Calculus optimization scenario

"A metal box company is asked to produce cylindrical metal tins, each of volume 535 cm^3. The base and top of each tin have to be thicker than the material used for the wall. This material costs twice as much per cm^2 as the thinner material. Find, in cm and correct to one decimal place, the base radius and height of each tin for the cost to be at a minimum"

Dunno how to do this, help me! Exams in 2 weeks and teacher is sick

Derive a formula for the cost:

Let the cost of the wall material be the unit cost, The wall cost is then the wall area,

Aw = 2*π*R*h

The base and top cost = 2 and each cost is 2*base area

Bw =2*π*R²

The cost for base and top is then 2*Bw = 4*π*R²

The total cost is then

T = 2*π*R*h + 4*π*R²

The volume V of the cylinder is given: V = π*R²*h

The substitute for either h using h = V/(π*R²) or R using R = √[V/π*h] to get the cost equation in terms of a single variable (either R or h). Then take the derivative of the cost with respect to that variable and set it to 0. Solve for the variable, then find the other.

Example: use h = V/(π*R²); then T = 2*π*R*V/(π*R²) + 4*π*R²
T = 2*V/R + 4*π*R²

From dT/dR = 0 solve for R

then from V = π*R²*h, get h = V/(π*R²) and use the R found above to get h

Let r = radius of base of cylinder
Let h = height of cylinder

V = 1/3 π r² h = 535
h = 1605/(π r²)

Area of base and top = 2 * π r² cm^2
Cost of material = 2k / cm^2
Cost for base and top = 4kπ r²

Lateral area = 2π r h cm^2
Cost of material = k / cm^2
Cost for lateral surface = 2kπ r h = 2kπ r * 1605/(π r²) = 3210k/r

Now we express cost as a function of r:
C(r) = 4kπ r² + 3210k/r

C'(r) = 8kπr - 3210k/r²

C'(r) = 0
8kπr - 3210k/r² = 0
8kπr = 3210k/r²
8kπr³ = 3210k
r³ = 3210k/(8kπ) = 1605/(4π)
r = ∛(1605/(4π)) ≈ 5.036030955

h = 1605/(π r²)
h = 1605/(π * 5.036030955²) ≈ 20.144123817

Base radius = 5.0 cm
Height = 20.1 cm

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