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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?

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$[280+180(2+\large\frac{4}{2})]$
$\qquad\qquad\qquad\qquad\;\;=Rs[280+720]$
$\qquad\qquad\qquad\qquad\;\;=Rs1000$
edited Dec 22, 2013