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.

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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.

Mathematics
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Area = LW Perimeter = L1 + L2 + 2W 180 = LW 2L(5) + W(5) + L(15) = cost
since 180 = LW, we can say that W = 180/L
that L(15) up there should be W(15) 2L(5) + (180/L)(5) + (180/L)(15) = Cost

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Other answers:

10L +900/L + 2700/L = cost
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
dP = L(2L) - (2L^2+360)(1) ------------------- L(L)
4L^2 - 2L^2 - 360 2L^2-360 ----------------- = --------- L^2 L^2 I adjusted for my stupidity :)
when the top is 0 we got a min or a max
sorry just a little confused because it 280 but it still works i can just go back and plug in my numbers
2(L^2 -180) = 0 L^2 - 180 = 0 L^2 = 180 L = +sqrt(180)
...... lol ..........
i forgot to adjust for all my stupidity lol
it ok you have it i see just a mix up of numbers is all...
L = sqrt(280) :)
W = 280/sqrt(280) W = sqrt(280) as well, so its a square I guess...
Well, that would maximise the area ..... but it doesnt help with costs :) lets see what costs gets us
2L(5) + (280/L)(5) + (280/L)(15) = Cost 10L + 900/L + 4200/L = cost
10L^2 + 900 + 4200 ------------------ = cost L
stop me when I get lost :)
10L^2 + 5100 ------------ = cost L L(20L) - 1(10L^2 + 5100) ----------------------- = cost' L^2
wait i am confused....it wants to know the dimensions...so why are we looking for cost
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?
yea...
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
Now figure out how much we need for the most expensive side :)
2(22.58)(5) + 12.40(5) + 12.40(15) =? 2(12.40)(5) + 22.58(5) + 22.58(15)? which is cheaper?
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
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...
280/L or 280/W doesnt matter, since its just a name for a side..right?
polpak is here.... hell know what I did wrong ;)
hit "post" lol
Ok, we have \[Area = LW = 280 \rightarrow W = 280/L\] \[Cost = 5(L + 2W) + 15L \] \[\rightarrow Cost = 20L + (5*2)(280)/L \]
Thanks figured it out...
\[\frac{d}{dL}Cost = 20 - \frac{2800}{L^2}\] The cost is a min when the derivative is 0 \[\rightarrow 0 = 20-\frac{2800}{L^2}\]
Ok cool.
so where did I miss it?
I can't tell, but you have a lot of cost functions and none of them look the same as the one I have.
:) they are similar.... I was just going over that to see if I made a mistake
I did 2L(5) + 180(5)/L + 280(15)/L for cost... but that should be correct either way....
make that 280 up there
If those are both 280 instead of 180 it should be ok.
It just means that your L is my W.
it is...they are :)
10L +900/L + 4200/L = cost
Ok then you would have \[Cost' = 10 -900/L^2 - 4200/L^2\]
right... you derived them seperately which is fine :)
10L^2 - 5100 = 0 L^2 - 510 = 0
L=sqrt(510).... is it right? the same as yours?
you get L = sqrt(140)
Yeah.
But my L is your W remember.
hmmm....... maybe thats right, havent checked that out yet :)
11.83 = you 12.40 = me.....
Hrm. Lemme check that cost again.
280*5 = 1400 ... not 900
Ah. That'll do it.
sqrt(560).... might be better :)
Yep... thats it.... I just forgot how to multiply integers LOL
Yep, much better.
Thats why I always save the plugging the numbers part for last.
Or as late as I can without making it look horrible.
yeah.... itda helped if I started out with the 280 and worked it, but..... you know :)

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