Use Lagrange multipliers to find the maximum value of f (x, y, z) = 6xyz subject to 4x + 3y + 4z = 144.
A) f (0, 0, 0) = 0
B) f (15, 17, 10) = 15,300
C) f (12, 16, 12) = 13,824
D) f (13, 14, 12) = 13,104

Question

Quizplus · Accepted Answer

To solve this problem using Lagrange multipliers, we set up the following equations based on the given function $f(x, y, z) = 6xyz$ and the constraint $4x + 3y + 4z = 144$. The Lagrange function $L$ is given by $L = 6xyz + \lambda(4x + 3y + 4z - 144)$. Taking partial derivatives and setting them equal to zero gives us a system of equations:1. $\frac{\partial L}{\partial x} = 6yz + 4\lambda = 0$2. $\frac{\partial L}{\partial y} = 6xz + 3\lambda = 0$3. $\frac{\partial L}{\partial z} = 6xy + 4\lambda = 0$4. $4x + 3y + 4z = 144$Solving this system of equations, we find that one of the solutions that satisfies the constraint is $x = 12$, $y = 16$, and $z = 12$. Substituting these values into the original function $f(x, y, z) = 6xyz$ gives $f(12, 16, 12) = 6 \times 12 \times 16 \times 12 = 13,824$, which is the maximum value under the given constraint.

Use Lagrange Multipliers to Find the Maximum Value of F