A tank with rectangular base and rectangular sides, open at the top is to the constructed so that its depth is $2 \mathrm{~m}$ and volume is $8 \mathrm{~m}^{3}$. If building of tank cost 70 per square metre for the base and Rs 45 per square matre for sides, what is the cost of least expensive tank?
Let $l, b$ and $h$ be the length, breadth and height of the tank, respectively.
Height, $h=2 \mathrm{~m}$
Volume of the tank $=8 \mathrm{~m}^{3}$
Volume of the tank $=l \times b \times h$
$\therefore l \times b \times 2=8$
$\Rightarrow l b=4$
$\Rightarrow b=\frac{4}{l}$
Area of the base $=1 b=4 \mathrm{~m}^{2}$
Area of the 4 walls, $A=2 h(l+b)$
$\therefore A=4\left(l+\frac{4}{l}\right)$
$\Rightarrow \frac{d A}{d l}=4\left(1-\frac{4}{l^{2}}\right)$
For maximum or minimum values of $A$, we must have
$\frac{d A}{d l}=0$
$\Rightarrow 4\left(1-\frac{4}{1^{2}}\right)=0$
$\Rightarrow l=\pm 2$
However, the length cannot be negative.
Thus,
$I=2 \mathrm{~m}$
$\therefore b=\frac{4}{2}=2 \mathrm{~m}$
Now,
$\frac{d^{2} A}{d l^{2}}=\frac{32}{l^{3}}$
$\mathrm{At} 1=2:$
$\frac{d^{2} A}{d l^{2}}=\frac{32}{8}=4>0$
Thus, the area is the minimum when $/=2 \mathrm{~m}$
We have
$l=b=h=2 \mathrm{~m}$
$\therefore$ Cost of building the base $=$ Rs $70 \times(l b)=$ Rs $70 \times 4=$ Rs 280
Cost of building the walls = Rs $2 h(l+b) \times 45=$ Rs $90(2)(2+2)=$ Rs $8(90)=$ Rs 720
Total cost $=$ Rs $(280+720)=$ Rs 1000
Hence, the total cost of the tank will be Rs 1000 .
Disclaimer: The solution given in the book is incorrect. The solution here is created according to the question given in the book.