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Differentiation and it's Applications — Applied Maths STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceApplied MathsDifferentiation and it's Applications3 Marks
Question
A tank with rectangular base and rectangular sides open at the top is to be constructed so that its depth is 3 m and volume is $75 m^3$. If building of tank costs ₹ 100 per square metre for the base and ₹ 50 per square metre for the sides, find the cost of least expensive tank.
✓

Answer

Let 1, b, h be the length, breadth and depth of the tank, respectively.
$\therefore \quad 1 \times b \times 3=75$
$\Rightarrow 1 \times b=25$
Let C be the cost, then
$C=100(l \times b)+50 \times 2[h(b+l)]$
$=100\left(l \times \frac{25}{l}\right)+300\left(\frac{25}{l}+l\right)$
$=2500+300\left(\frac{25}{l}+l\right)$
Differentiating w.rt. l,
$\frac{d C}{d l}=0+300\left(\frac{-25}{l^2}+1\right)$
For maximum and minimum cost,
$\frac{d C}{d l}=0$
$\Rightarrow \quad 300\left(-\frac{25}{l^2}+1\right)=0$
$\Rightarrow \quad l^2=25$ or $l= \pm 5$
Getting $\left(\frac{d^2 C}{d l^2}\right)=300\left(\frac{50}{l^3}\right)$
$\Rightarrow\left(\frac{d^2 C}{d l^2}\right)_{a t l=5}=\frac{15000}{125}>0$
i.e., C is minimum, when l = 5
$\Rightarrow \quad b=5$
$\therefore \quad C=100(25)+300(10)$
= 2 ,500+3,000
= 5 ,500
Hence, the minimum cost is ₹5,500.

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