Question

Let \( f(x, y) = -x^2 - y^2 + 2x + 4y + 5 \). Let \( (x^*, y^*) \) denote the solution to the following optimization problem: \[ {Maximize } f(x, y) \] subject to \[ x \geq 0, y \geq 0, 2x + y \leq 6 \] Then the value of \( f(x^*, y^*) \) is ________________ (in integer).

Hint:

In constrained optimization problems, always check both the critical points and the boundary points to ensure you find the maximum or minimum value.

Answer 1

Step 1: Define the objective function and constraints. The objective function is: \[ f(x, y) = -x^2 - y^2 + 2x + 4y + 5 \] The constraints are: \[ x \geq 0, \quad y \geq 0, \quad 2x + y \leq 6 \] Step 2: Find the critical points of the objective function.
We need to take the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \), and set them equal to zero. The partial derivative with respect to \( x \) is: \[ \frac{\partial f}{\partial x} = -2x + 2 \] Setting this equal to zero: \[ -2x + 2 = 0 \quad \Rightarrow \quad x = 1 \] The partial derivative with respect to \( y \) is: \[ \frac{\partial f}{\partial y} = -2y + 4 \] Setting this equal to zero: \[ -2y + 4 = 0 \quad \Rightarrow \quad y = 2 \] Thus, the critical point is \( (x^*, y^*) = (1, 2) \). 
Step 3: Check the feasibility of the critical point.
We need to check if the point \( (1, 2) \) satisfies all the constraints:
\( x = 1 \geq 0 \) (satisfied),
\( y = 2 \geq 0 \) (satisfied),
\( 2x + y = 2(1) + 2 = 4 \leq 6 \) (satisfied).
Since all constraints are satisfied, the point \( (1, 2) \) is feasible. 
Step 4: Evaluate the objective function at the critical point.
Substitute \( x = 1 \) and \( y = 2 \) into the objective function: \[ f(1, 2) = -(1)^2 - (2)^2 + 2(1) + 4(2) + 5 \] \[ f(1, 2) = -1 - 4 + 2 + 8 + 5 = 10 \] Step 5: Check the boundary points.
We also need to check the boundary points, given the constraints \( x \geq 0 \), \( y \geq 0 \), and \( 2x + y \leq 6 \). 1. When \( x = 0 \), the constraint \( 2x + y \leq 6 \) becomes \( y \leq 6 \). The objective function becomes: \[ f(0, y) = -(0)^2 - y^2 + 2(0) + 4y + 5 = -y^2 + 4y + 5 \] This is a quadratic function in \( y \), and its maximum occurs at the vertex. The vertex occurs at: \[ y = \frac{-4}{-2} = 2 \] Substituting \( y = 2 \) into the objective function: \[ f(0, 2) = -(0)^2 - (2)^2 + 2(0) + 4(2) + 5 = -4 + 8 + 5 = 9 \] 2. When \( y = 0 \), the constraint \( 2x + y \leq 6 \) becomes \( 2x \leq 6 \), so \( x \leq 3 \). The objective function becomes: \[ f(x, 0) = -x^2 - (0)^2 + 2x + 4(0) + 5 = -x^2 + 2x + 5 \] This is a quadratic function in \( x \), and its maximum occurs at the vertex. The vertex occurs at: \[ x = \frac{-2}{-2} = 1 \] Substituting \( x = 1 \) into the objective function: \[ f(1, 0) = -(1)^2 - (0)^2 + 2(1) + 4(0) + 5 = -1 + 2 + 5 = 6 \] Step 6: Compare the values.
We have the following values for the objective function:
\( f(1, 2) = 10 \),
\( f(0, 2) = 9 \),
\( f(1, 0) = 6 \).
The maximum value of the objective function is \( 10 \), which occurs at \( (x^*, y^*) = (1, 2) \). Thus, the value of \( f(x^*, y^*) \) is \( \boxed{10} \).