Expert: Paul Klarreich - 12/5/2006

Question
I don't know if you do optimization problems, but I've got a couple
that I canNOT figure out:(
1)Find the dimensions (radius and height) of the right circular
cylinder of greatest volume that can be inscribe in a right circular cone
having a radius of 3 feet and a height of 5 feet.  Hint : Secondary
equation comes from similar triangles.
The only part of the problem i might have is the primary equation,
which i think could be Volume = pi*r^2h, but i don't know if that's even
right.

2) An orchard has an average yield of 25 bushels per tree when there
are at most 40 trees per acre.  When there are more than 40 trees per
acre, the average yield per tree decreases by one-half bushel per tree.  
Find the number of trees per acre that will provide the maximum yield
per acre.  Hint : write yield as a function of t, the number of trees per
acre more than 40.  Also, the primary equation can be set up as a
function of one variable so no secondary is needed.
I'm completely at a loss with this one:(

Answer
Questioner:  Lily
Category:  Calculus
 
Subject:  Classic optimization problems
Question:  I don't know if you do optimization problems, but I've got a couple that I cannot figure out:(

1)Find the dimensions (radius and height) of the right circular cylinder of greatest volume that can be inscribe in a right circular cone having a radius of 3 feet and a height of 5 feet.  Hint : Secondary equation comes from similar triangles.
The only part of the problem i might have is the primary equation, which i think could be Volume = pi*r^2h, but i don't know if that's even right.

2) An orchard has an average yield of 25 bushels per tree when there are at most 40 trees per acre.  When there are more than 40 trees per acre, the average yield per tree decreases by one-half bushel per tree.  
Find the number of trees per acre that will provide the maximum yield per acre.  Hint : write yield as a function of t, the number of trees per acre more than 40.  Also, the primary equation can be set up as a function of one variable so no secondary is needed.
I'm completely at a loss with this one:(
..........................................................
Hi, Lily,

I corrected the spelling on your subject line.  A well-brought-up calculus student such as yourself would never use such language, I'm sure.

1) One of the great classics of the max-min literature.  For this, you need to look up a volume formula and recall some of your high school geometry, as in similar triangles.

As you think about it, you realize that you could have a skinny cylinder.  Then it reaches way up and you get good height.  But it is so skinny that you get lousy volume.  

But if you flatten it way out, you get a nice big radius, but such a fat cylinder won't reach high up.  With a small height, again you get no volume.  So the ideal is somewhere in between.

Yes, the volume of a cylinder is  V = pi r^2 h, but since you want to maximize V, you must express it in terms of one variable.  Therefore you need a relation between r and h to eliminate one of them.  

Your teacher calls that a 'secondary equation'.  I don't like that term -- it really doesn't tell you anything.  

To get this relation, you have to draw the picture --

Make an isosceles triangle with its vertex at the top and draw the altitude.  Something like this:

USE COURIER FONT TO VIEW THIS, OF COURSE.
     A
    / \
   / | \
  /  |  \
 /  5|   \
/    |    \
B-----+-----C
  3  D  3   

This represents a cross-section of the cone.

Now put the cylinder inside it.  You will draw only the cross-section, of course, just as you did for the cone.  (Too small for me -- you will have to use your own paper.)  
You will see that the top of the cylinder hits AC at point E, and the bottom of the cylinder lies along DC, extending to point F.

Now mark:

DF = r, the radius of the cylinder.
EF = h, the height of the cylinder.

Also mark

FC = 3 - r, since  DF + FC = 3

Now you have similar triangles:

CEF ~ CAD, and

EF   AD
-- = --
CF   CD

putting some things in there:

EF(h)     AD(5)
------- = ------
CF(3-r)   CD(3)

OK, get rid of those letters -- they served their purpose.

h     5
--- = ---
3-r    3

Solve:
   5(3 - r)
h = --------
      3

   15 - 5r
h = --------
      3

We're in business.  There's our relation between h and r.  Now we use it to eliminate h in the volume formula, which was:

V = pi r^2 h
           15 - 5r
V = pi r^2 (-------)
              3

   pi r^2(15 - 5r)
V = -----------------
         3

   pi(15r^2 - 5r^3)
V = -----------------
         3
Good. Now we can differentiate that, set the derivative to zero, solve, etc.
dV   pi(30r - 15r^2)
-- = ----------------
dr         3

Set that (just the top, really) equal to zero:

30r - 15r^2 = 0

2r - r^2 = 0
r(2 - r) = 0
That gives  r = 0  (obviously a minimum) and r = 2.  

OK. You can put that back and get your value of h and finish up now.
...........................................................
2) An orchard has an average yield of 25 bushels per tree when there are at most 40 trees per acre.  

When there are more than 40 trees per acre, the average yield per tree decreases by one-half bushel per tree.  

>> I think that sentence should say "decreases by one-half bushel per EXTRA tree."

Find the number of trees per acre that will provide the maximum yield per acre.  

Hint : write yield as a function of t, the number of trees per acre more than 40.  

>> yes, that looks like a good thing to do.

Also, the primary equation can be set up as a
function of one variable so no secondary is needed.
......................................
OK, let's set it up:

Once again, you realize that you could:
A. Leave the trees alone, just planting 40/acre.  They will be happy trees and make lots of oranges each -- 25 bushels each, in fact.  But there are only 40 trees, so...

B. You could plant them really close to each other.  Then there will be a whole lot of trees, but they will be so unhappy that they will make hardly any oranges each.  You end up drinking tomato juice.
Surely the best is somewhere in between.

Let  t = number of trees MORE than 40.
    a = average yield per tree.
    Y = total yield.

What about that 'yield per tree'?  I think that the y.p.t., a = 25 (the original, happiest) minus 1/2 for each extra tree, so  

a = 25 - t/2  

is the number of bushels for 40 + t trees in an acre.

And the total yield, Y is:

Y = number of trees * bushels per tree.

Y = (40 + t)(25 - t/2)

That's it -- you have your function.

Y = 1000 - 20t + 25t - t^2/2

Y = 1000 + 5t - t^2/2

Y' = 5 - t

Set that equal to zero.

5 - t = 0

That gives you t = 5, so you should plant an extra 5 trees per acre.