Question

Cal 1 Question (Optimization)

Answer 1

The base of the box is square, so if the sides of the base are
`x` and the box has height
`y`, then the volume of the box is


`V=x^2y`

We are maximizing this subject to the constraint of surface area,


`72=\underbrace{5x^2}_{\text{glass top}}+\underbrace{4xy}_{\text{lateral faces}}+\underbrace{x^2}_{\text{base}}`

`72=6x^2+4xy`

Solving for
`y` gives


`y=(72-6x^2)/(4x)`

and substituting into the volume equation gives


`V=x^2(72-6x^2)/(4x)`

`V=18x-\frac32x^3`

Differentiating, we get


`V'=18-\frac92x^2`

which has roots at
`x=\pm2`, but we omit the negative root. So
`x=2`.

This means we must have


`y=(72-6(2)^2)/(4(2))=6`