anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
Area = LW Perimeter = L1 + L2 + 2W 180 = LW 2L(5) + W(5) + L(15) = cost
amistre64
  • amistre64
since 180 = LW, we can say that W = 180/L
amistre64
  • 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
  • amistre64
10L +900/L + 2700/L = cost
amistre64
  • 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
  • amistre64
dP = L(2L) - (2L^2+360)(1) ------------------- L(L)
amistre64
  • amistre64
4L^2 - 2L^2 - 360 2L^2-360 ----------------- = --------- L^2 L^2 I adjusted for my stupidity :)
amistre64
  • amistre64
when the top is 0 we got a min or a max
anonymous
  • anonymous
sorry just a little confused because it 280 but it still works i can just go back and plug in my numbers
amistre64
  • amistre64
2(L^2 -180) = 0 L^2 - 180 = 0 L^2 = 180 L = +sqrt(180)
amistre64
  • amistre64
...... lol ..........
amistre64
  • amistre64
i forgot to adjust for all my stupidity lol
anonymous
  • anonymous
it ok you have it i see just a mix up of numbers is all...
amistre64
  • amistre64
L = sqrt(280) :)
amistre64
  • amistre64
W = 280/sqrt(280) W = sqrt(280) as well, so its a square I guess...
amistre64
  • amistre64
Well, that would maximise the area ..... but it doesnt help with costs :) lets see what costs gets us
amistre64
  • amistre64
2L(5) + (280/L)(5) + (280/L)(15) = Cost 10L + 900/L + 4200/L = cost
amistre64
  • amistre64
10L^2 + 900 + 4200 ------------------ = cost L
amistre64
  • amistre64
stop me when I get lost :)
amistre64
  • amistre64
10L^2 + 5100 ------------ = cost L L(20L) - 1(10L^2 + 5100) ----------------------- = cost' L^2
anonymous
  • anonymous
wait i am confused....it wants to know the dimensions...so why are we looking for cost
amistre64
  • 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
  • anonymous
yea...
amistre64
  • 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
  • amistre64
Now figure out how much we need for the most expensive side :)
amistre64
  • 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
  • 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
  • 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
  • amistre64
280/L or 280/W doesnt matter, since its just a name for a side..right?
amistre64
  • amistre64
polpak is here.... hell know what I did wrong ;)
amistre64
  • amistre64
hit "post" lol
anonymous
  • anonymous
Ok, we have \[Area = LW = 280 \rightarrow W = 280/L\] \[Cost = 5(L + 2W) + 15L \] \[\rightarrow Cost = 20L + (5*2)(280)/L \]
anonymous
  • anonymous
Thanks figured it out...
anonymous
  • 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
  • anonymous
Ok cool.
amistre64
  • amistre64
so where did I miss it?
anonymous
  • 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
  • amistre64
:) they are similar.... I was just going over that to see if I made a mistake
amistre64
  • amistre64
I did 2L(5) + 180(5)/L + 280(15)/L for cost... but that should be correct either way....
amistre64
  • amistre64
make that 280 up there
anonymous
  • anonymous
If those are both 280 instead of 180 it should be ok.
anonymous
  • anonymous
It just means that your L is my W.
amistre64
  • amistre64
it is...they are :)
amistre64
  • amistre64
10L +900/L + 4200/L = cost
anonymous
  • anonymous
Ok then you would have \[Cost' = 10 -900/L^2 - 4200/L^2\]
amistre64
  • amistre64
right... you derived them seperately which is fine :)
amistre64
  • amistre64
10L^2 - 5100 = 0 L^2 - 510 = 0
amistre64
  • amistre64
L=sqrt(510).... is it right? the same as yours?
amistre64
  • amistre64
you get L = sqrt(140)
anonymous
  • anonymous
Yeah.
anonymous
  • anonymous
But my L is your W remember.
amistre64
  • amistre64
hmmm....... maybe thats right, havent checked that out yet :)
amistre64
  • amistre64
11.83 = you 12.40 = me.....
anonymous
  • anonymous
Hrm. Lemme check that cost again.
amistre64
  • amistre64
280*5 = 1400 ... not 900
anonymous
  • anonymous
Ah. That'll do it.
amistre64
  • amistre64
sqrt(560).... might be better :)
amistre64
  • amistre64
Yep... thats it.... I just forgot how to multiply integers LOL
anonymous
  • anonymous
Yep, much better.
anonymous
  • anonymous
Thats why I always save the plugging the numbers part for last.
anonymous
  • anonymous
Or as late as I can without making it look horrible.
amistre64
  • amistre64
yeah.... itda helped if I started out with the 280 and worked it, but..... you know :)

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