Question

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.)

Answer 1

Answer:

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