If x and y are two positive numbers such that x+y=32, then the maximum value of `x^2+y^2` is ,

Options

A. 500
B. 256
C. 1024
D. 512

Answer

To find the maximum value of \( x^2 + y^2 \) given that \( x + y = 32 \), we can follow these steps:

### Step 1: Express \( y \) in terms of \( x \)
Since we know that \( x + y = 32 \), we can express \( y \) as:
\[ y = 32 - x \]

### Step 2: Substitute \( y \) in the expression \( x^2 + y^2 \)
Now, we substitute \( y \) into the expression \( x^2 + y^2 \):
\[ x^2 + y^2 = x^2 + (32 - x)^2 \]

### Step 3: Expand the expression
Next, we expand \( (32 - x)^2 \):
\[ (32 - x)^2 = 32^2 - 2 \cdot 32 \cdot x + x^2 = 1024 - 64x + x^2 \]
Thus, we have:
\[ x^2 + y^2 = x^2 + (1024 - 64x + x^2) = 2x^2 - 64x + 1024 \]

### Step 4: Differentiate the expression
To find the maximum value, we differentiate \( x^2 + y^2 \) with respect to \( x \):
\[ \frac{d}{dx}(2x^2 - 64x + 1024) = 4x - 64 \]

### Step 5: Set the derivative to zero
Setting the derivative equal to zero to find critical points:
\[ 4x - 64 = 0 \]
Solving for \( x \):
\[ 4x = 64 \implies x = 16 \]

### Step 6: Find \( y \)
Now, substitute \( x = 16 \) back into the equation for \( y \):
\[ y = 32 - x = 32 - 16 = 16 \]

### Step 7: Calculate \( x^2 + y^2 \)
Now we calculate \( x^2 + y^2 \):
\[ x^2 + y^2 = 16^2 + 16^2 = 256 + 256 = 512 \]

### Step 8: Verify maximum using second derivative test
To confirm that this is a maximum, we can check the second derivative:
\[ \frac{d^2}{dx^2}(2x^2 - 64x + 1024) = 4 \]
Since \( 4 > 0 \), this indicates a minimum. However, since we are constrained by \( x + y = 32 \) and both \( x \) and \( y \) are positive, we are indeed at a maximum for the values of \( x^2 + y^2 \).

### Conclusion
The maximum value of \( x^2 + y^2 \) is:
\[ \boxed{512} \]