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?

1 Answer
Jun 21, 2015

The optimum dimensions are r~~5.2r5.2 inch and h~~3.3h3.3 inch

Explanation:

We have to minimize the function P(r,h)=2pi*r^2*2+2pi*r*h*6P(r,h)=2πr22+2πrh6
P(r,h)=4pir^2+12pi*r*hP(r,h)=4πr2+12πrh
P(r,h)=4*pi*r(r+3h)P(r,h)=4πr(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*hV=πr2h and V=300V=300, so 300=pi*r^2*h300=πr2h, so we can calculate, that:

h=V/(pi*r^2)h=Vπr2

h=300/(pi*r^2)h=300πr2

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

P(r)=4*pi*r*(r+3*(300/(pi*r^2)))P(r)=4πr(r+3(300πr2))

P(r)=4*pi*r*(r+900/(pi*r^2))P(r)=4πr(r+900πr2)

P(r)=4pi*r^2+3600/rP(r)=4πr2+3600r

Now to find the value of rr with tht minimal price P(r)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.