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!

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.

http://www.mathpotd.org

So that would be how u do it in vertex form?

what does ^ this mean?

Thanks

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

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

wait i think i do....so the answer is 40 by 40 or do i have 2 do another step?

If the original problem required the vertex form, we have to do another step.
Math Problem of the Day

[Scion]

what is vertex? u mean angles blah i dony know

[Scion]

Originally Posted by zwu_ca

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.

http://www.mathpotd.org

plz dont adveertise other sites