A cylinder shaped can needs to hold 550 cm^3 of soup. The sides of the can costs 0.03 cents per cm^2. The top and bottom of the can costs 0.06 cents per cm^2. Minimize the production cost.

Question

Find the dimensions for the can that will minimize the production cost.

Radius=?

Height=?

Minimum Cost=?

Answer 1

The surface area (SA) of the can = SA_sides + SA_top/bottom = 2(pi)(r)(h) + 2(pi)(r²)

The Volume = (pi)(r²)(h) = 550 therefore h = 550/[(pi)(r²)] Plugging that into the SA equation for h, we get:

SA(r) = 2(pi)(r)(550/[(pi)(r²)]) + 2(pi)(r²)

The cost is .03 cents for the sides and .06 cents for the top/bottom so the total cost (C) (in cents) is:

C(r) = (0.03)2(pi)(r)(550/[(pi)(r²)]) + (0.06)2(pi)(r²) = 33/r + 0.12(pi)(r²)

To minimize, take the derivative and set it equal to zero.

C'(r) = -33/r² + 0.24(pi)(r)

-33/r² + 0.24(pi)(r) = 0

33/r² = 0.24(pi)(r)

r³ = 43.768

r = 3.524 cm

plugging that back into h = 550/[(pi)(r²)] gives:

h = 14.096 cm

To find the minimum cost, plug r = 3.524 cm into the cost equation C(r) = 33/r + 0.12(pi)(r²) = 14.046 cents

Answer 2

Surface area =2πrh V=550cm³ Cost = .03(2πrh) + .06(2πr²) h= 550/πr² C=.06(550)/r +.12πr²

dC/dr= -33/r² + .24πr and d²C/dr² = 66/r³ + .24r >0 so since the second derivative is positive then there is a minimum. Solving for r gives 0 = -33 + .24πr³ where r=3.5197 cm and then solving for h gives h= 14.13 cm The minimum cost is gotten from the above equation for C which gives appx 14 cents.