- #1
courtrigrad
- 1,236
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Hello all
For this problem:
3. A manufacturer wishes to construct a cylindrical can to hold 100 pi in^3. The top and bottom of the can are to be stronger than the sides. The tin used in making the top and bottom will cost 2.5 cents per square inch while the metal used in making the sides will cost 1.35 cents per square inch. What dimensions should be used to minimize the cost of the metal?
pi*r^2 = 100pi
Area of sides = 2pi*r*h
Area of top and bottom: 2pi*r^2
h = 100/ r^2
2pi*r(100/r^2) + 2*pi*r^2
Now do I multiply the cost of the sides by the coefficients?
Any help is appreciated
Thanks!
courtrigrad is online now Edit/Delete Message
For this problem:
3. A manufacturer wishes to construct a cylindrical can to hold 100 pi in^3. The top and bottom of the can are to be stronger than the sides. The tin used in making the top and bottom will cost 2.5 cents per square inch while the metal used in making the sides will cost 1.35 cents per square inch. What dimensions should be used to minimize the cost of the metal?
pi*r^2 = 100pi
Area of sides = 2pi*r*h
Area of top and bottom: 2pi*r^2
h = 100/ r^2
2pi*r(100/r^2) + 2*pi*r^2
Now do I multiply the cost of the sides by the coefficients?
Any help is appreciated
Thanks!
courtrigrad is online now Edit/Delete Message