Question
A cylinder shaped can needs to be constructed to hold 300 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 0.07 cents per square centimeter. Find the dimensions for the can that will minimize production cost.
Helpful information:
h : height of can, r : radius of can
Volume of a cylinder: V=πr^2h
Area of the sides: A=2πrh
Area of the top/bottom: A=πr^2
To minimize the cost of the can:
Radius of the can: ___
Height of the can: ___
Minimum cost: ___cents
Helpful information:
h : height of can, r : radius of can
Volume of a cylinder: V=πr^2h
Area of the sides: A=2πrh
Area of the top/bottom: A=πr^2
To minimize the cost of the can:
Radius of the can: ___
Height of the can: ___
Minimum cost: ___cents
Answers
cost = .07(top + bottom) + .04(side)
= .07(2πr^2) + .04(2π r h)
now we know that πr^2 h = 300
h = 300/(πr^2)
cost = .14π r^2 + .08π r (300/πr^2)
= .14π r^2 + 24/r
d cost / dr = .28π r - 24/r^2 = 0 for a min of cost
.28π r^3 = 24
r^3 = 24 / .28π = 27.2837....
r = 27.2837....^(1/3) = 3.04047...
h = 10.5366...
round to whatever accuracy you need
plug r = .... into the cost equation
cost = .14π r^2 + 24/r
= .07(2πr^2) + .04(2π r h)
now we know that πr^2 h = 300
h = 300/(πr^2)
cost = .14π r^2 + .08π r (300/πr^2)
= .14π r^2 + 24/r
d cost / dr = .28π r - 24/r^2 = 0 for a min of cost
.28π r^3 = 24
r^3 = 24 / .28π = 27.2837....
r = 27.2837....^(1/3) = 3.04047...
h = 10.5366...
round to whatever accuracy you need
plug r = .... into the cost equation
cost = .14π r^2 + 24/r
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