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.

Question

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

i forgot to adjust for all my stupidity lol

anonymous · Answer

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

amistre64 · Answer

L = sqrt(280) :)

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

stop me when I get lost :)

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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 · Answer

yea...

amistre64 · Answer

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 · Answer

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

amistre64 · Answer

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

anonymous · Answer

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 · Answer

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 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

hit "post"  lol

anonymous · Answer

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

anonymous · Answer

Thanks figured it out...

anonymous · Answer

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

anonymous · Answer

Ok cool.

amistre64 · Answer

so where did I miss it?

anonymous · Answer

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 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

make that 280 up there

anonymous · Answer

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

anonymous · Answer

It just means that your L is my W.

amistre64 · Answer

it is...they are :)

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

you get L = sqrt(140)

anonymous · Answer

Yeah.

anonymous · Answer

But my L is your W remember.

amistre64 · Answer

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

amistre64 · Answer

11.83 = you

12.40 = me.....

anonymous · Answer

Hrm. Lemme check that cost again.

amistre64 · Answer

280*5 = 1400 ... not 900

anonymous · Answer

Ah. That'll do it.

amistre64 · Answer

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

amistre64 · Answer

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

anonymous · Answer

Yep, much better.

anonymous · Answer

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

anonymous · Answer

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

amistre64 · Answer

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