A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is $2m$ and volume is $8{m^3}$. If building of tank costs $Rs.70$ per sq. metre for the base and $Rs.45$ per square metre for sides. What is the cost of least expensive tank?

Let us assume that the dimensions of the base be $x$ metres and $y$ metres, the total expenses be $Rs.S$ and the volume of the tank be

$V{m^3},$ then
$V = 2xy \Rightarrow 8 - 2xy = 0 \Rightarrow y = \cfrac{4}{x}$ …(i)

Also, area of the base $ABCP = xy = x \times \cfrac{4}{x} = 4{m^2}$ (from (i))

and total area of the four sides $(ABSQ,PCDR,BCDS,APRQ)$
$= (2x + 2x + 2y + 2y){m^2}$

$= \left( {4x + 4\left( {\cfrac{4}{x}} \right)} \right){m^2}$ (from (i))
$= 4\left\{ {x + \cfrac{4}{x}} \right\}{m^2}$

Therefore $S = 4 \times 70 + 45 \times 4\left\{ {x + \cfrac{4}{x}} \right\} = 280 + 180\left\{ {x + \cfrac{4}{x}} \right\}$ ...(ii)

Differentiating (ii) w.r.t. $x$, we get $\cfrac{{dS}}{{dx}} = 180\left( {1 - \cfrac{4}{{{x^2}}}} \right)$ …(iii)

For maximum/minimum expenses, $\cfrac{{dS}}{{dx}} = 0$
$\Rightarrow 180\left( {1 - \cfrac{4}{{{x^2}}}} \right) = 0 \Rightarrow {x^2} = 4 \Rightarrow x = 2$

Differentiating (iii) w.r.t. $x$, we get

$\cfrac{{{d^2}S}}{{d{x^2}}} = 180\left( {0 + \cfrac{8}{{{x^3}}}} \right) = \cfrac{{1440}}{{{x^3}}}$ …(iv)
Now ${\left( {\cfrac{{{d^2}S}}{{d{x^2}}}} \right)_{x = 2}} = \cfrac{{1440}}{{{2^3}}} > 0$

Therefore $S$ is least when $x = 2.$

Thus, cost of least expensive tank $= Rs.\left\{ {280 + 180\left( {2 + \cfrac{4}{2}} \right)} \right\}$ (from (ii)) $= Rs.(280 + 720) = Rs.1000$