Class 12th Maths Miscellaneous Exercise on Chapter 6 – Question-6

Question 6

A tank with rectangular base and rectangular sides, open at the top, is to be constructed so that its depth is 2 m and volume is 8 m³. If building the tank costs Rs 70 per m² for the base and Rs 45 per m² for the sides, what is the least cost of constructing the tank?

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Solution

Let the rectangular base have dimensions:

$$l=\text{length},w=\text{width},h=\text{depth}=2\text{ m}$$

Given: Volume

$$l\cdot w\cdot h=8$$
$$lw\cdot 2=8\Rightarrow lw=4$$

So:

$$w=\frac{4}{l}$$

Cost calculation

Base area

$$\text{Area}=lw=4{\text{ m}}^2$$

Cost of base:

$$C_{\text{base}}=70\times 4=280\text{ Rs}$$

Area of 4 rectangular side walls

$$\text{Total side area}=2lh+2wh=2l(2)+2\left(\frac{4}{l}\right)(2)=4l+\frac{16}{l}$$

Cost of sides:

$$C_{\text{sides}}=45(4l+\frac{16}{l})=180l+\frac{720}{l}$$

Total Cost

$$C=C_{\text{base}}+C_{\text{sides}}=280+180l+\frac{720}{l}$$

To minimize cost, differentiate $C$ w.r.t. $l$:

$$C^{\mathrm{\prime }}(l)=180-\frac{720}{l^2}$$

Set $C^{\mathrm{\prime }}(l)=0$:

$$180=\frac{720}{l^2}$$

$$l^2=\frac{720}{180}=4\Rightarrow l=2\text{ m}$$

Then:

$$w=\frac{4}{l}=\frac{4}{2}=2\text{ m}$$

Minimum Cost

$$C=280+180(2)+\frac{720}{2}=280+360+360=1000$$

Final Answer

$$\boxed{{\displaystyle \text{The least cost of constructing the tank is Rs 1000.}}}$$