A grain silo has the shape of a right circular cylinder surmounted by a hemisphere. If the silo is to have a volume of 505π ft3, determine the radius and height of the silo that requires the least amount of material to build. Hint: The volume of the silo is πr2h + 2 3 πr3, and the surface area (including the floor) is π (3r2 + 2rh). (Round your answers to one decimal place.)

radius x = 3 ft

height h = 23,8 ft

Step-by-step explanation:

From problem statement

V (t) = V (cylinder) + V (hemisphere)

let x be radius of base of cylinder (at the same time radius of the hemisphere)

and h the height of the cylinder, then:

V (c) = π*x²*h area of cylinder = area of base + lateral area

A (c) = π*x² + 2*π*x*h

V (h) = (2/3) * π*x³ area of hemisphere A (h) = (2/3) * π*x²

A (t) = π*x² + 2*π*x*h + (2/3) * π*x²

Now A as fuction of x

total volume 505 = π*x²*h + (2/3) * π*x³

h = [505 - (2/3) * π*x³ ] / 2 * π*x

Now we have the expression for A as function of x

A (x) = 3π*x² + 2π*x*h A (x) = 3π*x² + 505 - (2/3) π*x³

Taking derivatives both sides

A' (x) = 6πx - 2πx² A' (x) = 0 6x - 2x² = 0

x₁ = 0 we dismiss

6 - 2x = 0

x = 3 and h = [505 - (2/3) * π*x³]/2 * π*x

h = (505 - 18.84) / 6.28*3

h = 23,8 ft