Question

A tank with rectangular base and rectangular sides, open at the top is to the constructed so that its depth is 2 m and volume is 8 m³. If building of tank cost 70 per square metre for the base and Rs 45 per square matre for sides, what is the cost of least expensive tank?

Answer 1

Let l, b and h be the length, breadth and height of the tank, respectively.

Height, h = 2 m

Volume of the tank = 8 m³

Volume of the tank = l × b × h

∴ l × b × 2 = 8

$$\begin{gathered} \Rightarrow lb=4 \\ \Rightarrow b=\frac{4}{l} \end{gathered}$$

Area of the base = lb = 4 m²

Area of the 4 walls, A= 2h (l + b)

$$\begin{gathered} \therefore A=4\left(l+\frac{4}{l}\right) \\ \Rightarrow \frac{dA}{dl}=4\left(1-\frac{4}{l^2}\right) \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}A,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dA}{dl}=0 \\ \Rightarrow 4\left(1-\frac{4}{{\mathrm{l}}^2}\right)=0 \\ \Rightarrow l=\pm 2 \end{gathered}$$

However, the length cannot be negative.

Thus,
l = 2 m

$$\begin{gathered} \therefore b=\frac{4}{2}=2\mathrm{m} \\ \mathrm{Now}, \\ \frac{d^{2}A}{d{l}^2}=\frac{32}{l^3} \\ \mathrm{At}\mathrm{l}=2: \\ \frac{d^{2}A}{d{l}^2}=\frac{32}{8}=4>0 \end{gathered}$$

Thus, the area is the minimum when l = 2 m

We have
l = b = h = 2 m

∴ Cost of building the base = Rs 70 × (lb) = Rs 70 × 4 = Rs 280

Cost of building the walls = Rs 2h (l + b) × 45 = Rs 90 (2) (2 + 2)= Rs 8 (90) = Rs 720

Total cost = Rs (280 + 720) = Rs 1000

Hence, the total cost of the tank will be Rs 1000.

Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.