Question

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 $8m^3$. If building of tank costs Rs 70 per sq metre for the base and Rs 45 per sq metre for sides. What is the cost of least expensive tank ?

Answer 1

Let the length and breadth of the tank be x metre and y metre, respectively. The depth (height) is 2m.
$\therefore$ Volume of tank $=2\times x\times y=2xy=8$
[Given volume of tank $=8m^3$]
$\Rightarrow xy=4$
$\Rightarrow y=\frac{4}{x}$ ...(i)

Also, area of the base $=xy=x\times \frac{4}{x}=4$
and area of the four sides $=2x+2y÷2x+2y=4x+4y=4(x+y)$
It is given that, the cost of construction on base is Rs 70 per sq. metre and for side is Rs 45 per sq metre
So, cost of construction, $C=Rs[70\times xy+45\times 4(x+y\left)\right]$
= Rs [70xy + 180(x+y)] ...(ii)
On putting value of y in Eq. (ii) from Eq. (i).we get
$C=70\times 4+180(x+\frac{4}{x})=280+180(x+\frac{4}{x})$ ...(iii)
On differentiating w.r.t.x, we get $\frac{dC}{dx}=0+180(1-\frac{4}{x^2})=180\left(\frac{x^2-4}{x^2}\right)$
For least expensive, put $\frac{dC}{dx}=0\Rightarrow 180\left(\frac{x^2-4}{x^2}\right)=0\Rightarrow x^2=4\Rightarrow x=\pm 2$
$\frac{dC}{dx}$ changes sign from negative to positive at x = 2.
$\therefore$ C is minimum at x = 2
[length of the tank cannot be negative. So, x = - 2 is not consider]
x = 2 and $y=\frac{4}{x}=\frac{4}{2}=2$
Thus, tank is a cube of side 2m.
Least cost of construction [from Eq. (iii)]
$=Rs[280+180(2+\frac{4}{2}\left)\right]=Rs[280+720]=Rs1000$