An open tank with a square bottom, to contain 4000 cubic cm . of liquid, is to be constructed. The…

An open tank with a square bottom, to contain 4000 cubic cm . of liquid, is to be constructed. The dimensions of the tank, so that the surface area of the tank is minimum, are
  1. side of square bottom $=40 \mathrm{~cm}$, height $=10 \mathrm{~cm}$.
  2. side of square bottom $=20 \mathrm{~cm}$, height $=10 \mathrm{~cm}$.
  3. side of square bottom $=10 \mathrm{~cm}$, height $=40 \mathrm{~cm}$.
  4. side. of square bottom $=5 \mathrm{~cm}$, height $=160 \mathrm{~cm}$.

Solution

Let $x$ be the length of the side of square bottom, h be the height, V be the volume and A be the surface area of open tank. Then, $\begin{aligned} & \mathrm{V}=x^2 \mathrm{~h}=4000...(i) \\ & \mathrm{~A}=x^2+4 x \mathrm{~h}...(ii) \end{aligned}$
From (i), $\mathrm{h}=\frac{4000}{x^2}$
Substituting the value of $h$ in (ii), we get $\begin{aligned} & \mathrm{A}=x^2+\frac{16000}{x} \\ \therefore \quad & \frac{\mathrm{dA}}{\mathrm{~d} x}=2 x-\frac{16000}{x^2} \end{aligned}$ $A$ is minimum, if $\frac{d A}{d x}=0$ $\begin{aligned} & \Rightarrow 2 x-\frac{16000}{x^2}=0 \\ & \Rightarrow x^3=8000 \\ & \Rightarrow x=20 \end{aligned}$
Now, $\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{~d} x^2}=2+\frac{32000}{x^3}$ $\Rightarrow\left(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{~d} x^2}\right)_{x=20}=6 \gt 0$ $\therefore \quad \mathrm{A}$ is minimum when $x=20 \mathrm{~cm}$. $\mathrm{h}=\frac{4000}{x^2}=\frac{4000}{400}=10 \mathrm{~cm}$

Asked in: MHT CET 2024 (15 May Shift 2)

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