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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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The cost of construction for an open tank is Rs. 200 per sq. unit for the base, and the cost for doing the walls is Rs. 150 per sq. unit. An open tank is to be constructed with rectangular base of height 2 units and volume of 8 cubic units.What is the cost of the least expensive open tank given these constraints?

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Toolbox:
  • Volume=$2bh$
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
  • Maxima & Minima $\Rightarrow f'(x)=0$
Step 1:
Let the length and breadth of the tank be $x$  and $y$ units respectively.The depth of it is 2 units
Volume of tank=$2\times x\times y$
$\qquad\qquad\quad\;\;=2xy$
Volume=$8 units^3$
$\Rightarrow 2xy=8$
$xy=4$-----(1)
Step 2:
Area of base=$xy$
Area of sides=$2.2(x+y)$
$\qquad\qquad\;=4(x+y)$
Cost of construction=Rs$[200xy+150\times 4(x+y)]$
$\qquad\qquad\qquad\;\;=Rs[200xy+600(x+y)]$
$\qquad\qquad\qquad\;\;=Rs[200xy+600(x+y)]$-----(2)
Step 3:
Put the value of $y$ in (2) from (1) we have
$xy=4$
$y=\large\frac{4}{x}$
$C=200\times4+600\times(x+\large\frac{4}{x})$
$\;\;\;=800+600(x+\large\frac{4}{x})$
Step 4:
Differentiating with respect to $x$ we get,
$\large\frac{dc}{dx}$$=0+600(1-\large\frac{4}{x^2})$
$\quad\;=600(\large\frac{x^2-4}{x^2})$
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 $2 units$
Step 6:
Least cost of construction=Rs$[800+600(2+\large\frac{4}{2})]$
$\qquad\qquad\qquad\qquad\;\;=Rs[800+2400]$
$\qquad\qquad\qquad\qquad\;\;=Rs.3200$

 

answered Dec 23, 2013 by balaji.thirumalai
edited Dec 24, 2013 by rvidyagovindarajan_1
 

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