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Answer:
Step-by-step explanation:
The cost of the ends of the can will be ...
c1 = 0.07(2πr²)
The cost of the side of the can will be ...
c2 = 0.04(2πrh)
The volume of the can will be ...
v = πr²h
We want the derivative of the total cost to be zero, and we want the volume to be 550 cm³. We can take the derivatives of both equations to find a relation between r and h.
d(c1 +c2) = 0.28πr·dr +0.08π(h·dr +r·dh) = 0
d(v) = 2πrh·dr +πr²·dh = 0
Solving the first equation for dh/dr gives ...
dh/dr = -π(0.28r +0.08h)/(π(0.08r)) = -(7r+2h)/(2r)
Solving the second equation for dh/dr gives ...
dh/dr = -2πrh/(πr²) = -2h/r
Equating these expressions, we get ...
-(7r +2h)/(2r) = -2h/r
7r +2h = 4h . . . . . . . multiply by -2r
h = 7/2r . . . . . . . . . . subtract 2h, divide by 2
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Now, we can find the can dimensions from the volume equation.
550 = πr²(7/2r)
r³ = 1100/(7π)
r ≈ ∛50.02 ≈ 3.685 . . . . . cm
h = 7/2(3.685 cm) = 12.896 cm
The can cost will be a minimum when the radius is 3.685 cm and the height is 12.896 cm.
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Additional comment
You may notice that the ratio of height to diameter is the same as the ratio of end cost to side cost: 7/4. This is the generic solution to this sort of problem.
