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Minimum and Maximum Problems Using Algebra


Date: 04/05/2000 at 16:06:47
From: Hadila
Subject: Algebra 2 word problems

Hello,

I tried, but these math teasers I cannot get. Please help. Thanks!

1) A wire 40 cm long is cut into two pieces, and each piece is bent 
into a square. Where should the wire be cut if the total area of the 
two squares is to be a minimum? (For this I got 20?)

2) A rectangular field adjacent to the straight bank of a river is to 
be fenced, but there is to be no fencing along the riverbank. If 180 m 
of fencing is available, what is the maximum area that can be 
enclosed?

3) A limousine shuttle service operating between an airport and the 
center of a city charges a fare of $10.00 and carries 300 persons per 
day. The firm estimates that business will decrease by 15 passengers 
per day for each increase of $1.00 in the fare. Find the most 
profitable fare to charge for the service. (I know that it will be the 
maximum on the parabola, but don't know where to start to get there.)

4) Four hundred people will attend a sidewalk display if tickets cost 
$1.00 each. Attendance will decrease by 20 people for each 10-cent 
increase in the ticket price. What ticket price will yield the maximum 
income, and what is the maximum income from the ticket sales?

Thank you very much for your help. We students need this support and 
as one of the students, I appreciate it!


Date: 04/06/2000 at 11:11:52
From: Doctor Jeremiah
Subject: Re: Algebra 2 word problems

Hi Hadila.

Word problems are the hardest thing, aren't they?

1) Let x = the length of the first piece of wire after its cut. Let 
y = the length of the second piece of wire after its cut.

Then x+y = 40 cm because before it was cut, the wire was 40 cm.

The perimeter of the square made by piece one is x (because the wire 
is bent into a square). The length of the side of this square is x/4 
and the area of the square made by bending the first piece into a 
square is (x/4)^2.

Since the second piece is also bent into a square the area of that 
square will be (y/4)^2.

So the total area will be A = (x/4)^2 + (y/4)^2

Now we have two equations:

         A = (x/4)^2 + (y/4)^2
     x + y = 40 

If we solve x+y = 40 for y we get:

     x + y = 40 
         y = 40 - x

And if we substitute y = 40 - x into A = (x/4)^2 + (y/4)^2 we get:

     A = (x/4)^2 + (y/4)^2
       = (x/4)^2 + ((40 - x)/4)^2
       = (x/4)^2 + (40/4 - x/4)^2

Then, if we rearrange this to have the form of a quadratic

     0 = a*x^2 + b*x + c

then we can easily figure out what the maximum area is. And if we know 
the maximum value of the area we can use the original equations to get 
the values of x and y.

I will let you solve the rest of this. The maximum area is 50 square 
centimeters. If you get stuck again mail me back.


2)
       |      x
       |-------------+
       |             |
     R |             |
     I |             |
     V |             | y
     E |             |
     R |             |
       |             |
       |-------------+
       |      x


So the three pieces of fence are 180 meters long x+y+x = 180, and the 
area inside the fence is A = x*y

Now we just have to do the same thing as last time:

     x+y+x = 180
      2x+y = 180
         y = 180 - 2x

     A = x*y
       = x*(180 - 2x)
       = 180x - 2x^2

One way to test for the maximum area is to find where the derivative 
of the area is zero. The derivative is the slope, so when the area's 
derivative is zero then the slope of the area has flattened out and 
that can only happen on a maximum or a minimum. Since the equation is 
quadratic, it only has one inflection and so it will be the maximum. I 
don't know if you know about derivatives or not...

          A  = 180x - 2x^2

     d/dx(A) = d/dx(180x - 2x^2)
             = d/dx(180x) - d/dx(2x^2)
             = 180*d/dx(x) - 2*d/dx(x^2)
             = 180*d/dx(x) - 2*2x*d/dx(x)
             = 180*1 - 2*2x*1
             = 180 - 4x

We want the point when d/dx(A) = 0 so:

      0 = 180 - 4x
     4x = 180
      x = 45

     y = 180 - 2x
       = 180 - 2*45
       = 90

So the maximum area is

     A = x*y
       = 45*90
       = 4050


3)   $10 * 300 = $3000
     $11 * 285 = $3135
     $12 * 270 = $3240

       x * (300 - 15*(x-10)) = $
or
     $ = x * (300 - 15*(x-10))

This equation seems to create the lines above. For example:

     $ = $11 * (300 - 15*(11-10))
     $ = $11 * (300 - 15)
     $ = $11 * 285

And when it is simplified we get:

     $ = x * (300 - 15*(x-10))
     $ = x * (300 - 15*x+150)
     $ = x * (-15*x + 450)
     $ = -15*x^2 + 450*x

Now that's a parabola. If we can find the maximum, then we will know 
the answer.


4) This is exactly the same type of question as the last one.

     $1   * 400 = $400
     $1.1 * 380 = $418

        x * (400-20*(x-1)) = $
or
     $ = x * (400-20*(x-1))

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

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