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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
A tank with rectangular base and rectangular sides, open at the top is to the constructed so that its depth is $2m$ and volume is $8m^3.$ 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?
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Answer

Let $l, b,$ and $h$ repersent the length, breadth, and height of the respectively.
Then, We have height $(h) = 2m^3$
Volume of the tank $= 8m^3$
Volume of the tank $= l \times b \times h$
$\therefore 8 = l \times b \times 2$
$\Rightarrow lb = 4$
$\Rightarrow \frac{4}{\text{l}}$
Now, area of the base$ \Rightarrow lb = 4$
Area of the Walls $(A) = 2h(l + b)$
$\therefore \text{A}=4\Big(\text{l}+\frac{4}{\text{l}}\Big)$
$\therefore \frac{\text{dA}}{\text{d}\text{l}}=4\Big(l-\frac{4}{\text{l}^{2}}\Big)$
Now, $\frac{\text{dA}}{\text{d}l}=0$
$\Rightarrow 1-\frac{4}{l^{2}}=0$
$\Rightarrow \text{l}^{2}=4$
$\Rightarrow \text{l}= \pm2$
However, the length cannot be negative.
Therefore, We have $ l = 4$
$\therefore \text{b}=\frac{4}{l}=\frac{4}{2}=2$
Now, $\frac{\text{d}^{2}\text{A}}{\text{d}l^{2}}=\frac{32}{l^{3}}$
When,$ l = 2, \frac{\text{d}^{2}\text{A}}{\text{d}l^{2}}=\frac{32}{8^{3}} =4>0$
Thus, by second derivative test, the area is the minimum, when l = 2.
We have $l = b = h = 2.$
$\therefore$ Cost of building the base $= Rs.\ 70 \times (lb) = Rs.\ 70 \times (4) = Rs.\ 280$
Cost of building the walls $= Rs. 2h(l + b) \times 45 = Rs. 90 \times (2) (2 + 2)$
$= Rs. 8(90) = Rs. 720$
Required total cost $= Rs. (280 + 720) = Rs. 1000$
Hence, the total cost of the tank will be $Rs. 1000.$

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