Results 1 to 8 of 8

Math Help - Calculus III maximum value problem

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    11

    Calculus III maximum value problem

    Find three positive numbers a,b,and c such that
    13*a+5*b+2*c=172 and such that the product of a,b, and is is a maximum
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    I would use your addition equation to solve for one parameter. Plug that equation into your product. Then what do you suppose you should do?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,713
    Thanks
    1472
    Another way to do this is to use "Lagrange multipliers".

    You want to maximize f(a,b,c)= abc subject to the constraint that g(a,b,c)= 13a+ 5b+ 2c= 172, a constant. At a minimum or maximum point, the gradients will be parallel:
    \nabla f= \lambda \nabla g for some constant \lambda.

    \nabla (abc)= bc\vec{i}+ ac\vec{j}+ ab\vec{k}= \lambda\nabla(13a+ 5b+ 2c)= 13\lambda\vec{i}+ 5\lambda\vec{j}+ 2\lambda\vec{k}
    so that we have bc= 13\lambda, ac= 5\lambda, and ab= 2\lambda where " \lambda" is the "Lagrange multiplier".

    Since the precise value of \lambda is not necessary for the solution of Lagrange multiplier problems, I find it usually best to start by eliminating \lambda by dividing one equation by another. For example, dividing the first equation by the second, \frac{bc}{ac}= \frac{b}{a}= \frac{13}{5} so that b= \frac{13}{5}a. Similarly, dividing the first equation by the third, \frac{bc}{ab}= \frac{c}{a}= \frac{13}{2} so that c= \frac{13}{2}a.

    Put those into the constraint, 13a+ 5b+ 2c= 172, to get an equation you can solve for a.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member mfetch22's Avatar
    Joined
    Feb 2010
    From
    Columbus, Ohio, USA
    Posts
    168
    Quote Originally Posted by Jessica11 View Post
    Find three positive numbers a,b,and c such that
    13*a+5*b+2*c=172 and such that the product of a,b, and is is a maximum
    Listen to Ackbeet.

    EDIT=You can also do it the way Hallsofivy showed above=EDIT

    Write out the equations, solve for a variable and substitute. If your in calc three then thats the easy part. This is what you have:

    P =ab

    you want to maximize P, and you know that you want to express it in terms of the other equation (this ensures that you've "mixed" the two restricting equations in your problem). We also have:

    13a+5b+2c=172

    Pick a or b and explicitly define one (in terms of b and c, or a and c; depending on which you pick) by moving around variables and numbers in the equaiton above. Substitute it into the P=ab equation, then simplify. I'm assuming since your in calc 3 your intructor might have shown you an example? Do you understand where to go next? Give the algebra a try first, post back if your truly still stuck
    Last edited by mfetch22; June 23rd 2010 at 11:02 AM. Reason: Additional Post in thread to adress
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2009
    Posts
    11
    i actually just realized i made a typo it was supposed to be the product of a,b, and c is a maximum so would that mean that p=abc instead of p=ab? if so i solved for b and got
    p=(ac/5)*(172-13*a-2*c) but i have no idea what to do next?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Take the partial derivative with respect to a, and the partial derivative with respect to c. Set them both equal to zero, and solve for a and c.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member mfetch22's Avatar
    Joined
    Feb 2010
    From
    Columbus, Ohio, USA
    Posts
    168
    Quote Originally Posted by Ackbeet View Post
    Take the partial derivative with respect to a, and the partial derivative with respect to c. Set them both equal to zero, and solve for a and c.
    Just alittle add on here, incase you didnt know. But this is the higher dimensional analog to finding the critical points of a single variable function. This is how you find the critical points of a 3-dimensional surface. At critical points, there exist four possible occurances (from my understanding).

    [1] There can be a minimum at a critical point

    [2] There may be a maximum at a critical point

    [3] There may be a saddle at a critical point (i.e. if you draw two perpendicular axes at this point, one axis would slope downward in both directions, and the other would slope upaward in both directions)

    [4] There may be nothing that you can determine about the behavior of the graph at that critical point (i.e. if the determinant of the square matrix of partial derivitives is equal to zero)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    All true. And I also forgot to add that you need to check the boundaries of the allowable region.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: December 4th 2011, 03:30 PM
  2. Replies: 3
    Last Post: February 27th 2011, 03:28 AM
  3. Applications of Calculus - maximum height
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 24th 2010, 03:42 AM
  4. Solve maximum problem without using calculus
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: June 28th 2009, 05:02 AM
  5. Maximum profit - Calculus (urgent!)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 31st 2008, 12:02 AM

Search Tags


/mathhelpforum @mathhelpforum