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.