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Maximizing Oranges

Date: 06/29/2002 at 09:27:37
From: Julie
Subject: maximizing


An orchard has 800 orange trees, each of which yields 120 oranges per 
season. For each new tree that is added to the orchard, the output  
per individual tree decreases by 2 oranges per season. Determine the 
number of trees that would maximize the orchard's output.

I have no idea where to begin!


Date: 06/30/2002 at 00:25:24
From: Doctor Achilles
Subject: Re: maximizing


Hi Julie,

Thanks for writing to Dr. Math.

This is a problem that requires you to use a lot of different skills, 
so it's really tough, including some calculus.  Let's take it step by 
step.

So you already know that you need to maximize the number of oranges.  
Let's call the total number of oranges "n".  So we want to maximize n 
(make it as big as we possibly can).

How do we find n?  In other words, what equation can we use to figure 
out what n is equal to?

Well, n is the total number of oranges.  The way the orchard started 
out, there are 800 trees and each tree yields 120 oranges.  With that 
set up, what is n?  Why, it's just 800 * 120 (the "*" is a 
multiplication symbol).

So we can find what n is with 800 trees, but the problem is trickier 
than that.  We can go changing the number of trees.  Since the number 
of trees is variable (it changes), let's call it "t".

It gets even more complicated though, because not only does the 
number of trees change, so does the yield of oranges per tree.  Let's 
call the yield per tree "y".

Remember how we found n.  We found it by multiplying the number of 
trees by the yield per tree.  So we can write down this equation:

  n = t * y

The values of n, t, and y might change, but the the equation is 
always going to be true.

The next thing to do is figure out how y changes when we change the 
number of trees.  To do that, let's just forget about n for a while 
and try to find an equation that relates y and t to each other.  
Here's what we know:

1)  y = 120 when t = 800

2)  When you add 1 to t, you have to subtract 2 from y.

So: y = 118 when t = 801

And y = 116 when t = 802

etc.

If we wanted to, we could graph the relationship between t and y:

                y

                |
                |
                |
                |
                |
                |
                |
  --------------|------------------ t
                |
                |
                |
                |
                |
                |

Where could we put a point?  Well we could put a point at (800, 120) 
and another at (801, 118) and another at (802, 116).  These points 
will all form a line.  Now, we need to find an equation for the line.

The first step is to find the slope of the line.  To do that, we take 
the rise (change in y) divided by the run (change in t).

       (116 - 120)     -4
  m = ------------- = ---- = -2
       (802 - 800)      2

So the slope is -2.  So our equation will look like:

  y = -2*t + b

...where b is the y-intercept.

To find b, we use the point (800, 120) and plug it in to the equation:

  120 = -2*800 + b

Which simplifies to:

  120 = -1600 + b

or:

  120 + 1600 = b

or:

  1720 = b

So our equation is:

  y = -2*t + 1720

We now have these two equations:

  n = t * y

  y = -2*t + 1720

We can use substitution on y.  In other words, we can substitute

  (-2t + 1720)

for y in the first equation.  That will give us:

  n = t(-2t + 1720)

If we multiply the t over the parentheses, we get:

  n = -2t^2 + 1720t

(The little ^2 means "squared", the "^" is for an exponent.)

So now we need to find the maximum value of the expression:

  -2t^2 + 1720t

There are a few different ways to do this.  Here are two I can think 
of:

1)  The easy way is to use a graphing calculator to graph the 
equation and find the maximum.

2)  Since the equation is quadratic, it will only have one point 
where the slope is equal to zero.  This will be the maximum of the 
function (if it has a maximum) or the minimum of the function (if it 
has a minimum).  You can see the way this works by looking at a graph 
of the equation on a graphing calculator.  The point where the slope 
goes from being positive (sloping up) to being negative (sloping 
down) is the maximum, at that point the slope is zero.

Remember that the way to find the slope of a function is to take the 
derivative.  If you want a quick review of derivatives, check out 
this page in the Dr. Math archive:

  Derivatives
    http://www.mathforum.com/library/drmath/view/53398.html 

The derivative of the function

  -2t^2 + 1720t

is

  -4t + 1720

We want to find a point where the slope is zero (and therefore we 
want a point where the derivative is zero).  So we want to solve this 
equation:

  0 = -4t + 1720

or

  4t = 1720

or

  t = 430

So the magic point is where t = 430 -- in other words, where the
number of trees is equal to 430.

Just to double check, you should go back and figure out how many 
oranges there will be when there are 430 trees.  To do that, first 
figure out how many oranges each tree will be making, then figure out 
what the total number of oranges will be.

Next, check how many oranges will be produced by 431 trees using the 
same procedure.  Then check how many will be produced by 429 trees.  
If 430 is the right answer, then both of these numbers should be 
smaller.

[NOTE: I did NOT check my work, so I could be wrong.]

Hope this helps.  If you have other questions about this or you're 
still stuck on any point, please write back.

- Doctor Achilles, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra
High School Calculus

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