SOLUTION: A soup can has a volume 54in^3. Determining the height and radius of the can that requires the least amount of cost to construct if the metal cost 50cents per sq inch and the pape

Algebra ->  Algebra  -> Surface-area -> SOLUTION: A soup can has a volume 54in^3. Determining the height and radius of the can that requires the least amount of cost to construct if the metal cost 50cents per sq inch and the pape      Log On

Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

   

Question 169775: A soup can has a volume 54in^3.
Determining the height and radius of the can that requires the least amount of cost to construct if the metal cost 50cents per sq inch and the paper label cost 10 cents per square in.
Thankyou so much.
Answer by Fombitz(13823) About Me  (Show Source):
You can put this solution on YOUR website!
The can is a cylinder.
The amount of metal is the top, bottom, and circular tube portion.
A=A(top)+A(bottom)+A(tube)
A%5Bm%5D=pi%2Ar%5E2%2Bpi%2Ar%5E2%2B2%2Api%2Ar%2Ah
A%5Bm%5D=2%2Api%2Ar%5E2%2B2%2Api%2Ar%2Ah
The amount of paper just covers the circular tube portion.
A%5Bp%5D=2%2Api%2Ar%2Ah
The total cost equation is then,
Cost=Area Cost of Metal*(Metal Area)+Area Cost of Paper*(Paper Area)
C=0.50%2A%28A%5Bm%5D%29%2B0.10%2A%28A%5Bp%5D%29
C=0.50%2A%282%2Api%2Ar%5E2%2B2%2Api%2Ar%2Ah%29%2B0.10%2A%282%2Api%2Ar%2Ah%29
C=pi%2Ar%5E2%2Bpi%2Ar%2Ah%2B0.2%2Api%2Ar%2Ah
C=pi%2Ar%5E2%2B1.2%2Api%2Ar%2Ah
The total equation has both r and h as variables.
.
.
.
The internal volume of the can is fixed.
V=pi%2Ar%5E2%2Ah=54
h=54%2F%28pi%2Ar%5E2%29
We can substitute this expression into the total cost equation and reduce it to a one variable equation.
C=pi%2Ar%5E2%2B1.2%2Api%2Ar%2Ah
C=pi%2Ar%5E2%2B%281.2%2Api%2Ar%2A54%29%2F%28pi%2Ar%5E2%29
C=pi%2Ar%5E2%2B64.8%2Fr
We can graph the function and look for a minimum.
+graph%28+300%2C+300%2C+-2%2C+5%2C+-5%2C+60%2C+3.14%2Ax%5E2%2B64.8%2Fx%29+
Looks like we have a minimum around r=2.
We can find the exact value by taking the derivative of C with respect to r and setting it equal to zero.
dC%2Fdr=2%2Api%2Ar-64.8%2Fr%5E2=0
2%2Api%2Ar=64.8%2Fr%5E2
r%5E3=64.8%2F%282%2Api%29
r%5E3=10.313
highlight%28r=2.177%29
Then from above,
h=54%2F%28pi%2Ar%5E2%29
highlight%28h=3.628%29