Question

HELP! WILLD MEDAL!

The backyard of a property is to be fenced off in a rectangular design. Fencing is only needed on three sides since the back of the house will make up the fourth side. There are 800 total feet of fencing to use. We need to find the dimensions that will maximize the area to be fenced in, and the maximum area that can be fenced in.

Answer 1

A square's area is typically greater than a rectangles so let's use a square

Answer 2

So you know you have 800 feet of fence BUT you only need 3 sides of a square, so what is the lenght of each side if all of them have to be equal

Answer 3

You don't use up fence on the house side, so I expect unequal side lengths to be optimal.

Answer 4

well yes but it would be a system of
3x=800
4x=?

Answer 5

Area = lw, fence = 2w+l

Answer 6

Oh right nvm, I remember that
Go on what @Redcan says

Answer 7

Lol it's been forever since Algebra 1

Answer 8

w=20? 0_o
hmm that seems way small

Answer 9

then the length would be l=760

Answer 10

true... must be an error

Answer 11

I remember how to do this type of problem the Calculus way,
I can't remember the Algebra way though lol

Answer 12

Can optimize this without Calculus?

Answer 13

Oh looks like Dolphin's previous questions are Calc related :)
So ya I guess that was the intended approach for this problem.

Answer 14

\[800 = 2w + l, \text{ and } Area = lw\]

Answer 15

Darn, I used x's and y's in mine XD lol

Answer 16

sry!

Answer 17

Did you figure it out dolphin? :)

Answer 18

not yet... i understand how you guys started, but was unsure as to where you were going with it...

Answer 19

So Perimeter is your constraint.
You're constrained to 800 feet of fencing, this is a total length, perimeter.\[\large\rm P=800=2y+x\]Solving for x gives us\[\large\rm x=800-2y\]We're going to come up with an area function,
and try to maximize it (using calculus).
This constraint will help us write our area function in terms of ONE VARIABLE.

Answer 20

\[\large\rm A=xy\]Using our information from the constraint,\[\large\rm A=(800-2y)y\]ya?

Answer 21

\[\large\rm A=800y-2y^2\]Now that our area function is in terms of y only,
it is much easier to differentiate.

Answer 22

Remember what you're trying to do.
You have an area function...
critical points / stationary points of this function tell us where maximums or minimums are located.

So we want to take a derivative and look for these special points, ya? :)

Answer 23

Oh your last response was an hour ago LOL
my bad :) so you might not be around

Answer 24

i understand!! i know how to take the derivative, so i will post my answer. i went to bed before you started answering, but if you come back, would you mind checking it for me? Thank you so much!

Answer 25

what did you get for your answer, @amenah8?