If you solve this problem generically, you find that the cost of the lateral area of the can will end up being double the cost of the end area of the can.* The ratio of lateral area to end area is h/r so this relation tells you
(0.04h)/(0.05r) = 2
h = 2.5r
Then the radius can be found from the volume equation
V = πr^2*h = 2.5πr^3 = 500 cm^3
r = ∛(200/π) cm ≈ 3.993 cm
h = 2.5r ≈ 9.982 cm
The can is about 4 cm in radius and 10 cm high.
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* In the case of a rectilinear shape, the costs of pairs of opposite sides (left-right, front-back, top-bottom) are all the same in the optimum-cost design. It is not too much of a stretch to consider the lateral area of the cylinder to be the sum of left-right and front-back areas, hence twice the cost of the top-bottom area.
(0.04h)/(0.05r) = 2
h = 2.5r
Then the radius can be found from the volume equation
V = πr^2*h = 2.5πr^3 = 500 cm^3
r = ∛(200/π) cm ≈ 3.993 cm
h = 2.5r ≈ 9.982 cm
The can is about 4 cm in radius and 10 cm high.
_____
* In the case of a rectilinear shape, the costs of pairs of opposite sides (left-right, front-back, top-bottom) are all the same in the optimum-cost design. It is not too much of a stretch to consider the lateral area of the cylinder to be the sum of left-right and front-back areas, hence twice the cost of the top-bottom area.