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Thread: Help with partial derivative word problem using multiple variables.

  1. #1
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    Help with partial derivative word problem using multiple variables.

    Question as is

    A grocery store carries two brands of frozen apple juice, a local brand that it obtains at the cost of 22 cents per can and a national brand that it obtains at the cost of 60 cents per can. The grocer estimates that if the local brad is sold for x cents per can and the national brand for y cents per can, approximately 70 - 5x+4y cans of the local brand and 80+6x-7y cans of the national brand will be sold each day. How should the grocer price each brand to maximize the profit from the sale of juice.

    So they've given us the cost of each brand and the demand function for each. The price for one effecting the other. I figured the first thing I should do is find the Total Revenue function with respect to both x and y. So I got

    R(x,y) = x(70-5x+4y) + y(80+6x-7y)

    I distribute and combined like terms.

    R(x,y) = -5x2+70x-7y2-80y+10xy

    now I think I need to take the partial derivative of the function with respect to x then again with respect for y then plug in the prices of .22 and .60 for x and y.

    Is this correct? Or do I need to take the second derivative of either of the partial derivatives then plug in the price?
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    Re: Help with partial derivative word problem using multiple variables.

    I don't know that I can replicate the method you are supposed to use but this is a standard maximization problem

    The grocer wants to maximize the profit. This is the total number of cans of each type of juice sold times their respective profits per can.

    $P(x,y) = (70-5x+4y)(x-22) + (80+6x-7y)(y-60) = -5 x^2+10 x y-180 x-7 y^2+412 y-6340$

    In order to find the maximum of this we set both partials to zero and solve for $x,y$

    $\dfrac{\partial P}{\partial x} =-10 x+10 y-180 = 0$

    $\dfrac{\partial P}{\partial y} = 10 x-14 y+412 = 0$

    solving we get

    $x = 40,~y=58$

    Now we have to check that this is in fact a maximum.

    We need to use the second partial test

    $H=\begin{pmatrix}P_{xx} &P_{xy} \\ P_{xy} &P_{yy} \end{pmatrix} = \begin{pmatrix}-10 &10 \\ 10 &-14 \end{pmatrix}$

    $D = Det(H) = 40 > 0$ so this point is actually either a minimum or a maximum

    $P_{xx}(40,58) = -10 < 0$ and thus this point is a maximum
    Thanks from XxXScorpionXxX
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    Re: Help with partial derivative word problem using multiple variables.

    I checked it against the answer key, you're correct. Thanks so much.
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    Re: Help with partial derivative word problem using multiple variables.

    Sorry one last question. But when I try to set them to zero and solve I end up canceling everything out. For

    -10x+10y-180=0

    I solved for y first and got y=x+18 then I plugged that it for why so that the problem was -10+10(x+18)-180=0 and it just cancels the entire thing out. How do I get x=40?
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    Re: Help with partial derivative word problem using multiple variables.

    okay well I got y somehow

    10x+10y=180
    10x+14y=412
    -----------------
    4y=232

    y=58

    but I'm not sure how to get x. Or if thats the correct way of getting y for that matter.
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    Re: Help with partial derivative word problem using multiple variables.

    Quote Originally Posted by XxXScorpionXxX View Post
    okay well I got y somehow

    10x+10y=180
    10x+14y=412
    -----------------
    4y=232

    y=58

    but I'm not sure how to get x. Or if thats the correct way of getting y for that matter.
    you messed up a sign

    $-10x + 10y = 180$

    $-10x +580 = 180$

    $10x=400$

    $x=40$
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