SOLUTION: A rectangular field is to enclosed with 600 m of fencing. What dimensions will produce a maximum area?

Area = length * width

Let y = the area
Let x = the width
Let L = the length

      y = x * L

Now since the perimeter is 600

      P = 2*length + 2*width

    600 = 2L + 2x

Divide through by 2

    300 = L + x

Solve for L by subtracting x from both sides:

300 - x = L

Substituting 300-x for L in

      y = x * L

      y = x * (300 - x)

      y = 300x - x2

Write in descending order of exponents

      y = -x2 + 300x 

Since the coefficient of  is negative,
the parabola opens downward and has a maximum
value at the vertex.  So we use the vertex
formula:

The vertex of the parabola whose equation is 

      y = ax2 + bx + c

has for its x coordinate 

and then its y-coordinate is found by substituting
that value for x in the equation and solving for
y.


Write your equation as

      y = -1x2 + 300x + 0

And a=-1, b=300 and c=0

So the vertex of the parabola whose equation is 

      y = -1x2 + 300x + 0

has for its x coordinate 

That's the width of the rectangle of maximum area.  We need to
find the length L from the equation above

300 - x = L

300 - 150 = L

150 = L

So we see that the largest area possible is when the rectangle
is chosen to be the square 150 m by 150 m

You weren't asked for the value of that maximum area. But if
you had been, then the y-coordinate of the vertex would have
given us that by substituting 150 for x in the equation and 
solving for y.

      y = -1x2 + 300x + 0
      y = -1(150)2 + 300(150) + 0
      y = -1(22500) + 45000 + 0 
      y = -22500 + 45000
      y = 22500

So that maximum area would be 22500 square meters.

The graph is:




Edwin