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July 23, 2014

July 23, 2014

Posted by **MathGuru** on Wednesday, January 17, 2007 at 8:38pm.

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 1:45amblah blah blah

- mathgebra -
**[PuPe]ING ----JOHN**, Friday, October 12, 2007 at 1:48amIM NOTHING

- mathgebra -

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