Question

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

Answer 1

Given a godown with square base.

So,Length of godown$=$Breadth of godown

Now,Volume of godown$=length\times breadth\times height$

Volume of godown$=l\times l\times h=l^{2}h$

$\Rightarrow h=\frac{v}{l^2}$ .........$\left(1\right)$

Here volume is given.So it is a constant.

Thus, we write height in terms of $v$ and $l$

Given that,it costs three times as much cost per square meter is incurred for constructing the roof as compared to the walls.

Let Rate of constructing the wall$=Rs.kper{m}^2$

So,Rate of constructing the roof$=Rs.3kper{m}^2$

Cost of constructing the walls$=$Rate of construction of wall$\times$Area of walls

$=k\times 4lh$

Putting value of $h$ from $\left(1\right)$ we get

Cost of constructing the walls$=k\times 4l\times \frac{v}{l^2}=\frac{4kv}{l}$

Cost of constructing the roof$=$Rate of construction of roof$\times$Area of roof

$=3k\times l\times l=3k{l}^2$

Now,cost of constructing roof and wall$=$cost of constructing roof$+$cost of constructing the wall

$=3k{l}^2+\frac{4kv}{l}$

$C=3k{l}^2+\frac{4kv}{l}$

Now, we need to minimize the cost $C$

Since length $l$ is a variable, we differentiate w.r.t $l$

$\frac{dC}{dl}=6kl+4kv\times \frac{-1}{l^2}$

$\frac{dC}{dl}=6kl-\frac{4kv}{l^2}$

Putting $\frac{dC}{dl}=0$

$\Rightarrow 0=6kl-\frac{4kv}{l^2}$

$\Rightarrow 6kl=\frac{4kv}{l^2}$

$\Rightarrow 6k{l}^3=4kv$

$\Rightarrow l^3=\frac{2v}{3}$

$\therefore l={\left(\frac{2v}{3}\right)}^{\frac{1}{3}}$ ......$\left(2\right)$

$\frac{dC}{dl}=6kl-\frac{4kv}{l^2}$

Differentiating w.r.t $l$ we get

Now,$\frac{d^{2}C}{d{l}^2}=6k-4kv\left(\frac{-2}{l^3}\right)=6k+\frac{8kv}{l^3}$

Put $l={\left(\frac{2v}{3}\right)}^{\frac{1}{3}}$ in

$\frac{d^{2}C}{d{l}^2}=6k+\frac{8kv}{\frac{2v}{3}}=6k+12k=18k>0$

Since $k$ is positive,$\frac{d^{2}C}{d{l}^2}>0$

$\therefore \frac{d^{2}C}{d{l}^2}>0$ when $l={\left(\frac{2v}{3}\right)}^{\frac{1}{3}}$

$C$ is minimum when $l={\left(\frac{2v}{3}\right)}^{\frac{1}{3}}$

Now, finding $h$

From $\left(1\right),$ we have

$h=\frac{v}{l^2}=\frac{v}{{\left(\frac{2v}{3}\right)}^{\frac{2}{3}}}$

$={\left(\frac{3}{2}\right)}^{\frac{2}{3}}{v}^{1-\frac{2}{3}}$

$={\left(\frac{9}{4}\right)}^{\frac{1}{3}}{v}^{\frac{1}{3}}$

$={\left(\frac{9v}{4}\right)}^{\frac{1}{3}}$

Thus,Length of godown$=l={\left(\frac{2v}{3}\right)}^{\frac{1}{3}}$

Breadth of godown$=l={\left(\frac{9v}{4}\right)}^{\frac{1}{3}}$