A fence is to be built to enclose a rectangular area of 280 square feet. The fence along three sides is to be made of material that costs 5 dollars per foot, and the material for the fourth side costs 15 dollars per foot. Find the dimensions of the enclosure that is most economical to construct.

- anonymous

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- amistre64

Area = LW
Perimeter = L1 + L2 + 2W
180 = LW
2L(5) + W(5) + L(15) = cost

- amistre64

since 180 = LW, we can say that W = 180/L

- amistre64

that L(15) up there should be W(15)
2L(5) + (180/L)(5) + (180/L)(15) = Cost

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## More answers

- amistre64

10L +900/L + 2700/L = cost

- amistre64

maybe we need to minimise the perimeter :)
P = 2L + 2W
P = 2L + 2(180/L)
P = 2L + 360/L
P = 2L^2 + 360
---------- take the derivative
L

- amistre64

dP = L(2L) - (2L^2+360)(1)
-------------------
L(L)

- amistre64

4L^2 - 2L^2 - 360 2L^2-360
----------------- = ---------
L^2 L^2
I adjusted for my stupidity :)

- amistre64

when the top is 0 we got a min or a max

- anonymous

sorry just a little confused because it 280 but it still works i can just go back and plug in my numbers

- amistre64

2(L^2 -180) = 0
L^2 - 180 = 0
L^2 = 180
L = +sqrt(180)

- amistre64

...... lol ..........

- amistre64

i forgot to adjust for all my stupidity lol

- anonymous

it ok you have it i see just a mix up of numbers is all...

- amistre64

L = sqrt(280) :)

- amistre64

W = 280/sqrt(280)
W = sqrt(280) as well, so its a square I guess...

- amistre64

Well, that would maximise the area ..... but it doesnt help with costs :)
lets see what costs gets us

- amistre64

2L(5) + (280/L)(5) + (280/L)(15) = Cost
10L + 900/L + 4200/L = cost

- amistre64

10L^2 + 900 + 4200
------------------ = cost
L

- amistre64

stop me when I get lost :)

- amistre64

10L^2 + 5100
------------ = cost
L
L(20L) - 1(10L^2 + 5100)
----------------------- = cost'
L^2

- anonymous

wait i am confused....it wants to know the dimensions...so why are we looking for cost

- amistre64

20L^2 -10L^2 -5100 = 0
10L^2 - 5100 = 0
10(L^2 -510) = 0
L^2 - 510 = 0
L=sqrt(510)
because..... the diminsions dont need to be "minimized for area, they need to be minimized for cost...makes sense?

- anonymous

yea...

- amistre64

lets plug sqrt(510) into the calulator to give us a roundabout:
22.58 is what I get for one side :)
280 = LW
280 = 22.58W
280/22.58 = W
12.40 = W

- amistre64

Now figure out how much we need for the most expensive side :)

- amistre64

2(22.58)(5) + 12.40(5) + 12.40(15) =?
2(12.40)(5) + 22.58(5) + 22.58(15)? which is cheaper?

- anonymous

my answer are not correct....thats what my website i have to submit my answer into is telling me...so i don't know if there was a mix up somewhere

- amistre64

maybe..... I could of thought it wrong as well. but to me, to minimize cost, we would derive the equation for Price and not Area, figure out what a side would be for that price then figure the rest out...

- amistre64

280/L or 280/W doesnt matter, since its just a name for a side..right?

- amistre64

polpak is here.... hell know what I did wrong ;)

- amistre64

hit "post" lol

- anonymous

Ok, we have
\[Area = LW = 280 \rightarrow W = 280/L\]
\[Cost = 5(L + 2W) + 15L \]
\[\rightarrow Cost = 20L + (5*2)(280)/L \]

- anonymous

Thanks figured it out...

- anonymous

\[\frac{d}{dL}Cost = 20 - \frac{2800}{L^2}\]
The cost is a min when the derivative is 0
\[\rightarrow 0 = 20-\frac{2800}{L^2}\]

- anonymous

Ok cool.

- amistre64

so where did I miss it?

- anonymous

I can't tell, but you have a lot of cost functions and none of them look the same as the one I have.

- amistre64

:) they are similar.... I was just going over that to see if I made a mistake

- amistre64

I did 2L(5) + 180(5)/L + 280(15)/L for cost... but that should be correct either way....

- amistre64

make that 280 up there

- anonymous

If those are both 280 instead of 180 it should be ok.

- anonymous

It just means that your L is my W.

- amistre64

it is...they are :)

- amistre64

10L +900/L + 4200/L = cost

- anonymous

Ok then you would have
\[Cost' = 10 -900/L^2 - 4200/L^2\]

- amistre64

right... you derived them seperately which is fine :)

- amistre64

10L^2 - 5100 = 0
L^2 - 510 = 0

- amistre64

L=sqrt(510)....
is it right? the same as yours?

- amistre64

you get L = sqrt(140)

- anonymous

Yeah.

- anonymous

But my L is your W remember.

- amistre64

hmmm....... maybe thats right, havent checked that out yet :)

- amistre64

11.83 = you
12.40 = me.....

- anonymous

Hrm. Lemme check that cost again.

- amistre64

280*5 = 1400 ... not 900

- anonymous

Ah. That'll do it.

- amistre64

sqrt(560).... might be better :)

- amistre64

Yep... thats it.... I just forgot how to multiply integers LOL

- anonymous

Yep, much better.

- anonymous

Thats why I always save the plugging the numbers part for last.

- anonymous

Or as late as I can without making it look horrible.

- amistre64

yeah.... itda helped if I started out with the 280 and worked it, but..... you know :)

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