Answer to Question #311829 in Calculus for Nur

Question

A cylindrical container has one end opened and the other end closed. It has a circular base of radius r cm. Given that the total surface area of the container is 200π cm2.

(a) Show that the volume of the container is V = 100πr - πr^3/2 cm3

(b) Find the maximum value of V.

Answer 1

Let height of the cylindrical container be h cm

So πr² + 2πrh = 200π

=> r² + 2rh = 200

=> h = $\frac{200-r²}{2r}$

So volume of the cylindrical container is , V =

$πr²h=(\frac{200-r²}{2r}) = πr(100-\frac{r²}{2})$

= $(100πr-\frac{πr³}{2})$ cm³

Now we find maximum volume....

$\frac{dV}{dr} = 100π - \frac{3πr²}{2}$

$\frac{dV}{dr} = 0 =>$ $100π - \frac{3πr²}{2}=0$

=> r² = $\frac{200}{3}$

=> r = $\frac{10\sqrt2}{\sqrt3}$

Now $\frac{d²V}{dr²} = - 3πr$

$[\frac{d²V}{dr²} ]_{r=\frac{10\sqrt2}{\sqrt3}}= - 3π*\frac{10\sqrt2}{\sqrt3} <0$

So volume is maximum when r= $\frac{10\sqrt2}{\sqrt3}$

So maximum volume is $(100πr-\frac{πr³}{2})$

= $\frac{π}{2}[200-r²]r$

= $\frac{π}{2}[200-\frac{200}{3}]\frac{10\sqrt2}{\sqrt3}$

= $\frac{π}{2}[\frac{400}{3}]\frac{10\sqrt2}{\sqrt3}$

= 1710.06644 cm³

= 1710.07 cm³ ( Correct up to two decimal places)