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Minima of a function

  1. Aug 30, 2011 #1
    1. The problem statement, all variables and given/known data
    This is question is from my test paper.
    Find the minimum value of 9tan2x+4cot2x.
    I wasn't able to solve it during my examination.
    But later i remembered that i can find its minima which unfortunately my teacher still hasn't taught. :(
    I have a very little knowledge of it so my attempts maybe wrong.


    2. Relevant equations



    3. The attempt at a solution
    I substituted y=tan2x

    [tex]\frac{d}{dy}(f(y))=9-\frac{4}{y^2}[/tex]
    [tex]\frac{d^2}{d^2y}(f(y))=\frac{8}{y^3}[/tex]

    [tex]9-\frac{4}{y^2}=0[/tex]
    I get y=2/3

    Substituting y=2/3 in the second derivative i get 27 but the answer is 12. :confused:
     
    Last edited: Aug 30, 2011
  2. jcsd
  3. Aug 30, 2011 #2

    NascentOxygen

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    Staff: Mentor

    This sub-forum is for problems that can be solved without calculus. Does your teacher expect you to arrive at the answer without using calculus?

    You seem to be saying the derivative of 9.tan2x = 9
    and the derivative of 4/(tan2x) = 4/(x2)

    Neither is correct. How did you calculate these?
     
    Last edited: Aug 30, 2011
  4. Aug 30, 2011 #3
    My teacher solved it without finding the minima but i still want to know where i am wrong. :confused:

    Please moderators,, move it to the right sub-forum. :smile:
     
  5. Aug 30, 2011 #4
    Didn't you see that i took tan2x=y?
     
  6. Aug 30, 2011 #5
    Bump :)
     
  7. Aug 30, 2011 #6

    Mark44

    Staff: Mentor

    I don't see any advantage in using the substitution you used. As far as I can tell, you need calculus techniques to find the minimum of the given function.

    You have y = 9tan2(x) + 4cot2(x)

    Take the derivative directly and set it to zero.

    You'll need to use a few basic identities, but otherwise the problem is pretty straightforward.
     
  8. Aug 30, 2011 #7
    I already tried that before but i was stuck after finding out the derivative:-
    [tex]\frac{dy}{dx}=18tan(x)sec^2(x)-8cot(x)cosec^2(x)[/tex]

    If i substitute the derivative to zero, then what i get is this:-
    [tex]18sec^4(x)-8cosec^4(x)[/tex]

    I don't understand what should i do? :confused:

    Could you please tell me where i am wrong in my method?
     
  9. Aug 30, 2011 #8

    Mark44

    Staff: Mentor

    The equation above is correct, but the one below is not.
    You are confused. tan(x) [itex]\neq[/itex] sec2(x) and cot(x) [itex]\neq[/itex] csc2(x). Those are derivative formulas (one of them has the wrong sign), but that's not what you're doing in this step.

    Rewrite all of the functions in terms of sines and cosines.

    Here's a start:
    18tan(x)/cos2(x) - 8/(tan(x)sin2(x)) = 0

    Now replace the tangent factors.
     
  10. Aug 30, 2011 #9
    I did as you said. :smile:
    What i now get is:-
    [tex]\frac{18sin^4(x)-8cos^4(x)}{cos^3(x)sin^3(x)}=0[/tex]

    Is it correct?
     
  11. Aug 30, 2011 #10

    Mark44

    Staff: Mentor

    Yes. You can simplify this to 18sin4(x) = 4cos4(x), and you can turn this into an equation involving the tan function.

    Notice that cos(x) cannot equal zero, nor can sin(x) = 0, but these are places where either tan(x) is undefined or cot(x) is undefined, so you're not going to get a minimum for your function at any of these points.

    One other thing: there are lots of solutions to the equation you're trying to solve.
     
  12. Aug 30, 2011 #11
    Is it 18sin4(x)=8cos4(x)?
    I get:-
    [tex]tan^4(x)=\frac{8}{18}[/tex].

    But then i would have to take the fourth root. Fourth root of 4 is [itex]\sqrt{2}[/itex] but then what is the fourth root of 3?
     
  13. Aug 30, 2011 #12

    Mark44

    Staff: Mentor

    Simplify to tan4(x) = 4/9, so tan2(x) = 2/3. (You can ignore the solution tan2(x) = -2/3.)

    So tan(x) = ? and hence x = ?
     
  14. Aug 30, 2011 #13
    tan(x)=[itex]\frac{\sqrt{2}}{\sqrt{3}}[/itex] and x=tan-1[itex]\frac{\sqrt{2}}{\sqrt{3}}[/itex]

    Right..?
     
  15. Aug 30, 2011 #14

    Mark44

    Staff: Mentor

    Partly. tan(x) = [itex]\pm \sqrt{2/3}[/itex]

    There is not just one solution, though - there are an infinite number of solutions.
     
  16. Aug 30, 2011 #15

    Ray Vickson

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    Homework Helper

    If you want to minimize f = 9y + 4/y without using calculus, here is how to do it: write f = w1*(9y/w1) + w2*((4/y)/w2), where w1, w2 > 0 and w1+w2=1. By the arithmetic-geometric inequality, w1*r1 + w2*r2 >= r1^w1 * r2^w2, with equality if and only if r1 = r2. So, with r1 = 9y/w1 and r2 = (4/y)/w2 we have f >= (9/w1)^w1*(4/w2)^w2*y^(w1-w2). Note that when w1 = w2 (so w1 = w2 = 1/2), we have f >= sqrt(18)*sqrt(8) = 12; that is, we *always* have f >= 12 for any y > 0. We get *equality* for r1 = r2, or 9y/w1 = (4/y)/w2, or 9y = 4/y (because w1 = w2 = 1/2). So, when 9y = 4/y (y = 2/3) we have f = 12, so that must be the minimum: f >= 12 for all y > 0 and we have achieved a value f = 12!

    In general, to minimize f = c1*y + c2/y with c1, c2 > 0 we must take both terms of f equal; that is, c1*y = c2/y, so y = sqrt(c2/c1), and f_min = 2*c1*sqrt(c2/c1) = 2*sqrt(c1*c2).

    Note: this type of thing is the basis of *Geometric Programming*, devised by Duffin, Peterson and Zener (yes, the Zener of diode fame).

    RGV
     
    Last edited: Aug 30, 2011
  17. Aug 31, 2011 #16
    tan(x)=[itex]\pm \sqrt{2/3}[/itex]. If i substitute tan(x)=[itex]\pm \sqrt{2/3}[/itex] in the given equation i get the answer as 12. Is this the minimum value?

    I don't get what do you mean by there are infinite solutions?
     
  18. Aug 31, 2011 #17

    Ray Vickson

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    Science Advisor
    Homework Helper

    How may values of x are there that satisfy the equation tan(x) = sqrt(2/3)? Note: there is only ONE minimum VALUE of f (namely, f_min = 12), but there may be more than one value of x that gives you that f-value.

    RGV
     
  19. Aug 31, 2011 #18
    More values of x include the x+2kpi, where k is an integer. :smile:
    Right..?
     
  20. Aug 31, 2011 #19

    Mark44

    Staff: Mentor

    Not quite. You're missing three-quarters of the values if you take multiples of 2[itex]\pi[/itex].
     
  21. Aug 31, 2011 #20
    So what should be it? :confused:
     
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