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