Want to ask us a question? Click here
Browse Questions
 Ad
0 votes

# 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?

Can you answer this question?

## 1 Answer

0 votes
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
edited Dec 22, 2013

0 votes
0 answers

0 votes
0 answers

0 votes
1 answer

0 votes
0 answers

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer