For a closed cylinder with radius r ⁢ cm and height h ⁢ cm, find the dimensions giving the minimum surface area, given that the volume is 40 cm3.

we know the surface area of the two ends of the closed cylinder is: 2 (r 2).

The surface are of the side of the cylinder is h (2r)

So the total surface area is: 2 (r 2) + h (2 r)

The volume of the cylinder is r 2 h = 40 cm3

now solve this last equation for h:

so, h = 40 / (*r 2)

Substitute the value for h into the expression for the total surface area:

2 (r 2) + (40 / (r 2)) (2 r)

Now Simplify the above expression we get:

2 (r 2) + (80 / r)

Now take the derivative:

4 r - 80 r - 2

Set it equal to zero and solve for r:

4 r - 80 r - 2 = 0

Divide by 4:

*r - 20 r - 2 = 0

Multiply by r2 we get:

r3 - 20 = 0

r3 = 20

r3 = 20/

Hence,

r = (20 / ) (1/3)

r 1.8534 cm

Substitute this value for r into h = 40 / (r 2) and evaluate h:

h = 40 / * (1.8534) 2

h 3.7067