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Old 12-05-2006   #1
MM92
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Default Algebra based question

Hi everyone, I am new to this site. I was wondering if anyone here can help me answer a question, in order for me to study correctly for my math test tomorrow.

Here's the question:

Melissa plans to put a fence around her rectangular garden. She has 150 feet of fencing material to make the fence. If there is to be a 10 foot opening left for an entrance on one side of the garden, what dimensions should the garden be for maximum area? This question has to be answered in vertex form.

If anyone can answer this and explain it step by step to me that would be great!
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Old 12-06-2006   #2
zwu_ca
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Default Solution:

Suppose one side of the rect is x, the other side is (160-2x)/2=80-x (taking account of 10 feet of the door).. The area is x(80-x) = 80x - x^2 = 1600 - (1600-80x+x^2) = 1600 - (x-40)^2. So maximum area is reached when x is 40. The dim is 40x40.

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Old 12-06-2006   #3
MM92
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So that would be how u do it in vertex form?

what does ^ this mean?

Thanks
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Old 12-06-2006   #4
zwu_ca
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Default x^2 = x.x, i.e., x times x.

The vertex form can be many, many. Here is a one:

(0,0), (0,10), (0,40), (40,40), (40,0).

Last edited by zwu_ca; 12-06-2006 at 06:51 PM..
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Old 12-06-2006   #5
Cookie Dough 7
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wait i think i do....so the answer is 40 by 40 or do i have 2 do another step?
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Old 12-07-2006   #6
zwu_ca
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Default We don't know what exactly the original problem is.

If the original problem required the vertex form, we have to do another step.
Math Problem of the Day
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Old 03-30-2007   #7
Scion

 
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Quote:
Originally Posted by zwu_ca View Post
Suppose one side of the rect is x, the other side is (160-2x)/2=80-x (taking account of 10 feet of the door).. The area is x(80-x) = 80x - x^2 = 1600 - (1600-80x+x^2) = 1600 - (x-40)^2. So maximum area is reached when x is 40. The dim is 40x40.

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plz dont adveertise other sites
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