Question

A cuboidal shaped godown with square base is to be constructed. Three times as much cost per square meter is incurred for constructing the roof as compared to the walls. Find the dimensions of the godown if it is to enclose a given volume and minimize the cost of constructing the roof and walls.

Answer 1

Let the length and breadth of the base be x and the height of the godown be y. Let C be the cost of constructing the godown and V be the given volume. So, $V=x^{2}y\Rightarrow xy=\frac{V}{x}....\left(i\right)$
And, $C=(3x^2+4xy)\times k$, where k is cost incurred for walls.
$\therefore C=(3x^2+\frac{4V}{x})\times k$ [By (i)]
Now $\frac{dC}{dx}=(6x-\frac{4V}{x^2})\times k$ and, $\frac{d^{2}C}{d{x}^2}=(6+\frac{8V}{x^3})\times k$
For maximum and/or minimum value of C, $\frac{dC}{dx}=0\Rightarrow 6x-\frac{4V}{x^2}=0$
$\Rightarrow x^3=\frac{2V}{3}\therefore x={\left(\frac{2V}{3}\right)}^{\frac{1}{3}}$
When $x^3=\frac{2V}{3}$, then $\frac{d^{2}C}{d{x}^2}=(6+8\times \frac{3}{2})\times k=18k>0(\because k$ is cost, so k > 0)
So, C is minimum when $x={\left(\frac{2V}{3}\right)}^{\frac{1}{3}}$
Now, $V=x^{2}y\Rightarrow \frac{3x^3}{2}=x^{2}y\Rightarrow x^2(\frac{3x}{2}-y)=0\Rightarrow \frac{3x}{2}=y\therefore y=\frac{3}{2}{\left(\frac{2V}{3}\right)}^{\frac{1}{3}}$
Hence the dimensions of the godown are ${\left(\frac{2V}{3}\right)}^{\frac{1}{3}}\times {\left(\frac{2V}{3}\right)}^{\frac{1}{3}}\times \frac{3}{2}{\left(\frac{2V}{3}\right)}^{\frac{1}{3}}$.