$
f(x, y) = \sin(\alpha + \beta)
$$
with the constraint that $ \alpha, \beta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Therefore, the sum $ \alpha + \beta \in [-\pi, \pi] $.

The sine function achieves its maximum value of 1 at $ \alpha + \beta = \frac{\pi}{2} $, which is within the allowed range. Thus, the maximum value of $ f(x, y) $ is:
$$
\sin\left(\frac{\pi}{2}\right) = 1
$$

---

### Step 3: Verifying the Maximum

We now verify that this maximum is achievable. For instance, take:
$$
\alpha = \frac{\pi}{2}, \quad \beta = 0
\Rightarrow x = \sin\left(\frac{\pi}{2}\right) = 1, \quad y = \sin(0) = 0
$$
Then:
$$
f(1, 0) = 1 \cdot \sqrt{1 - 0^2} + 0 \cdot \sqrt{1 - 1^2} = 1
$$
Similarly, taking $ \alpha = 0, \beta = \frac{\pi}{2} $, we get $ f(0, 1) = 1 $.

Also, if $ x = \cos \theta $, $ y = \sin \theta $, then:
$$
f(x, y) = \cos \theta \cdot \sqrt{1 - \sin^2 \theta} + \sin \theta \cdot \sqrt{1 - \cos^2 \theta} = \cos^2 \theta + \sin^2 \theta = 1
$$
So, the function achieves the value 1 on the unit circle $ x^2 + y^2 = 1 $, and this is the maximum possible value.

---

### Step 4: Boundary and Critical Point Analysis

We also consider the boundaries of the domain $ -1 \leq x, y \leq 1 $. For example:
- If $ x = 1 $, then $ f(1, y) = \sqrt{1 - y^2} $, which is maximized at $ y = 0 $, giving $ f(1, 0) = 1 $
- If $ y = 1 $, then $ f(x, 1) = \sqrt{1 - x^2} $, which is maximized at $ x = 0 $, giving $ f(0, 1) = 1 $

Using calculus, we find that the critical points of $ f(x, y) $ occur when $ x^2 + y^2 = 1 $, and on this curve, the function again evaluates to 1.

---

### Step 5: Inequality Approach (Cauchy-Schwarz)

We can also apply the Cauchy-Schwarz inequality:
$$
f(x, y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2} \leq \sqrt{(x^2 + y^2)\left((1 - y^2) + (1 - x^2)\right)} = \sqrt{(x^2 + y^2)(2 - x^2 - y^2)}
$$
Let $ S = x^2 + y^2 $. Then the expression becomes:
$$
\sqrt{S(2 - S)}
$$
This is maximized when $ S = 1 $, giving:
$$
\sqrt{1 \cdot (2 - 1)} = \sqrt{1} = 1
$$
Thus, the maximum value of $ f(x, y) $ is again confirmed to be 1.

---

### Final Conclusion

All approaches—trigonometric substitution, calculus, and inequality—lead to the same conclusion. The function achieves its maximum value of 1 when $ x = 1, y = 0 $, $ x = 0, y = 1 $, or when $ x^2 + y^2 = 1 $ with appropriate angle relations.

---

### Final Answer

$$
\boxed{1}
$$

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 $\boxed{1}.$