Question

A covered box of volume $72{\text{cm}}^3$ and the base sides in a ratio of $1:2$ is to be made. The length of all sides so that the total surface area is the least possible is

• A. $6,3,4$
• B. $8,4,3$
• C. $2,4,9$
• D. $3,6,2$

Answer 1

The correct option is A $6,3,4$

Let $l,b,h$ be the dimension, $l=2b,$ so $V=lbh=2b^{2}h$
$\Rightarrow 72=2b^{2}h$
$\Rightarrow h=\frac{36}{b^2}$
The surface area $S=2(lb+bh+hl)=2(2b^2+b\left(\frac{36}{b^2}\right)+2b\times \frac{36}{b^2})=2(2b^2+\frac{108}{b})=4(b^2+\frac{54}{b})$
$\frac{dS}{db}=4(2b-\frac{54}{b^2}),\frac{dS}{db}$ is zero if $b=3$ and $\frac{d^{2}S}{d{b}^2}=4(2+\frac{108}{b^3})>0$
Hence, $S$ is minimum when $b=3.$
So, the dimensions are $6,3,\frac{36}{9}=6,3,4$