We need to build a rectangular aquarium with a volume of 12 cubic feet. The material for the bottom costs 6 dollars per sq foot, and the sides cost 2 dollars per square foot; it has no top. find the dimensions of the least expensive aquarium we can build.

Optimization problem - aquarium

How do they get to $8=l^3$ from $6l-\frac{48}{8/l^2}$=0 ?

  • $\begingroup$ Please don’t include critical parts of your question as an image. It is neither searchable nor accessible to those using a screen reader. $\endgroup$
    – amd
    Mar 20, 2017 at 3:22
  • $\begingroup$ This was, of course, an exercise in partial derivatives of multivariable functions, but in practice, you could’ve simplified the problem after deriving $C(l,w)$. Observe that this function is symmetric in its arguments, so you’re going to end up with $l=w$ and can reduce this to a single-.variable problem. $\endgroup$
    – amd
    Mar 20, 2017 at 6:31

1 Answer 1


$6l - \frac{48}{(8/l^2)^2}=0$

$6l - \frac{48}{64/l^4}=0$

$6l - \frac{48l^4}{64}=0$

$24l - 3l^4=0$

$l=0 $ or $l^3=8$


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