Question

Prove that between cylinder and cube ,cube has greater surface area.

Answer 1

Let $V_1=$ Volume of cube $=a^3$

Let $S_1=$ Surface area of cube $={6a}^2$

Let $V_2=$ Volume of cylinder $=\pi r^{2}h$

let $S_2=$ Surface area of cylinder $=2\pi rh$

Where $r$ is radius and $h$ is height

Assume volume $V_1=V_2$

then $a^3=\pi r^{2}h$

$a=\frac{1}{3}\pi r^{2}h$ $a^2=\frac{2}{3}\pi r^{2}h$

Now $S_1={6a}^2=6\left(\frac{2}{3}\pi r^{2}h\right)$ and $S_2=2\pi rh$

Comparing $S_1$ and $S_2$

We have $6\left(\frac{2}{3}\pi r^{2}h\right)$ compared to $2\pi rh$

Dividing both sides by $6$ and cubing both sides

${\left(\pi r^{2}h\right)}^2$ compared to ${\left(\frac{2\pi rh}{3}\right)}^3$

Dividing right by left

$1$ compared to $\left(\frac{\pi }{27}\right)\times \left(\frac{h}{r}\right)=0.11\times \left(\frac{h}{r}\right)$

If $\frac{h}{r}>=9$ then $S_2>S_1$ if $h/r<9$ then $S_1>S_2$.