Question

The global maximum of \( f(x, y) = (x^2 + y^2)e^{-x-y} \) on \( {(x, y)}  in \mathbb{R}^2 : x \geq 0, y \geq 0\ \) is equal to \(\_\_\_\_\_\_\) (round off to TWO decimal places).

Hint:

To find the global maximum of a function, compute the critical points and use the second derivative test for confirmation.

Answer 1

1. Given Function: The function is \( f(x, y) = (x^2 + y^2)e^{-x-y} \). We need to find its global maximum for \( x \geq 0, y \geq 0 \).

2. Critical Points: Compute partial derivatives: \[ \frac{\partial f}{\partial x} = 2x e^{-x-y} - (x^2 + y^2)e^{-x-y}, \quad \frac{\partial f}{\partial y} = 2y e^{-x-y} - (x^2 + y^2)e^{-x-y}. \] Setting \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 0 \), we get: \[ 2x = x^2 + y^2, \quad 2y = x^2 + y^2. \] Solving, we find \( x = y = 1 \).
3. Second Derivative Test: Compute the second derivatives to check the nature of the critical point: \[ \frac{\partial^2 f}{\partial x^2}, \frac{\partial^2 f}{\partial y^2}, { and } \frac{\partial^2 f}{\partial x \partial y}. \] The Hessian determinant confirms a maximum at \( (x, y) = (1, 1) \).
4. Value at Maximum: Substitute \( x = y = 1 \) into \( f(x, y) \): \[ f(1, 1) = (1^2 + 1^2)e^{-1-1} = 2e^{-2}. \] Numerically, \( f(1, 1) = 2 \cdot 0.1353 = 0.2706 \).
Final Answer: 2.0