Calculus BC - Optimization

Question

Find the dimensions of a cylinder with volume of 12π and minimal surface area.

Answer 1

V=πR²h

12π=πR²h

R²h=12 so h = 12/R²

Surface area

S = 2πRh + 2πR²

Substitute h with 12/R²

S = 24π/R + 2πR²

To find the minimum value, take a derivative and make it =0

S'= - 24π/R² + 4πR

S'=0

- 24π/R² + 4πR=0

4πR=24π/R²

R³=6

R = cube root(6) - critical point.

Since it's the only critical point, it has to be the local minimum

Plugging into S

S = 24π/cube root(6) + 2π cube root(6)² Find common denominator:

S = (24π +2π *6)/cube root(6)

S=36π/cube root(6) Multiplying both top and bottom by cube root(6²)

S=36π * cube root(36) /6

S=6π * cube root(36)