Name:  Roberta Jones

Project:  Strategies  for solving problems.

Description:  This lesson models several different ways of solving an application  problem.    The problem is one typically found in an Algebra I text.   These include using a table, graphing with two variables, graphing with one variable, and solving algebraically.   Examples  are given of how to extend the problem in several interesting ways.

Problem Solving Using    
             Several Techniques

Here's a Problem

A farmer with 2986 meters of fencing wants to enclose a rectangular field that borders on a straight highway which has an existing fence.  So the farmer needs to use his fence for three sides of the field.  He also wants the area of the field to be 832,000 square meters.  Find the dimensions of the field to the nearest whole number.

Some people may think there is one and only one way to solve this problem.  That is not the case.  The purpose of this exercise is to point out different ways of solving a problem.  Please keep that in mind as you read the following pages.

First, let's draw a picture to see what the situation is.

Solve by Using a Table

We could simply make a table.  In the first column, put the width.  In the second column put the other width.  Then find the difference (2986 - twice the width) and that would be the length.   Then multiply length times width and get area.   Sounds complicated, but really it isn't.  Let's look at the table.

Width     Width       Length        Area

Now we can look at the table and see the area closest to 832 000.   The approximate solution is 370 by 2246.

Solve by graphing two equations

If we used two variables to represent the problem, say x for the width and y for the length, the two equations would be: 
xy = 832 000          and         y + 2x   =  2986
Let's graph those two equations.

Clear[x,f,g]
f[x_]=832000/x
g[x_]=2986-2x
Plot[{f[x],g[x]}, {x,.0001,2000},PlotStyle->{RGBColor[1,0,0],RGBColor[0,1,0]}]

What do these two lines represent?


Well, they represent our two constraints.  The red line shows all the points where the area is 832 000 square meters.  You write down what the green line represents.  Consult with your partner.

We now need to see where both constraints have been met.  This graph doesn't help us see that very well.   Why?

Did you decide that the range and domain aren't appropriate for seeing where the intersection point(s) are?  You're right.

Let's change them.

Plot[{f[x],g[x]}, {x,1,1200},PlotRange->{0,3500},PlotStyle->{RGBColor[1,0,0],RGBColor[0,1,0]}]

We see now that there appear to be two possible solutions, one where x = approximately 370 and y = around 2190 and another where x = about 1130 and y = about 700.
Why did we get two solutions in this case?  Do they both work?

Solve by graphing one equation

What if we wanted to express the relationships in the problem using one equation?  Can you write one equation in terms of x to do this?

It would be x( 2986 - 2x)  =  832 000


Now let's look at the picture of this function.

Clear[h]
h[x_]=832000-x(2986-2x)
Plot[h[x], {x,0,1500},PlotRange->{-280000,50000},PlotStyle->{RGBColor[1,0,0]}]

What does this graph represent and how can we use it to get a solution to our problem?

Solve one equation algebraically

Mathematica can also solve the problem for us algebraically.

eqn = x(2986-2x)==832000
N[Solve[eqn]]

Is this the complete solution.  If not, what must be done?

An extension --maximum area

Why would we want to look at the graph when all we have to do is solve the equation?

Sometimes solving the problem can bring up more questions which a graph might help answer.  For instance, what if the farmer wondered not just what the dimensions should be to get 832 000 square meters but what the dimensions would be to maximize the area of the field.   So one equation  one might come up with is related to the equation we just solved.  Can you determine what it is?

It is    y = x(2986 - 2x)    where y is the area.

Let's look at the picture of this.

Clear[j]
j[x_]=x(2986-2x)
Plot[j[x], {x,0,1500},PlotRange->{0,1500000},PlotStyle->{RGBColor[1,0,0]}]

What does this parabola represent?  Is our orginal problem represented in this picture?  Can you give an estimate of the maximum area the farmer could have with the fencing he has?

Further Applications

Did you notice that the previous two functions gave us pictures of parabolas?    It's probably true that we can vary the other variable (the length of the fence)  and get other equations which will graph a variable.  Let's look at some and think about what they mean.

Suppose the farmer has half as much fencing.  what would the equation be?

It would be         y = x(1493 - 2x).   Let's graph this equation with the previous one.   What is the maximum area now?

Clear[k]
k[x_]=x(1493 - 2x)
Plot[{j[x],k[x]}, {x,-400,3000},PlotRange->{-1500000,1500000},PlotStyle->{RGBColor[1,0,0]}]

Looking at this graph, can you estimate what the dimensions of the rectangle would be that would give the greatest area?

So you see that we can generate a whole bunch of parabolas based on this problem by changing the length of the fencing;  the parabolas gives us the areas for different lengths and widths.  Isn't that rad?

You Try It

Now you try a problem.  See if you can think of different ways to solve it.  After solving it, think of other questions you could ask about the problem.  This could be fun!

In Physics, an important formula gives the height of an object which has been propelled in the air if you know the velocity (v), the initial height (hº) and the number of seconds (t) which have elapsed.

It is   h = -1/2(9.8)t¤  +  vt  +  hº              (In case you're curious,  9.8 is one measure for gravity,   g = 9.8 m/sec¤.)

A juggler throws a ball from her hand at a height of 1 meter with an initial velocity of 10 meters per second.  How high is ball after 4 seconds?