$\begin{array}{1 1} 3200 \\ 2400 \\ 800 \\ 1600 \end{array} $

- 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$[800+600(2+\large\frac{4}{2})]$

$\qquad\qquad\qquad\qquad\;\;=Rs[800+2400]$

$\qquad\qquad\qquad\qquad\;\;=Rs.3200$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...