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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 be constructed so that its depth is $2m$ and volume is $8m^3.$ If building of tank costs Rs. $70$ per square metre for the base and Rs.$ 45$ per square metre for the sides, what is the cost of least expensive tank?
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Answer

Let $l, b $and h be the length, breadth and height of the tank, respectively.
Height,$ h = 2m$
Volume of the tank $= 8m^3$
Volume of the tank $= l \times b \times h$
$\therefore\ \text{l}\times\text{b}\times2 = 8$
$\Rightarrow\text{lb}=4$
$\Rightarrow\text{b}=\frac{4}{\text{l}}$
Area of the base $= lb = 4m^2$​​​​​​​
Area of the 4 walls, $A = 2h(l + b)$
$\therefore\text{A}=4\Big(\text{l}+\frac{4}{\text{l}}\Big)$
$\Rightarrow\frac{\text{dA}}{\text{dl}}=4\Big(1-\frac{4}{\text{l}^2}\Big)$
For maximum or minimum values of A, we must have
$\frac{\text{dA}}{\text{dl}}=0$
$\Rightarrow4\Big(1-\frac{4}{\text{l}^2}\Big)=0$
$\Rightarrow\text{l}=\pm2$
However, the length cannot be negative
Thus,
$\text{l}=2\text{m}$
$\therefore\text{b}=\frac{4}{2}=2\text{m}$
Now,
$\frac{\text{d}^2\text{A}}{\text{dl}^2}=\frac{32}{\text{l}^3}$
$\text{At l}=2:$
$\frac{\text{d}^2\text{A}}{\text{dl}^2}=\frac{32}{8}=4>0$
Thus, the area is the minimum when l = 2m
We have
$l = b = h = 2m$
$\therefore\ $Cost of building the base $= Rs. 70 \times (lb) = Rs. 70 \times 4 = Rs. 280$
Cost of building the walls $= Rs. 2h(l + b) \times 45 = Rs. 90(2)(2 + 2) = Rs. 8(90) = Rs. 720$
Total cost $= Rs. (280 + 720) = Rs. 1000$
Hence, the total cost of the tank will be $Rs. 1000.$

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