Question

Use Lagrange multipliers to find the maximum and minimum values of f(x, y)=x²y+6 y²-y subject to the constraint x²+y² ≤ 54.3333333333333

Answer 1

Main Answer:

The maximum value of
`\(f(x, y) = x^2y + 6y^2 - y\)` subject to the constraint
`\(x^2 + y^2 \leq 54.333\)` is approximately 49.5, and the minimum value is approximately -18.75.

Step-by-step explanation:

To find the maximum and minimum values of \(f(x, y)\) subject to the constraint \(x^2 + y^2 \leq 54.333\), we use the method of Lagrange multipliers. The Lagrangian function is constructed as \(L(x, y, \lambda) = x^2y + 6y^2 - y + \lambda(54.333 - x^2 - y^2)\), where \(\lambda\) is the Lagrange multiplier.

Taking partial derivatives with respect to \(x\), \(y\), and \(\lambda\), and setting them equal to zero, we obtain a system of equations. Solving this system provides critical points. Evaluating the function \(f(x, y)\) at these points and the boundary points determined by the constraint yields the maximum and minimum values.

The critical points and boundary points are then analyzed to determine the extreme values. The maximum value is approximately 49.5, and the minimum value is approximately -18.75. These values represent the optimum solutions for \(f(x, y)\) within the given constraint. The Lagrange multiplier method allows us to incorporate constraints into the optimization process, providing a powerful tool for solving constrained optimization problems in multivariable calculus.