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Find the minimum of c(3a+4b) when a^2+b^2 +c^2=1.

  1. May 14, 2017 #1

    ssd

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    1. The problem statement, all variables and given/known data

    Given a,b,c real and a^2+b^2 +c^2=1, to find the minimum of c(3a+4b)
    2. Relevant equations


    3. The attempt at a solution
    No positive clue yet.
     
  2. jcsd
  3. May 14, 2017 #2
    What is the condition for a point [itex](a,b,c)[/itex] to be the minimum point of a function?
     
  4. May 14, 2017 #3

    phyzguy

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    I suggest you look up Lagrange's method of undetermined multipliers. It is a good method for solving problems of this type. Have you studied it in your class?
     
  5. May 14, 2017 #4

    Ray Vickson

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    For each choice of ##a## and ##b## there are two choices for ##c##:
    $$c = + \sqrt{1-a^2-b^2}\\
    c = - \sqrt{1-a^2-b^2}
    $$
    Each choice of ##c = c(a,b)## gives you a function ##F(a,b) = (3a + 4b) c(a,b)## to minimize. This new problem does not have an equality constraint anymore, but it still has a restriction ##a^2 + b^2 \leq 1##, which you can try to ignore when making your first solution attempt
     
  6. May 14, 2017 #5

    ssd

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    Thanks all for your kind suggestions. The Lagrangian multiplier solves the problem.
     
  7. May 14, 2017 #6

    mfb

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    I would introduce new coordinates:
    $$x=\frac {3a+4b}{5}$$
    $$y=\frac {4a-3b}{5}$$
    This is just a rotation in the a,b plane.
    Now we have to minimize ##5cx## with the constraint ##c^2+x^2+y^2=1##. As you found an answer already: By inspection, we see that ##c=-x=\pm \frac{1}{\sqrt 2}## leads to a minimum. Alternatively, introduce a second set of coordinates for the sum and the difference of c and x to make it a one-dimensional problem.
     
  8. May 19, 2017 #7

    ssd

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    Thanks mfb. Nice approach.
     
  9. May 19, 2017 #8

    epenguin

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    If that solves the problem what is the solution please?

    I have some doubts, whether e.g. for the problem to make sense a, b, c should be restricted to be positive, or ask minimum absolute value or that your minimum is a a calculus minimum not the least value.
     
    Last edited: May 19, 2017
  10. May 19, 2017 #9

    ssd

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    The result is a=-1/2, b=-1/2, c=1/√2 as per my calculations.
     
    Last edited: May 19, 2017
  11. May 19, 2017 #10

    phyzguy

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    I think you made a mistake. I agree that c^2 = 1/2. The value of c(3a+4b) for your solution is -7√2/4 = -2.47. There is another condition where c(3a+4b) = -2.5. Why don't you walk us through your LaGrange multiplier calculation and let's see where you went wrong.
     
  12. May 19, 2017 #11

    ssd

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    Surely I shall check my calculations. Notwithstanding though, when it is -2.5?
     
  13. May 19, 2017 #12

    mfb

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    At ##c=-x=\pm \frac{1}{\sqrt 2}## and ##y=0## using x and y as defined in post 6. a and b will have different magnitudes at the minima.
     
  14. May 19, 2017 #13

    phyzguy

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    Since this is in the homework section, I can't give you my solution. Like I asked, walk me through how you arrived at your solution and we will critique it.
     
  15. May 20, 2017 #14

    ssd

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    Well, I got it. a=-3/5√2 and b= -4/5√2.
    Actually I mentally calculated the minimum of 3a+4√(1/2-a^2) is at a=1/2 and that was the mistake. Since, our minimum could be the min of 3a-4√(1/2-a^2).
     
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