Find the dimensions of a cylinder of volume 61m^3 of minimal cost if the top and bottom are made of material that costs twice as much as the material for the

Question

Find the dimensions of a cylinder of volume 61 cubic meters of minimal cost if the top and bottom are made of material that costs twice as much as the material for the side.

 

r= _____m

h= _____m

Answer 1

In order to optimize you need to set up an equation that represents the total cost, then find when the derivative is equal to zero

 

Surface area of the top and bottom = 2*pi*r²

Surface area of the sides = 2*pi*r*h

let's say the cost of material per square meter for the sides is "c" and for the top and bottom is "2c"

 

Therefore the total cost is given by: (2*pi*r²)*2c + (2*pi*r*h)*c

 

Volume of cylinder = pi*r²*h = 61 m³ -->if you rearrange and solve for h, you get h = 61/(pi*r²)

substitute this in for the cost equation, and you get: 

 

total cost  = 4*pi*c*r² + 2*pi*c *(61/pi*r)

 

if you find the derivative with respect to r, you get: cost'(r) = 8*pi*cr - 122cr^-2

set this equal to zero to find the minimum: 

 

8*pi*cr = 122cr^-2

r³ = 122/(8*pi)

r is about equal to 4.85 m

 

plug that back into the equation: h = 61/(pi*r²) and you get that h is about equal to 0.824 m