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Matrices & word problem

  1. Jan 11, 2007 #1
    Hootenanny, I have uninstalled nexus skin...I guess it did contribute to the fact I couldn't use the site's template!

    1. The problem statement, all variables and given/known data
    Note: This problem comprises many parts, but I will just skip to the last one, since it's the one I'm struggling to answer!

    An economist producing x thousand items attempts to model a profit function as a quadratic model P(x) = ax²+bx+c thousand dollars.

    Given info: a=t, b=3-5t, c=5+4t = possible solutions for the system

    part c asks us to find the "actual profit function" when the profit for producing 2500 items is 19,750 dollars (Note: 19750 becomes 19.75 when we plug into the function)...

    Then part d asks us what is the max profit to be made and what level of production is needed to acheive it?

    2. Relevant equations



    3. The attempt at a solution

    After substitution, the function for part c is as follows:
    p(x)= (-29/9)x²+(172/9)x-(71/9) thousand dollars
    such that 19.75= (-29/9)x²+(172/9)x-(71/9) thousand dollars
    I'm guessing we have to use this formula to answer part d...I have solved x, where, in this case, x=0.44 or 5.48
    Yet, I don't think these x-values will help me solve part d?!
     
  2. jcsd
  3. Jan 11, 2007 #2

    berkeman

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    Staff: Mentor

    Have you learned how to use differentiation to find the max and min of a function? BTW, what is "t"?
     
  4. Jan 11, 2007 #3
    a=t=(-29/9)...hmm we have done some calculus...but we haven't applied it yet to functions in math. In physics, I have applied calculus when dealing with acceleration and velocity. I'm assuming it's the same idea.
     
    Last edited: Jan 11, 2007
  5. Jan 11, 2007 #4

    berkeman

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    The derivative of a function is the slope of the function at each point. So for example, if y = -2x^2, then dy/dx = -4x. So if you plot y(x) and dy/dx, you will see that the function y(x) is an upside-down parabola centered on the origin, and the slope function dy/dx is positive for -x and negative for +x, and it is zero at the origin where the function is maximum.

    So the general technique for finding maxima and minima of a function is to take the deriviate of the function and set that equal to zero. Solving that equation gives you all the values of x where the function has a max or min. You then either have to plot the function to see if it's a max or min, or else take the second derivative to see if the funtion has positive curvature (like a cup shape) or negative curvature (like an umbrella shape) at each max/min. Make sense?
     
  6. Jan 11, 2007 #5
    This part is rather confusing "So if you plot y(x) and dy/dx, you will see that the function y(x) is an upside-down parabola centered on the origin, and the slope function dy/dx is positive for -x and negative for +x, and it is zero at the origin where the function is maximum"
    But I understood the rest!
    Based on your comment, I can simply graph the quadratic equation (0= (-29/9)x²+(172/9)x-(995/36)) then find the coordinates of the peak which represent the max. profit. I plug in the values which will enable me to find the max. profit! Using differentiation, the equation will be: (-58/9)x+(172/9)...right?
     
  7. Jan 11, 2007 #6
    Thank you I found the answer graphically, where the number of products equals about 2970 and max. profit= 20,448 dollars. But can you please elaborate on the differentation concept?
     
  8. Jan 11, 2007 #7

    berkeman

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    Sure. Check out "Using Derivitaves to Graph Functions" at this wikipedia.org page.

    http://en.wikipedia.org/wiki/Derivative
     
  9. Jan 11, 2007 #8

    berkeman

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    Looks like the number you got graphically also. Good job.
     
  10. Jan 11, 2007 #9
    While calculus is a good tool for solving this; it seems as if he's not learning Calculus at the moment. Rather, Algebra or Pre-Calculus.

    Thus, we must use matrices and other forms to solve the answer =).

    One way I'm thinking is to set it up in a matrix and then row reference to solve for the variables.

    There's also another way in which you plot two graphs and the intersection of these two graphs produces the maximums =).
     
    Last edited: Jan 11, 2007
  11. Jan 11, 2007 #10

    berkeman

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    I'm not familiar with those methods. Can you provide an example or a web pointer? I'd be interested in seeing how they work.
     
  12. Jan 11, 2007 #11
    The simplest way, especially if youre not familliar with calc, is to find the axis of symmetry and plug that in to get the y coordinate of the vertex. The calc comes in handy for higher degree polynomials
     
  13. Jan 12, 2007 #12

    HallsofIvy

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    Since the question in question is a parbola, you don't need to use calculus: complete the square to find the vertex of the parabola.
     
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