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April 24, 2014

April 24, 2014

Posted by **Marysvoice** on Friday, February 15, 2008 at 5:07pm.

what dimentions would yield the maximum area?

What dimentions would yeild the minimum area?

- Quadratic equations -
**Reiny**, Friday, February 15, 2008 at 5:09pmA rectangle of largest area is a square.

So divide your perimeter by 4.

To build a room of minimum area is a rather silly question.

- Quadratic equations -
**Marysvoice**, Friday, February 15, 2008 at 6:44pmI apologize for my silly question. I followed your advice, revised my question and found my answer. Thank you.

- Quadratic equations -
- Quadratic equations -
**Damon**, Friday, February 15, 2008 at 6:07pmx = length

y = width

A = x y

x+y=140

A = x(140-x)

A = 140 x - x^2

look at shape of parabola by completing the square

x^2 -140 x = -A

x^2 - 140 x + 4900 = -A + 4900

(x-70)^2 = -A + 4900

the vertex will be at x = 70 and A = 4900 and the -A means that the parabola opens down (sheds water) so that vertex is the maximum where x = y =70 as Reiny told you.

since a negative x or y does not make much sense, the area may not get any smaller than zero when x --> 0 or y -->0 0

- Quadratic equations -
**Marysvoice**, Friday, February 15, 2008 at 6:44pmI apologize for my silly question. I followed your advice, revised my question and found my answer. Thank you.

- Quadratic equations -

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