## anonymous 5 years ago 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.

1. amistre64

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

2. amistre64

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

3. amistre64

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

4. amistre64

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

5. 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

6. amistre64

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

7. amistre64

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

8. amistre64

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

9. anonymous

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

10. amistre64

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

11. amistre64

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

12. amistre64

i forgot to adjust for all my stupidity lol

13. anonymous

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

14. amistre64

L = sqrt(280) :)

15. amistre64

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

16. amistre64

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

17. amistre64

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

18. amistre64

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

19. amistre64

stop me when I get lost :)

20. amistre64

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

21. anonymous

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

22. 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?

23. anonymous

yea...

24. 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

25. amistre64

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

26. amistre64

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

27. 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

28. 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...

29. amistre64

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

30. amistre64

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

31. amistre64

hit "post" lol

32. anonymous

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

33. anonymous

Thanks figured it out...

34. anonymous

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

35. anonymous

Ok cool.

36. amistre64

so where did I miss it?

37. 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.

38. amistre64

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

39. amistre64

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

40. amistre64

make that 280 up there

41. anonymous

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

42. anonymous

It just means that your L is my W.

43. amistre64

it is...they are :)

44. amistre64

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

45. anonymous

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

46. amistre64

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

47. amistre64

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

48. amistre64

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

49. amistre64

you get L = sqrt(140)

50. anonymous

Yeah.

51. anonymous

But my L is your W remember.

52. amistre64

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

53. amistre64

11.83 = you 12.40 = me.....

54. anonymous

Hrm. Lemme check that cost again.

55. amistre64

280*5 = 1400 ... not 900

56. anonymous

Ah. That'll do it.

57. amistre64

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

58. amistre64

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

59. anonymous

Yep, much better.

60. anonymous

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

61. anonymous

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

62. amistre64

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