Of all the closed cylindrical cans (right circular), which enclosed a given volume 100 cubic centimeters, find the dimensions of the minimum surface area.
Let r cm be the radius of base and h cm be the height of the cylindrical can. Let its volume be V and S its total surface area. Then, V=100cm3
⇒πr2h=100 (given)
⇒h=100πr2 …(i)
Also, S=2πr2+2πrh …(ii)
In Eq. (ii), there are two independent variables r and h. We eliminate h, by putting the value of h in Eq. (ii) from Eq. (i), we get
S=2πr2+2πr(100πr2)=2πr2+200r …(iii)
On differentiating Eq. (iii) w.r.t.r, we get dSdr=4πr−200r2 …(iv)
Now, for maxima or minima put dSdr=0⇒4πr=200r2
⇒r3=2004π⇒r=(50π)13
On differentiating Eq. (iv) w.r.t. r, we get d2Sdr2=4π+400r3
AT r=(50π)13,d2Sdr2=400(50π)+4π=40050×π+4π=8π+4π=12π>0
∴ By second derivative test. the surface area is minimum when the radius of the cylinder is (50π)13.
Hence, S is minimum when r=(50π)13
On putting value of r is Eq. (i), we get
h=100π(50π)23=2(50π)(50π)−23=2(50π)13 cm