If `x+y =k`, where `x, y epsilon N ` then `xy` is maximum when `(x, y)-=..`

Options

A. `((k)/(3), (2k)/(3))`
B. `((k)/(4), (3k)/(4))`
C. `((k)/(2), (k)/(2))`
D. `((2k)/(3), (2k)/(3))`

Answer

To solve the problem of maximizing the product \( xy \) given the constraint \( x + y = k \) where \( x, y \in \mathbb{N} \), we can follow these steps:

### Step 1: Express \( y \) in terms of \( x \)
From the equation \( x + y = k \), we can express \( y \) as:
\[ y = k - x \]

### Step 2: Write the product \( P \)
The product \( P = xy \) can be rewritten using the expression for \( y \):
\[ P = x(k - x) = kx - x^2 \]

### Step 3: Differentiate \( P \) with respect to \( x \)
To find the maximum value of \( P \), we need to differentiate \( P \) with respect to \( x \):
\[ \frac{dP}{dx} = k - 2x \]

### Step 4: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ k - 2x = 0 \]
Solving for \( x \) gives:
\[ 2x = k \quad \Rightarrow \quad x = \frac{k}{2} \]

### Step 5: Check the second derivative
To confirm that this point is a maximum, we find the second derivative:
\[ \frac{d^2P}{dx^2} = -2 \]
Since \( \frac{d^2P}{dx^2} < 0 \), this indicates that \( P \) has a maximum at \( x = \frac{k}{2} \).

### Step 6: Find \( y \)
Substituting \( x = \frac{k}{2} \) back into the equation for \( y \):
\[ y = k - x = k - \frac{k}{2} = \frac{k}{2} \]

### Conclusion
Thus, the values of \( x \) and \( y \) that maximize the product \( xy \) are:
\[ (x, y) = \left(\frac{k}{2}, \frac{k}{2}\right) \]