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Brent0423

  • 2 years ago

Find two positive numbers whose sum is 100 and the sum of whose squares is a minimum.

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  1. Brent0423
    • 2 years ago
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    Is it just me or is this problem pretty complex?

  2. Brent0423
    • 2 years ago
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    this is what all of my problems are like... FML

  3. ChmE
    • 2 years ago
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    I don't understand squares is a minimum... hmmm

  4. Algebraic!
    • 2 years ago
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    x+y=100 x^2+y^2= something x^2 + (100-x)^2 = something differentiate set equal to zero

  5. Brent0423
    • 2 years ago
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    i know how to do the problem now.. i just feel like these problems are really complex

  6. RadEn
    • 2 years ago
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    x+y=100 ----> y=100-x k=x^2+y^2 k=x^2+(100-x)^2 k=x^2+10,000-200x+x^2 k=2x^2-200x+10,000 u can use derivative like @Algebraic! said

  7. Brent0423
    • 2 years ago
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    A rancher with 750ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. a. Find a function that models the total area of the four pens. b. Find the largest possible total area of the four pens.

  8. RadEn
    • 2 years ago
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    the 2nd problem ?

  9. Brent0423
    • 2 years ago
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    ya......

  10. RadEn
    • 2 years ago
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    |dw:1359869626673:dw| from the figure we can maka a models the total area of the four pens : 6x+4y=750 or 3x+2y=375 2y=375-3x y=(375-3x)/2 ...(1) the total area (A) = 4xy ....(2) subt (1) to (2) we get a function : A=4x(375-3x)/2 A=-6x^2+750x

  11. Brent0423
    • 2 years ago
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    and your trying to tell me that this is easy....... -_-

  12. RadEn
    • 2 years ago
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    ups, sorry mistake from m the total area (A) = 3xy A=3x(375-3x)/2 A=562.5x-4.5x^2

  13. RadEn
    • 2 years ago
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    it just a model only :) for solution u can use like the first problem

  14. RadEn
    • 2 years ago
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    btw, the first has solved ?

  15. akash_809
    • 2 years ago
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    the correct way is to diffentiate and equate it to zero ...But i think there is an easier method , since \[(x+y)^2=x^2+y^2+2xy\] so \[10000=x^2+y^2+2xy\] now from A.M \[\ge\] GM , so \[(x^2+y^2)/2 \ge \sqrt{x^2.y^2}\] and for \[x^2+y^2\] to be minimum A.M=G.M i.e x=y , so the numbers would be 50,50 ....

  16. akash_809
    • 2 years ago
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    @Brent0423 do u get it

  17. akash_809
    • 2 years ago
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    and do you have theanswer , is it 50,50 i gave

  18. akash_809
    • 2 years ago
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    you gave me the medal so i assume the answer is 50, 50 right ?

  19. Brent0423
    • 2 years ago
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    yes

  20. RadEn
    • 2 years ago
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    waw, AM-GM is one of the math olympiads material, but i like it i think for basic, we just do it by use simple method... any way, may be u can use this formula : |dw:1359871521108:dw|

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