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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is $2 m$ and volume is $8 m^3$. If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?

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Toolbox:
  • Volume=$2bh$
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
  • Maxima & Minima $\Rightarrow f'(x)=0$
Step 1:
Let the length and breadth of the tank be $x$ metre and $y$ metro respectively.The depth of it is 2m
Volume of tank=$2\times x\times y$
$\qquad\qquad\quad\;\;=2xy$
Volume=$8m^3$
$\Rightarrow 2xy=8$
$xy=4$-----(1)
Step 2:
Area of base=$xy$
Area of sides=$2.2(x+y)$
$\qquad\qquad\;=4(x+y)$
Cost of construction=Rs$[70xy+45\times 4(x+y)]$
$\qquad\qquad\qquad\;\;=Rs[70xy+180(x+y)]$
$\qquad\qquad\qquad\;\;=Rs[70xy+180(x+y)]$-----(2)
Step 3:
Put the value of $y$ in (2) from (1) we have
$xy=4$
$y=\large\frac{4}{x}$
$C=70.4+180(x+\large\frac{4}{x})$
$\;\;\;=280+180(x+\large\frac{4}{x})$
Step 4:
Differentiating with respect to $x$ we get,
$\large\frac{dc}{dx}$$=0+180(1-\large\frac{4}{x^2})$
$\quad\;=180(\large\frac{x^2-4}{x^2})$
Step 5:
For maxima or minima $\large\frac{dc}{dx}$$=0$
$x^2-4=0$
$x^2=4$
$x=\pm 2$
$\large\frac{dc}{dx}$ changes sign from -ve to +ve at $x=2$
$\therefore c$ is maximum at $x=2$[length of the tank cannot be negative]
$\Rightarrow x=2$ and $y=\large\frac{4}{x}$
$y=\large\frac{4}{2}$$=2$
Thus tank is a cube of side $2m$
Step 6:
Least cost of construction=Rs$[280+180(2+\large\frac{4}{2})]$
$\qquad\qquad\qquad\qquad\;\;=Rs[280+720]$
$\qquad\qquad\qquad\qquad\;\;=Rs1000$
answered Aug 11, 2013 by sreemathi.v
edited Dec 22, 2013 by balaji.thirumalai
 

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