A cylinder shaped can needs to be constructed to hold 500 cubic centimeters of soup. the material for the sides of the can costs
0.04 cents per square centimeter. the material for the top and bottom of the can need to be thicker, and costs 0.05 cents per square centimeter. find the dimensions for the can that will minimize production cost.
1 answer:
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.
_____
* 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.
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Answer:
Step-by-step explanation:
we know that
The volume of a trough is equal to
where
B is the area of equilateral triangle
L is the length of a trough
step 1
Find the area of equilateral triangle B
The area of a equilateral triangle applying the law of sines is equal to
where
substitute
step 2
Find the volume of a trough
we have
substitute