# How do you find the dimensions of the aquarium that minimize the cost of the materials if the base of an aquarium with volume v is made of slate and the sides are made of glass and the slate costs five times as much (per unit area) as glass?

Mar 6, 2015

Without a specific value for the volume, $v$, only a generalized solution can be developed.

We need to develop an expression for the cost of materials based on one of the dimensions.

Let $w =$ width of the base;
$d =$ depth of the base; and
$h =$ height of the aquarium
$C =$ cost

$C =$ cost of base $+$ cost of walls
$= \left(5\right) \left(w \times d\right) + \left(1\right) \left(2 \left(w + d\right) \times h\right)$

Note that the minimum will be achieved when the width and the depth are equal. (I think this should be obvious but re-post if it isn't).
So we are only interested in the case (substituting $w$ for $d$)

$C = 5 {w}^{2} + 4 w h$

The volume in this case can be expressed as
$v = {w}^{2} h$

So if we re-write our Cost equation as
$C = 5 {w}^{2} + \frac{4 {w}^{2} h}{w}$

This becomes
$C = 5 {w}^{2} + 4 v {w}^{- 1}$

To find the minimum cost, we take the derivative of this and set the result to zero.

$\frac{d C}{\mathrm{dw}} = 10 w - 4 v {w}^{- 2}$

Once a value has been established for the volume ($v$) this equation can be solved for $w$ and via substitution for $d$ and $h$.