Question

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?

Answer 1

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
`= (5)( w xx d) +(1)( 2(w+d) xx h)`

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 = 5w^2 + 4wh`

The volume in this case can be expressed as
`v = w^2h`

So if we re-write our Cost equation as
`C = 5w^2 + (4w^2h)/w`

This becomes
`C = 5w^2 + 4vw^(-1)`

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

`(d C)/(dw) = 10w - 4vw^(-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`.