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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?

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  1. anonymous
    • one year ago
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    |dw:1433649074172:dw|

  2. anonymous
    • one year ago
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    Let A = xy 2x + 2y = 1600

  3. anonymous
    • one year ago
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    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
    • one year ago
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    why does it became 2x +2y sir?

  5. anonymous
    • one year ago
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    @ganeshie8

  6. anonymous
    • one year ago
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    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
    • one year ago
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    rectangle with given area but with minimum perimeter: |dw:1433659554258:dw|

  8. anonymous
    • one year ago
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    A=xy since x=y A=y*y 1600=y^2 y=40=x

  9. freckles
    • one year ago
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    how do you know y=x will give maximum? are you allowed to assume that?

  10. anonymous
    • one year ago
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    from the perimeter: 2x+2y=P 2(40)+2(40)=P P=160 therefore P=160 is the answer

  11. freckles
    • one year ago
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    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
    • one year ago
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    the min*

  13. anonymous
    • one year ago
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    by differentiating you can get y=x

  14. freckles
    • one year ago
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    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
    • one year ago
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  16. anonymous
    • one year ago
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    just see the attached file

  17. freckles
    • one year ago
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    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
    • one year ago
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    and then once I found x from that equation plug into xy=1600 to find y

  19. anonymous
    • one year ago
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    yeah you can also do that

  20. anonymous
    • one year ago
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    we came up with the same answer

  21. anonymous
    • one year ago
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    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
    • one year ago
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    @freckles thanks for your clear explanation

  23. freckles
    • one year ago
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    hmm it looks like he got the area and perimeter mixed up

  24. freckles
    • one year ago
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    yep that is exactly what happened

  25. anonymous
    • one year ago
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    oooh yeah..

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

  27. anonymous
    • one year ago
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    ooh thank you for this sir.. it will be a great help for me

  28. freckles
    • one year ago
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    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
    • one year ago
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    ooh thanks

  30. freckles
    • one year ago
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    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
    • one year ago
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    that second equation came from differentiating 500=x+2y

  32. freckles
    • one year ago
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    w.r.t. x

  33. freckles
    • one year ago
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    anyways neat

  34. anonymous
    • one year ago
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    oooh yes i got it

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