A silo is a storage building shaped like a cylinder with half a sphere on top. A farmer has 1000 square feet of material with which to build a silo.

Question

a. The formula for surface area is S=4(pie)r^2+2(pie)rh where r is the radius of the cylinder and h is the height of the cylinder. Substitute 1000 for S and solve for h.

 

b. The formula for the volume of the silo is V=(2/3)(pie)r^3+(pie)r^2h. Write an equation for the volume V of the silo as a polynomial function of r by substituting the expression for h from part (a) into the volume formula.

 

c. Use the function from part (b) to find te maximum volume of a silo with a surface area of 1000 square feet. What are the silo's dimensions.

Answer 1

a)

 

S = 4pi*r^2 + 2pi*r*h

 

1000 = 4pi*r^2 + 2pi*r*h

 

1000 - 4pi*r^2 = 2pi*r*h

 

h = (1000 - 4pi*r^2)/(2pi*r)

 

b)

 

V = 2/3*pi*r^3 + pi*r^2*h

 

h = (1000 - 4pi*r^2)/(2pi*r)

 

V = 2/3*pi*r^3 + pi*r^2*(1000 - 4pi*r^2)/(2pi*r)

 

V = 2/3*pi*r^3 + 1/2*r*(1000 - 4pi*r^2)

 

V = 2/3*pi*r^3 + 500*r -2*pi*r^3

 

V = -4/3*pi*r^3 + 500*r

 

c)

 

In order to find the maximum volume, we need to find dV/dr and set it equal to 0.

 

 

V' = -4*pi*r^2 + 500 = 0

 

r^2 = 125/pi

 

r = 5*√(5)/√(pi), -5*√(5)/√(pi)

 

Since we can't have a radius of negative length, we can disregard the second solution.

 

Your optimal radius will be 5*√(5)/√(pi)

 

Since the question asked for the optimal volume, not the optimal radius, we need to plug your optimal r into your volume equation.

 

 

 

V = -4/3*pi*r^3 + 500*r

 

V = -4/3*pi*(125/pi)*5*√(5)/√(pi) + 500*5*√(5)/√(pi)

 

After some cancelation/ combining of terms:

 

V = -2500/3*√(5)/√(pi) + 2500*√(5)/√(pi)

 

V = 5000/3*√(5)/√(pi) cubic feet.

 

This is the maximum volume you can achieve with the allotted surface area.