I've got two equations but i cant replace anything in one for the other, how do i figure this out? A carpenter is building a rectangular room with a fixed perimeter of 356 ft. What dimensions would yield the max. area? What is the max area?

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I've got two equations but i cant replace anything in one for the other, how do i figure this out? A carpenter is building a rectangular room with a fixed perimeter of 356 ft. What dimensions would yield the max. area? What is the max area?

Mathematics
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i think the maximum area is a square
There are several ways of doing this, what course is this from?
Algebra 2

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Assuming jimmyrep is correct then the sides would be 89 (356/4)
then my area would be 7921?
Yes (assuming jimmyrep is correct)
okay thank you. im just terrible with story problems they catch me off guard
I would wait and see what Nancy Lam has to say
will do
P= (L+W)2 365= (L+W)2 365/2= L+W 182.5 = L+W 182.5/2 =91.25 \[91.25^{2}=8.327?\]
Nancy don't you mean 356?
ok i was wron number, any way use 356 instate 365 my equation is right?
I was with you until the 5th line. what is that about???
Oh, I see you are letting L=W so it would be square. Seems like a burdensome method when you coluld just divide 356 by 4
P= (L+W)2 356= (L+W)2 356/2= L+W 178 = L+W 178/2 =89 \[89^{2}=7.921\]
It appears that Nancy Lam agrees with the 89 answer.
check: P= (L+W)2 356=(89+89)2 356=178*2 356=356
Thank
You realize niki 150 that this is true, if and only if, the maximum area is a square. I hope jimmyrep is right.
yeah i know. but there is no other way to figure it out so it's gotta be a square. if there was another way there isnt enough information to do so.
Agree, good luck with your studies. Have a good weekend
thanks.
u r we com

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