Answer: Option B
:
B
Conventional method
Volume of a cylinder, V = $\pi$r${}^2$h ( r = radius and h = height)
Given that r+h=6. Hence, h = 6-r
Hence, V = $\pi$r${}^2$(6-r)
V = 6$\pi$r${}^2$ - $\pi$r${}^3$
For maximizing volume, we need to differentiate V with respect to r and equate it to 0.
$\frac{dv}{dr}$ = 0
$\frac{dv}{dr}$ = 12pr - 3$\pi$r${}^2$= 0
Hence, r = 4
R + h is given as 6, hence h = 2
V = $\pi$r${}^2$h = $\pi$$\times$4${}^2$$\times$2 = 32$\pi$
Alternate Method:
Volume of a cylinder = $\pi$r${}^2$h ( r =radius and h= height)
Given that r+h=6
To maximize $\pi$r${}^2$h, we need to maximize r${}^2$h which happens when $\frac{r}{2}$ = $\frac{h}{1}$,
$\Rightarrow$ r = 4 and h =2
Maximum volume = $\pi$$\times$4${}^2$$\times$2 = 32$\pi$
Points to remember
1. If a+b=constant, ab will be maximum when a=b
2. If ab=constant, the minimum value of a+b will be obtained at a=b
3. If a+b+c is a constant, then a${}^m$. b${}^n$ .c${}^p$ is maximum when $\frac{a}{m}$ = $\frac{b}{n}$ = $\frac{c}{p}$ (the above question is an example of this)