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Solve problems with limits or ?

  1. Mar 6, 2017 #1
    1. The problem statement, all variables and given/known data
    I have a series of problems
    1.problem. When this number is summed with its square. The summ is the least possible.
    2.problem Find positive number which substraction with its third power would be the greatest.
    3.problem Find to numbers which x+y=5 but multiplication is the least
    2. Relevant equations
    1)x+x^2=
    2)x-x^3=
    3)x+y=5, x*y=

    3. The attempt at a solution
    I suppose I should you use limits but cant figure out how ( what kind of limit i should use) Some-kind of help ,please?
     
  2. jcsd
  3. Mar 6, 2017 #2
    Do you know/ can you use derivatives?
     
  4. Mar 6, 2017 #3

    Mark44

    Staff: Mentor

    These problems have nothing to do with limits. Problems 1 and 3 can be done without calculus. Problem 2 can be done using calculus, or you can find an approximate solution by graphing an equation.

    You posted this in the Precalc section, so our assumption is that you are in a math class that isn't at the calculus level. What class are you currently in?
     
  5. Mar 6, 2017 #4
    Yes,I can
     
  6. Mar 6, 2017 #5
    I rely do not understand the distinction between precalc an calculus. I just assumed that this is not "advanced" mathematics.

    I know derivatives, integration and differential solutions. I just dont understand how to solve this kind of problems.
     
  7. Mar 6, 2017 #6

    Mark44

    Staff: Mentor

    Precalc includes algebra and trigonometry - math areas that are usually taught before ("pre-") calculus.

    Do you know how to use derivatives to find the minimum or maximum value of a function?

    I am moving this thread to the Calculus & Beyond section.
     
  8. Mar 7, 2017 #7
    Like Mark44 said this is then a simple maxima minima problem.
     
  9. Mar 7, 2017 #8
    Yes, thank you for hint and already solved these problems,but i have another one which is a bit confusing.

    I have to find 2 positive numbers x+y=10 in such way x^2/2 +y^3=0 is the least.
    I plotted both functions and they have a cross point only when y is negative (15;-5). Should i plot x+y=10 and derivative function of x^2/2 +y^3=0 ?
     
  10. Mar 7, 2017 #9

    Dick

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    Science Advisor
    Homework Helper

    I'm assuming the problem is to minimize ##x^2/2+y^3## (no ##=0##, that wouldn't make much sense). Since you know ##x+y=10## then you also know ##y=10-x##. Substitute that into the quadratic to get a function of the single variable ##x##.
     
    Last edited: Mar 7, 2017
  11. Mar 7, 2017 #10
    If i do this, I end up getting x =14,7 that means y= -4,7 which does not comply with the condition that both numbers should be positive.
     
  12. Mar 7, 2017 #11

    Dick

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    I don't think that is correct. Can you show your work?
     
  13. Mar 7, 2017 #12

    Mark44

    Staff: Mentor

    When you solve a quadratic equation, there will be two solutions. In this case, neither of the solutions is x = 14.7, so I concur with Dick's statement.
     
  14. Mar 7, 2017 #13
    You mean x(-x2+61/2 x+700)?
    If I solve -x2+61/2 x+700 I also get x values which do not comply with the condition
     
  15. Mar 7, 2017 #14

    Mark44

    Staff: Mentor

    Where did this come from?

    It would be helpful if you showed us all of your work.
    -x2+61/2 x+700 is not an equation, so you can't solve it. An equation has = between two expressions.
     
  16. Mar 7, 2017 #15

    Ray Vickson

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    That is not the solution. Start again.
     
  17. Mar 7, 2017 #16

    Mark44

    Staff: Mentor

    I'm pretty sure you have not given the problem correctly. Here's what I believe the problem actually is:

    Find two positive numbers x and y, whose sum is 10, for which ##\frac{x^2}2 + y^3## is the smallest.
    What you wrote -- "in such a way x^2/2 +y^3=0 is the least." -- is not possible. If x and y are positive numbers, then there is no way that ##\frac{x^2}2 + y^3## could be equal to 0.
     
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