Surface area calculus question

a close cylindrical can is to hold 250pi cm^3 of liquid . What dimensions will minimize amount of material needed to produce can?

please show with steps. Many thanks in advance!!

V = Πr^2 h

=> Πr^2 h = 250Π

h = 250/ r^2

Surface area, A = 2Πr^2 + 2Π r h

A = 2Πr (r + h)

A = 2Πr (r + 250/r^2)

A = 2Π r^2 + 500Π/r

differentiate and equate it to zero

=> 4Π r - 500 Π/r^2 = 0

4 r^3 = 500

r^3 = 125

r = 5 cm

h = 250 /r^2

h = 250/25

h = 10 cm

The volume of a cylinder is pi times r^2 times h, in this case =250 pi. Solve for h: h=250 pi/(p r^2); cancel pi. so h=250/r^2. The surface area of a cylinder is 2 pi r h (the sides) + 2 pi r ^2 (the top and bottom). Plug your expression for h in this so you only have one variable. The surface area =2 pi r(250/r^2) +2 pi r^2. simplify: that's 500 pi/r +2 pi r^2. Take the first derivative and set equal to 0. so it's -500 pi/r^2 + 4 pi r. Multiply by r^2 to clear the fraction. -500pi +4 pi r^3 =0. or 4 pi r^3 =500 pi.
Divide out 4 pi and get r^3 =125. so r=5 cm. the height was 250/r^2 so 250/25 =10. The radius should be 5 cm and the height 10 cm.

Assume can has height h and end radius r.

Volume V = pi*(r^2)*h = 250*pi ----> (r^2)*h = 250 ----> h = 250/(r^2)

Area A = 2*pi*r*h (curved side) + 2*pi*r^2 (circular ends)

Substituting h = 250/(r^2) you get

A = 2*pi*r*250/(r^2) + 2*pi*r^2 = 500*pi / r + 2*pi*r^2

You differentiate that to get dA/dr and set it equal to 0. Solve the equation for the value of r which minimises area. From the r value calculate the h value as above. Done correctly you will find that the cylinder has equal height and diameter (not radius).