## anonymous one year ago a rectangular lot is to have an area of 1600 sq. m. find the least amount of fence that could be used to enclose the area?

1. anonymous

|dw:1433649074172:dw|

2. anonymous

Let A = xy 2x + 2y = 1600

3. anonymous

Since we are looking for the "least" amount of fencing, we have to differentiate (or even twice) to see that the outcome is a minimum

4. anonymous

why does it became 2x +2y sir?

5. anonymous

@ganeshie8

6. anonymous

2x + 2y is the perimeter of the rectangle. So lets make 'x' the subject hence -> x = 1600 - 2y 2 Now we resub the x value into A = xy -> sub that in and start differentiating -> Double differentiate if you have to find that the value is indeed a minimum.

7. anonymous

rectangle with given area but with minimum perimeter: |dw:1433659554258:dw|

8. anonymous

A=xy since x=y A=y*y 1600=y^2 y=40=x

9. freckles

how do you know y=x will give maximum? are you allowed to assume that?

10. anonymous

from the perimeter: 2x+2y=P 2(40)+2(40)=P P=160 therefore P=160 is the answer

11. freckles

that seems to be correct but are you allowed to assume the max is given by y=x are you do you have to show that?

12. freckles

the min*

13. anonymous

by differentiating you can get y=x

14. freckles

ok your answer seems to be fine as long as you also show in your paper how you came to the y=x part

15. anonymous

16. anonymous

just see the attached file

17. freckles

I know I would have done this: $xy=1600 \\ \text{ so that means we can write } y=\frac{1600}{x} \\ \text{ and we also have \to minmize } P=2x+2y \\ P=2x+2(\frac{1600}{x}) \\ P=2x+\frac{3200}{x} \\ \text{ then I would differentiate and set } P'=0 \text{ and solve for } x$

18. freckles

and then once I found x from that equation plug into xy=1600 to find y

19. anonymous

yeah you can also do that

20. anonymous

we came up with the same answer

21. anonymous

actually i am the one who asked this question because this is our assignment in school. unfortunately i didnt understand the explanation said by the first person. that's why I searched along the net. and luckily got the right answer :-)

22. anonymous

@freckles thanks for your clear explanation

23. freckles

hmm it looks like he got the area and perimeter mixed up

24. freckles

yep that is exactly what happened

25. anonymous

oooh yeah..

26. freckles

http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx I really like paul's notes he does other things besides optimization problems

27. anonymous

ooh thank you for this sir.. it will be a great help for me

28. freckles

also that way you showed is an awesome way I have never done optimization problem the way it was done in that pdf

29. anonymous

ooh thanks

30. freckles

doing Paul's first problem like that: we have: $A'=y+xy' \\ 0=1+2y' \\ \text{ so } y'=\frac{-1}{2} \\ \text{ so } A'=y+x(\frac{-1}{2}) \\ \text{ but } 2y=500-x \\ \text{ and multiplying } \frac{1}{2} \text{ on both sides gives } y=\frac{500}{2}-\frac{1}{2}x \\ \text{ so } A'=\frac{500}{2}-\frac{1}{2}x+x(\frac{-1}{2}) \\ A'=\frac{500}{2}-x \\ A'=0 \text{ when } x=\frac{500}{2}=250$

31. freckles

that second equation came from differentiating 500=x+2y

32. freckles

w.r.t. x

33. freckles

anyways neat

34. anonymous

oooh yes i got it