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### Calculus Cylinder-Cone Problem

```
Date: 07/13/99 at 04:53:17
From: Ted Drew
Subject: Calculus Cylinder-Cone Problem (How does one "see" the
simple solutions?)

Based on your archives, I was able to complete the problem below,
determining the relation between x and y as follows (see

Maximizing the Volume of a Cylinder
http://mathforum.org/library/drmath/view/53471.html   ) :

Find the altitide of the cylinder of maximum volume that can be
inscribed in a cone having a radius of 7 and a height of 12.

_
/\                  |
/  \                 |
/    \                |
/      \               |
/________\              |12
/|        |\             |
/ |        | \            |
/  |       y|  \           |
/   |        |   \          |
*******************        _|
|-x-|

|________|
7

(7X12) = 84 = area of the whole big triangle
(7-x)y = combined area of the triangle on the lower sides
(12-y)x = area of the top triangle

The area of the square is going to be the area of the big triangle,
less the areas of the little ones surrounding the square. So,

2xy = 84 - (7-x)y - (12 - y)x

which, solving for y, gives you

y = (84-12x)/7

We put this back into V = pix^2y and find the maximum of it using the
zero point(s) of dV/dx. So that was easy.

Now my real question (but first a little more).

I started this problem with a different tactic for relating x and y.
More specifically, looking at your same triangle diagram, I noted a
Pythagorean.

7^2 + 12^2 = the hypotenus (H) of one side of the big triangle,
or H = 193

_
/\                  |
/  \                 |
/    \ h2             |
/      \               |
/________\  H=193       |12
/|        |\             |
/ |        | \  h1        |
/  |       y|  \           |
/   |        |   \          |
*******************        _|
|-x-|
|________|
7

But then, one should be able to define the hypotenuse segments (h1 and
h2), as follows:

(7-x)^2 + y^2 = (h1)^2
x^2 + (12-y)^2 = (h2)^2

And since the hypotenuse segements add up the the hypotenuse overall,
you should be able to do

(h1)^2 + (h2)^2 = H = 193

[(7-x)^2 + y^2] + [x^2 + (12-y)^2] = 193      (1)

So, if we _could_ isolate y out of (1), then we ought to be able to do
the same plug-in operation into V = pix^2y as with the other simpler
relation above, and get the same answer.

Of course, I got hopelessly bogged down in trying to solve (1) for y,
and because of the squares, I think you even end up with multiple
solutions for y. So I just gave up.

Obviously, _your_ way is a lot easier. But now for my questions:

1.  Would my way be feasible, and assuming you could isolate y from
(1) and plug it into the cylinder volume formula and find the
maxima, would it work?
^^^^^^^^^^^^^
Is there a way of isolating y in (1) and doing it my way?

2.  More philosophically, how do you know when to bail on a proposed
method of solving a problem, and look for a simpler one? I am
getting the idea that something is wrong with me because I did not
"see" your solution for relating x and y first.

I mean, if I had never come across your suggestion for relating
x to y, I might have tried hopelessly and in vain with my approach
for hours on end.

In this case (because it was a workbook problem) I was relatively
certain there was going to be a reasonable way of solving the problem.
But if it had been a problem in the real world, how would I know?
What guidance would I have to tell me whether to keep pursuing a
difficult method or to look for a simpler one? Conversely, how would I
realize that a simpler method just didn't exist, and my only recourse
was to keep working away at the difficult one?
```

```
Date: 07/13/99 at 10:19:14
From: Doctor Rob
Subject: Re: Calculus Cylinder-Cone Problem (How does one "see" the
simple solutions?)

Thanks for writing to Ask Dr. Math!

Your method will work if you use the correct equations. The
Pythagorean Theorem says a^2 + b^2 = c^2.  You seem to be using
a^2 + b^2 = c, instead. The hypotenuse should be sqrt(193), not 193,
and the equation you get using your method would be

sqrt(193) = h1 + h2,
= sqrt((7-x)^2 + y^2) + sqrt(x^2 + (12-y)^2).

If you bring one of the square roots from the right side to the left
and square both sides, you will reduce the number of roots in the
equation to one. Isolate that one one side of the equation and square
again, and you will have a polynomial equation relating x and y. This
turns out to be of the form (12*x+7*y-84)^2 = 0, so you are led back
to the equation 12*x + 7*y - 84 = 0, or

y = (84-12*x)/7

as we saw above. So whichever way you do the algebra, the equation
relating x and y will turn out the same. The rest of the problem is
identical.

When you find the equation 12*x + 7*y - 84 = 0, which is the equation
of a straight line, you should realize that there is a simpler way.
You can identify the line as the diagonal line with negative slope in
the above diagram, if you know coordinate geometry. The two-intercept
form x/7 + y/12 = 1 might be the simplest way to see that.

If you are working from first principles, you should say to yourself,
"What if x = 0? The value of y would have to be 12 then. What if y =
0? The value of x would be 7. How does y vary with x? When x goes up
by X, what happens to y?" This should lead you to the same linear
equation relating x and y, again.

If the problem is from a textbook, probably the methods in the section
immediately preceding it will have some bearing on the solution path
you choose. If the equations start to get more and more complicated,
without any prospect of simplification, then you probably are on the
wrong path. After all, these kind of problems are designed to be
exercises to give you experience in using the methods of that section,
and shouldn't be too terrible in terms of the burden of computation.

If the problem arises in some other context, then you can't have that
kind of confidence when the calculations start to get messy.

In general, when faced with triangles, I try to use similarity and its
accompanying proportionality first. Then I try using the areas.
Finally I try the Pythagorean Theorem. I use this order because the
Pythagorean Theorem usually gives me a square root, which is harder to
deal with than the equations from the other two methods. Sometimes it
is the best way, or even the only way, but I try the other two attacks
first. I may even try starting all three, more or less in parallel,
and see which one looks the most promising after a few steps.

I hope this helps!  If not, write again.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Triangles and Other Polygons

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