Question

Use Lagrange multipliers to find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the given plane. x 3y 4z

Answer 1

I assume the given plane has equation

x + 3y + 4z = d

for some constant d ≠ 0.

You want to maximize the function V(x, y, z) = xyz subject to the above constraint. The Lagrangian would be

L(x, y, z, λ) = xyz - λ (x + 3y + 4z - d)

Find its critical points:

∂L/∂x = yz - λ = 0

∂L/∂y = xz - 3λ = 0

∂L/∂z = xy - 4λ = 0

∂L/∂λ = x + 3y + 4z - d = 0

The first three equations tell you that λ = yz = xz/3 = xy/4. Then, for instance,

yz = xz/3   →   3yz - xz = z (3y - x) = 0   →   z = 0   or   3y - x = 0

If z = 0, we get zero volume which is not useful. So 3y = x. Similarly, you would find x = 4z and 3y = 4z, so a critical point occurs when x = 3y = 4z.

In the fourth equation, we get upon substituting

x + 3y + 4z = 3x = d   →   x = d/3, y = d/9, z = d/12

Then the largest volume of this box would be

V(d/3, d/9, d/12) = d ³ / 324