Posted by MathGuru on Wednesday, January 17, 2007 at 8:38pm.
You have two formulas to use:
A = lw >area = length times width
P = 2l + 2w >perimeter
You know the perimeter, which is 200m of fencing.
Let length = x
Now let's solve the perimeter equation for w, using what we know:
200 = 2x + 2w
200  2x = 2w
(200  2x)/2 = w
2(100  x)/2 = w
100  x = w
Therefore:
A = x(100  x) = 100x  x^2
x^2 + 100x is the area of the rectangle.
To place the quadratic equation in vertex form, you will need to complete the square:
A = x^2 + 100x
A = (x^2  100x)
A = (x^2  100x + 2500  2500) >to complete the square, take 1/2 times the middle coefficient and square it (you'll need to add and subtract it as well)
A = (x  50)^2 + 2500 >vertex form
If x = 50, then 100  x = 50.
To maximize the area, take (50)(50) = 2500 square meters.
I hope this will help with other problems of this type.
Ms. Brown has 200m of fencing with which she intends to construct a rectangular enclosure along a river where no fencing is needed. She plans to divide the enclosure into two parts.
What should be the dimensions of the enclosure if its area is to be a maximum?
Show a complete algebraic solution.
thank you

mathgebra  Whistling, Friday, October 12, 2007 at 1:45am
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