Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?

Explanation

Let r be the radius of base and h be the height of cylindrical box. Again let S be its surface area and V be its volume. Then `V= 100 cm^(3)` (given)
` rArr pi r^(2) h = 100`
` rArr h= (100)/(pi r^(2))` …(1)
and ` S = 2pi r^(2) + 2 pi rh` ...(2)
put the value of h from equation (1) to equation (2),
` S = 2pi r^(2) + 2 pi r((100)/(pi r^(2))) = 2 pi r^(2) + 200/r ` ...(3)
Differentiate equation (3) w.r.t. r,
` (dS)/(dr) = 4 pi r - 200/r^(2)` ....(4)
Now, for maxima/minima` (dS)/(dr) = 0`
` rArr 4pi r = (200)/r^(2)`
`rArr r^(3) = (200)/(4pi) rArr r=((50)/pi)^(1/3)`
Differentiate equation (4) w.r.t.r,
`(d^(2)S)/(dr^(2)) = 4 pi + (400)/r^(3)`
at ` r = ((50)/pi)^(1/3)`,
`(d^(2)S)/(dr^(2))=400/(((50)/pi))+4pi = 400/50 xx pi + 4 pi`
` = 8 pi + 4 pi = 12pi gt 0 `
`:. ` Surface area is minimum when radius of the cylinder is `(50/pi)^(1/3)`.
`:. r = (50/pi) ^(1/3)`
put the value of r in equation (1)
`h = (100)/(pi(50/pi)^(2/3)) = 2 (50/pi)(50/pi)^((-2)/3)`
` = 2(50/pi)^(2/3)` cm