# Thread: Optimization problem

1. ## Optimization problem

A closed box with a square base is designed to have a surface area of 216 sq. ft. Find the dimensions of the box if the volume is to be a max.

Pardon my elementary approach to this.. but this is how I have been doing these sort of problems.. i make up 2 equations ...

6lw = 216 sq ft (surface area)
and
l^3 = maximum (volume of a cube, square on all sides)

then i solve for one of the variables, in this case L (length)
so 216/6w = 36/w = l

then i substitute

(36/w)^3 = maximum

do you guys follow? in this particular case, i get some weird numbers in the end, so i think my approach didn't work in this problem?

2. Originally Posted by alny
A closed box with a square base is designed to have a surface area of 216 sq. ft. Find the dimensions of the box if the volume is to be a max.

Pardon my elementary approach to this.. but this is how I have been doing these sort of problems.. i make up 2 equations ...

6lw = 216 sq ft (surface area)
and
l^3 = maximum (volume of a cube, square on all sides)

then i solve for one of the variables, in this case L (length)
so 216/6w = 36/w = l

then i substitute

(36/w)^3 = maximum

do you guys follow? in this particular case, i get some weird numbers in the end, so i think my approach didn't work in this problem?
Try drawing the net of your box first.

I think you'll find that if the box has a square base, then it's side lengths are $l$, but the height of the box is not necessarily $l$. We'll call it something else, $h$ (for height).

So $TSA = 2l^2 + 4lh$

And since $TSA = 216\,\textrm{ft}^2$, this means

$2l^2 + 4lh = 216$

$4lh = 216 - 2l^2$

$h = \frac{54}{l} - \frac{l}{2}$.

Now, the volume of the box is

$V = l^2h$

$= l^2\left(\frac{54}{l} - \frac{l}{2}\right)$

$= 54l - \frac{l^3}{2}$.

To maximise the volume, find its derivative, set equal to 0, solve for the length, solve for the height.