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Mathematics · 2023 · 4 marks
CBSE 2023 · Region 3 · Set 2 · Q36
In order to set up a rain water harvesting system, a tank to collect rain water is to be dug. The tank should have a square base and a capacity of…
In order to set up a rain water harvesting system, a tank to collect rain water is to be dug. The tank should have a square base and a capacity of $\displaystyle 250 \mathrm{~m}^{3}$. The cost of land is ₹ $\displaystyle 5,000$ per square metre and cost of digging increases with depth and for the whole tank, it is $\displaystyle ₹ 40,000 \mathrm{~h}^{2}$, where h is the depth of the tank in metres. $\displaystyle \mathrm{x}$ is the side of the square base of the tank in metres. ELEMENTS OF A TYPICAL RAIN WATER HARVESTING SYSTEM
Based on the above information, answer the following questions :
(i)
Find the total cost C of digging the tank in terms of x .
(ii)
Find $\displaystyle \frac{\mathrm{dC}}{\mathrm{dx}}$.
(iii)
(a)
Find the value of x for which cost C is minimum.
OR
(b)
Check whether the cost function $\displaystyle \mathrm{C}(\mathrm{x})$ expressed in terms of x is increasing or not, where $\displaystyle \mathrm{x}>0$. Case Study - $\displaystyle 2$
Marking-scheme solution
$$\begin{aligned}
& \text { (i) } \mathrm{C}=$\displaystyle 40000$ \mathrm{~h}^{$\displaystyle 2$}+$\displaystyle 5000$ \mathrm{x}^{$\displaystyle 2$}
& \text { as } \mathrm{x}^{$\displaystyle 2$} \mathrm{~h}=$\displaystyle 250$
& \Rightarrow \mathrm{C}=\frac{$\displaystyle 40000$($\displaystyle 250$)^{$\displaystyle 2$}}{\mathrm{x}^{$\displaystyle 4$}}+$\displaystyle 5000$ \mathrm{x}^{$\displaystyle 2$}
\end{aligned}
$$(ii) $\displaystyle \frac{\mathrm{dC}}{\mathrm{dx}}=\frac{-160000(250)^{2}}{\mathrm{x}^{5}}+10000 \mathrm{x}$
(iii)(a)
For minimum cost $\displaystyle \frac{d \mathrm{C}}{d \mathrm{x}}=0$
$$\Rightarrow $\displaystyle 10000$ \mathrm{x}^{$\displaystyle 6$}=$\displaystyle 250$ \times $\displaystyle 250$ \times $\displaystyle 160000$
$$$\Rightarrow \mathrm{x}=10$
showing $\displaystyle \frac{d^{2} \mathrm{C}}{d \mathrm{x}^{2}}>0$ at $\displaystyle \mathrm{x}=10$
∴ cost is minimum when $\displaystyle \mathrm{x}=10$
$$\begin{aligned}
& \text { (iii)(b) } \frac{d \mathrm{C}}{d \mathrm{x}}=\frac{-$\displaystyle 160000$($\displaystyle 250$)^{$\displaystyle 2$}}{\mathrm{x}^{$\displaystyle 4$}}+$\displaystyle 10000$ \mathrm{x}
& \frac{d \mathrm{C}}{d \mathrm{x}}=$\displaystyle 0$ \text { gives } \mathrm{x}=$\displaystyle 10$
& \frac{d \mathrm{C}}{d \mathrm{x}}>$\displaystyle 0$ \text { in }($\displaystyle 10$, \infty) \text { and } \frac{\mathrm{dC}}{\mathrm{dx}}<$\displaystyle 0$ \text { in }$\displaystyle (0,10)$ .
\end{aligned}
$$Hence, cost function is neither increasing nor decreasing for $\displaystyle \mathrm{x}>0$
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