SOLUTION: It is desired to enclose a rectangular field adjacent to the straight bank of a river. No fence is needed along the river bank. If 160 meters of fence is to be used, what are the

Hi, there--

The Problem:
It is desired to enclose a rectangular field adjacent to the straight bank of a river. No fence is 
needed along the river bank. If 160 meters of fence is to be used, what are the dimensions of 
the rectangle which will enclose the maximum area, and what is the maximum area?

A Solution:
Let W be the width of the rectangular field.
Let L be the length of the rectangular field.
Let A be the area of the rectangular field.

We need to write two equations using the information in the problem.

The enclosure uses is 160m of fence. The length along the river is not fenced. We see that 160m
of fence will span two widths and one length. (The other length is the riverbank.) In algebra, 
we write this relationship as 2W + L = 160.

The formula for the area of the field is A = L * W.

Now we have two equations with three variables. We will use substitution to beginning solving this 
system. Rewrite the first equation in "L=..." form.
2W + L = 160
L = 160 - 2W

We see that L is equivalent to 160-2W. Make this substitution in the second equation.
A = L * W
A = (160-2W) * W

This is a quadratic equation for the area of the rectangular fence in square meters. The graph of a
quadratic equation is a parabola. The vertex of this parabola is the maximum point on the curve. We
will find the vertex of the parabola and use this information to find the maximum possible area of 
the enclosure. 

(NOTE: I'm not sure of you math level. If you are in a more advanced math class, you can also solve 
this problem by finding the first derivative of A = (160-2W)*W. If you would like to see that 
solution method, please email me.)

Simplify the area equation by clearing the parentheses.
A = 160W - 2W^2

Re order the terms in descending order by power.
A = -2W^2 + 160W

Now we'll find the vertex. This quadratic is now in standard form, y = ax^2 +bx +c, where a is the
coefficient of the squared term, and b is the coefficient of the first degree term. In your equation, 
a = -2 and b = 160.

In quadratic equations, if the leading coefficient a is negative, the parabola opens downward, 
and the vertex is the maximum point. This means the vertex in this equation will give us a maximum 
area because the coefficient of the W-squared term is -2.

The formula for the x term (or W in your equation) of the vertex is -b/2a.
W = -b/2a = -160/((2)(-2)) = 40

In the context of your problem W = 40 means that we have the maximum area when the width is 
40m. To find the area when the width is 40, we substitute 40 for W in the equation.
A = -2W^2 + 160W
A = -2(40)^2 + 160(40)
A = -2(1600) + 6400
A = -3200 + 6400
A = 3200

In the context of your problem A = 3200 means that the maximum area is 3,200 square meters.

The final step is to find the length with the width is 40m. We substitute 40 for W in our very 
first equation.
2W + L = 160
2(40) + L = 160
80 + L = 160
L = 80

So the length is 80m when we maximize the area.

That's it! Please email me if you have questions or comments about the solution. I would appreciate 
the feedback.

Ms.Figgy
math.in.the.vortex@gmail.com