## anonymous 5 years ago A rancher wants to fence in an area of 2693400 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use?

1. anonymous

hmm.. the "fence down the middle" confuses me.... but here's what I think.. The minimum perimeter is a square... so the length = sqrt (2693400) = 1641.158128 There are a total of 5 equal lengths of the fence due to the middle fence... so the total fence = 1641.158128 x 5 = 8205.790638 Just round it to whatever your teacher needs...

2. anonymous

You have to take into account the fence down the middle.

3. anonymous

yeah, that's what confuses me...

4. anonymous

is the fence down the middle just the same as one of the width of the rectangular fence

5. anonymous

Yes.

6. anonymous

Set up an equation for the area of the field. Set that equal to 2693400. Set up the equation for the perimeter of the fence. Use the equation of the area to make the perimeter equation have only one variable instead of 2. Use the derivative to find the local mins of the perimeter function.

7. anonymous

ok thanks i will do that if i get stuck i will ask for more help :) but that seems to clarify things

8. anonymous

If you get stuck, post here what equations you're using for area and perimeter.

9. anonymous

A=LW; P=2L+2W I get P=2L+2(2693400/L)

10. anonymous

That perimeter is not quite right because you need to take into account the fence down the middle. So P = 2L + 3W

11. anonymous

ok yea i don't know why i am leaving it out lol

12. anonymous

so my P'=2+8050200/L^2

13. anonymous

Close, but $\frac{d}{dx} (\frac{1}{x}) = \frac{-1}{x^2}$ So you've got a sign problem in P'

14. anonymous

yea i saw it so i get 2/8050200=L^2 after i fix sign error then i get sqrt(2/8050200)=L

15. anonymous

uh.. I think you're playing a bit fast/loose here.. $P' = 2-\frac{8050200}{L^2}$ $P'=0 \rightarrow \frac{8050200}{L^2} = 2$ $\rightarrow \frac{8050200}{2} = L^2$

16. anonymous

You have it the other way around.

17. anonymous

yea cause its like saying 8050200(1/L^2)

18. anonymous

ok so will L be my answer..

19. anonymous

I am getting a wrong answer...

20. anonymous

The L you find by setting P' to 0 will be the length at which your perimeter function will be at its minimum value. You then need to plug in this L to your normal perimeter function to find how much fence the farmer needs.

21. anonymous

my math is wrong it was 8080200 not 8050200

22. anonymous

I got the right answer...thanks for your help sorry i was just making minor errors but you helped me see them

23. anonymous

could you help me with another If 3072 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

24. anonymous

Same type deal. Set up equation for surface area, and volume. Eliminate a variable, take the derivative of the volume function and find a local max (where V' = 0). Plug in that dimension back into V to find the max Volume

25. anonymous

err 2 equations, one for surface area, one for volume.

26. anonymous

is it surface area because we know the amount of material that we use

27. anonymous

it's surface area because they give us the material yes (in square centimeters)

28. anonymous

V=w^2*H and my surface area would be of a cube?

29. anonymous

Yes with the top missing.

30. anonymous

so its just five sides SA=5x^2 or is it SA=2w^2+4wh

31. anonymous

That last one. The length and width is the same because the bottom is square, but it doesn't stipulate that the sides are square. Although it should only be w^2 not 2w^2 since the top is open.

32. anonymous

ok

33. anonymous

i am kinda lost with the derivative...

34. anonymous

What's the equation you have for Volume?

35. anonymous

V=w^2(3072-w^2/4w)

36. anonymous

Shouldn't that be $V = \frac{w^2(3072-w^2)}{4w}$

37. anonymous

could you eliminate the w from the bottom and one w from w^2

38. anonymous

yes.

39. anonymous

$V=\frac{1}{4} * [3072w - w^3]$

40. anonymous

V'=1/4[3072-3w^2]

41. anonymous

yep.

42. anonymous

Thanks again i got it again... i am starting to understand optimization

43. anonymous

A rectangle is inscribed with its base on the x -axis and its upper corners on the parabola y=2-x^2 What are the dimensions of such a rectangle with the greatest possible area? width= height= i figured out height was 4/3 cause i thought width was $\sqrt{2/3}$ and plugged it in y=2-x^2