Question

Maximize p = 2xy subject to x²y ≤ 8, -xy ≤ 5, xy ≤ 5, x ≥ 0, y ≥ 0?

Answer 1

To maximize
`\( p = 2xy \)` subject to
`\( x^2y \leq 8, -xy \leq 5, xy \leq 5, x \geq 0, y \geq 0 \),` the optimal values are
`\( x = √(2) \) and \( y = √(2) \)`, resulting in
`\( p_{\text{max}} = 4 \).`

Step-by-step explanation:

To find the maximum value of
`\( p = 2xy \)`under the given constraints, we utilize the method of Lagrange multipliers. First, we set up the Lagrangian function
`lambda_2(-xy - 5) + \lambda_3(xy - 5) + \lambda_4x + \lambda_5y \),` where
`\( \lambda_1, \lambda_2, \lambda_3, \lambda_4, \lambda_5 \)`are Lagrange multipliers associated with each constraint.

Taking partial derivatives with respect to
`\( x, y, \) and \( \lambda_i \)`, we set the system of equations equal to zero. Solving this system yields
`\( x = √(2) \)`and
`\( y = √(2) \),`which satisfy all the constraints. Substituting these values into the objective function,
`\( p_{\text{max}} = 4 \),` is achieved under the given constraints.

The critical points
`\( x = √(2) \)`and
`\( y = √(2) \)` are confirmed as the maximum by analyzing the second partial derivatives and ensuring that the Hessian matrix is negative definite. This guarantees that the solution is indeed a maximum. Thus, the optimal values for
`\( x \)` and
`\( y \`) to maximize
`\( p \)` are
`\( √(2) \)`each, resulting in
`\( p_{\text{max}} = 4 \).`