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Math Help - Elasticity question: Finding maximized price.

  1. #1
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    Elasticity question: Finding maximized price.

    The consumer demand curve for cases of gadgets is given by q = (90 - p)^2, where p is the price per case and q is the demand in weekly sales.

    a) If the gadgets cost $30 per case to produce, use calculus to determine the price that should be charged per case to maximize the weekly profit.

    I know how to find revenue but the $30 in the equation's throwing me off. On my answer key the answer is $50 per case but I keep coming up with $45. Here's what I am doing:

    R=pq
    p(90-p)
    (90p-p^2)
    (90-2p)
    -2p+90=0
    -2p=-90
    p=45

    I am sure it's something silly considering I am neglecting to incorporate the $30 into the picture but that's what 6 hours of studying at 3am will do to you!
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  2. #2
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    What you want to do is find the profit, subject to there being a positive demand (price<90) which would be

    Profit=Revenue per week - Cost per week

    <br />
P=pq-30q<br />

    <br />
P=p(90-p)^2-30(90-p)^2<br />

    <br />
P=p^3-210p^2+13500p-243000<br />

    Find stationary values by taking the first derivative of the profit function and setting it to 0, remembering the above condition then confirm which of the answers is a maximum by finding which would make the second derivative negative
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  3. #3
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    Quote Originally Posted by thelostchild View Post
    What you want to do is find the profit, subject to there being a positive demand (price<90) which would be

    Profit=Revenue per week - Cost per week

    <br />
P=pq-30q<br />

    <br />
P=p(90-p)^2-30(90-p)^2<br />

    <br />
P=p^3-210p^2+13500p-243000<br />

    Find stationary values by taking the first derivative of the profit function and setting it to 0, remembering the above condition then confirm which of the answers is a maximum by finding which would make the second derivative negative
    Ugh, it's late. Okay, so if I take P'(x) I get 3p^2-410p+13500 = 0, correct? Forgive me, but refresh me on how exactly to find the specific stationary values. Also, if I take the derivative, again, of the above profit function it would be 6p-410=0; is that right?
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  4. #4
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    It must be late, you differentiated it incorrectly it's

    <br />
P'(p)=3p^2-420p+13500=0

    <br />
p^2-140p+4500=0

    <br />
(p-90)(p-50)=0

    so the stationary values would be p=90 and p=50

    if you plug them into the second differential P''(p)=6p-420

    you find that its negative for p=50 and positive for p=90

    so it's maximised for p=50
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