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Maxima and Minima — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMaxima and Minima4 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 8m3. 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) = 2m3
Volume of the tank = 8m3
Volume of the tank = l × b × h
$\therefore$ 8 =  l × b × 2
⇒ lb = 4
$\Rightarrow \frac{4}{\text{l}}$
Now, area of the base ⇒ 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 × (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
Required total cost = Rs. (280 + 720) = Rs. 1000
Hence, the total cost of the tank will be Rs. 1000.

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