Question

Define the function \(f: \R^2 → \R\) by
f(x, y) = 12xy e^-(2x+3y-2).
If (a, b) is the point of local maximum of f, then f(a, b) equals

• A. 2
• B. 6
• C. 12
• D. 0

Answer 1

The Correct Option is A

To find the point \( (a, b) \) of local maximum, we need to calculate the critical points by finding the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \), and setting them equal to zero. 
1. First, calculate the partial derivatives: \[ \frac{\partial f}{\partial x} = 12y e^{-(2x + 3y - 2)} - 24xy e^{-(2x + 3y - 2)} = 12y e^{-(2x + 3y - 2)} (1 - 2x) \] \[ \frac{\partial f}{\partial y} = 12x e^{-(2x + 3y - 2)} - 36xy e^{-(2x + 3y - 2)} = 12x e^{-(2x + 3y - 2)} (1 - 3y) \] 2. Set these partial derivatives equal to zero to find the critical points: From \( \frac{\partial f}{\partial x} = 0 \), we get: \[ y = 0 \quad \text{or} \quad 1 - 2x = 0 \quad \Rightarrow \quad x = \frac{1}{2}. \] From \( \frac{\partial f}{\partial y} = 0 \), we get: \[ x = 0 \quad \text{or} \quad 1 - 3y = 0 \quad \Rightarrow \quad y = \frac{1}{3}. \] Thus, the critical point occurs at \( (x, y) = \left( \frac{1}{2}, \frac{1}{3} \right) \). 3. Evaluate \( f(a, b) \) at this critical point: Substitute \( x = \frac{1}{2} \) and \( y = \frac{1}{3} \) into the function \( f(x, y) \): \[ f\left( \frac{1}{2}, \frac{1}{3} \right) = 12 \times \frac{1}{2} \times \frac{1}{3} \times e^{-(2 \times \frac{1}{2} + 3 \times \frac{1}{3} - 2)} = 2 \times e^{-(1 + 1 - 2)} = 2 \times e^0 = 2. \] Thus, the correct answer is (A): 2.