Question #311829

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.


1
Expert's answer
2022-03-16T02:22:35-0400

Let height of the cylindrical container be h cm

So πr² + 2πrh = 200π

=> r² + 2rh = 200

=> h = 200r22r\frac{200-r²}{2r}

So volume of the cylindrical container is , V =

πr2h=(200r22r)=πr(100r22)πr²h=(\frac{200-r²}{2r}) = πr(100-\frac{r²}{2})

= (100πrπr32)(100πr-\frac{πr³}{2}) cm³

Now we find maximum volume....

dVdr=100π3πr22\frac{dV}{dr} = 100π - \frac{3πr²}{2}

dVdr=0=>\frac{dV}{dr} = 0 => 100π3πr22=0100π - \frac{3πr²}{2}=0

=> r² = 2003\frac{200}{3}

=> r = 1023\frac{10\sqrt2}{\sqrt3}

Now d2Vdr2=3πr\frac{d²V}{dr²} = - 3πr

[d2Vdr2]r=1023=3π1023<0[\frac{d²V}{dr²} ]_{r=\frac{10\sqrt2}{\sqrt3}}= - 3π*\frac{10\sqrt2}{\sqrt3} <0


So volume is maximum when r= 1023\frac{10\sqrt2}{\sqrt3}

So maximum volume is (100πrπr32)(100πr-\frac{πr³}{2})

= π2[200r2]r\frac{π}{2}[200-r²]r

= π2[2002003]1023\frac{π}{2}[200-\frac{200}{3}]\frac{10\sqrt2}{\sqrt3}

= π2[4003]1023\frac{π}{2}[\frac{400}{3}]\frac{10\sqrt2}{\sqrt3}

= 1710.06644 cm³

= 1710.07 cm³ ( Correct up to two decimal places)


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