Question

A rectangular box has a square base. if the sum of the height and the perimeter of the square base is 12 in, what is the maximum possible volume?

Answer 1

Let the square base has length $b$ and the height of the box be $h$ . It is given that $4b+h=12\\ h=4(3-b)$

The volume of the box is $V=b^2h=4b^2(3-b)$.

For maximum volume $\frac{dV}{db} =0\\ \frac{d(4b^2(3-b))}{db} =0\\ 6b-3b^2=0\\ b=2$

The maximum possible volume occurs when $b=3 in$.

$V_{max} =4(2)^2(3-2) =16 in^3$