A cylinder has a volume of 300 cubic inches. The top and bottom parts of the cylinder cost $2 per square inch. And the sides of the cylinder cost $6 per square inch. What are the dimensions of the Cylinder that minimize cost based on these constraints?

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Answer 1 · 2015-06-21T10:18:26

The optimum dimensions are #r~~5.2# inch and #h~~3.3# inch

We have to minimize the function #P(r,h)=2pi*r^2*2+2pi*r*h*6#
#P(r,h)=4pir^2+12pi*r*h#
#P(r,h)=4*pi*r(r+3h)#

To get rid of one of the variables we use the fact, that the volume of the cylinder is given.

#V=pi*r^2*h# and #V=300#, so #300=pi*r^2*h#, so we can calculate, that:

#h=V/(pi*r^2)#

#h=300/(pi*r^2)#

Now we can write #P# as a function of only one variable:

#P(r)=4*pi*r*(r+3*(300/(pi*r^2)))#

#P(r)=4*pi*r*(r+900/(pi*r^2))#

#P(r)=4pi*r^2+3600/r#

Now to find the value of #r# with tht minimal price #P(r)# we have to calculate #P'(r)#

#P'(r)=8pi*r+((-3600)/r^2)#
#P'(r)=8pi*r-3600/r^2#

Now we have to find #r# for which #P'(r)=0#

#8pir-3600/r^2=0#

#(8pir^3-3600)/r^2=0#

#8pir^3-3600=0#

#r^3=3600/(8pi)#

#r~~5.2# inch

Now we have to calculate #h# using the formula:

#h=300/(pir^2)#

#h~~3.3# inch.

So finally we get the dimensions with the minimal price:
#r~~5.2# inch and #h~~3.3# inch.