Optimization question

an open storage bin w/ a square base and vertical sides is gonna be constructed from 108 ft of material. What are the dimensions that would have the maximum volume?

how do you do this?

You must mean that it will be constructed from 108 square feet of material. Let the square base have side length x and the height be y. The volume is

V = x²y.

The surface area is

S = x² + 4xy.

That gives a base and four sides but no top (open). Okay, we know that S = 108, so

x² + 4xy = 108 ==> y = (108 - x²)/(4x).

This makes the volume a function of x alone

V(x) = x²(108 - x²)/(4x) = ¼ x(108 - x²) = ¼(108x - x^3).

Take the derivative to find the critical value(s).

V'(x) = ¼(108 - 3x²) ==> V'(x) = 0 when x² = 36 ==> x = 6.

We can ignore negative solutions since x is a length. Note that

V''(6) = -6(6)/4 < 0, so this is a maximum.

Getting y

y = (108 - 6²)/(4∙6) = 3.

The box should be 6' x 6' x 3'.

square base gives a hint, and turns the problem into a single variable calculus problem, rather than a multi-variable one.

let x be the length of the base legs, let y be the height

x^2 + 4xy = 108

V = y*x^2

(108-x^2)/4x = y

V = (108-x^2)*(x/4)

dV/dx = 0 = (108/4)-(3/4*(x^2)) =0

x = 6, y = 3.