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?

Question

9142 views around the world

You can reuse this answer
Creative Commons License

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.