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Math Help - Word Problem#3

  1. #1
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    Word Problem#3

    A successful bake shop buys a very rich whipped-cream-covered cake from a supplier at $6.00 per cake and then sells the cakes for $10.00 each. The bake shop sells 200 cakes per week. Research shows that they can increase their profit if they increase the selling price. However, for every $0.50 increase, the sales will drop by five. Determine the selling price that will maximize profit.
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  2. #2
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    Quote Originally Posted by Chrissy
    A successful bake shop buys a very rich whipped-cream-covered cake from a supplier at $6.00 per cake and then sells the cakes for $10.00 each. The bake shop sells 200 cakes per week. Research shows that they can increase their profit if they increase the selling price. However, for every $0.50 increase, the sales will drop by five. Determine the selling price that will maximize profit.
    Hello,

    let x be the times which the price is increased. Then you get:

    new price: 10 + .5*x
    number of sold cakes: 200 - 5*x
    money to be paid for the cakes: (200-5*x)*6
    profit with respect to x : p(x)

    Now you can form an equation

    p(x) = (200-5*x)*(10+.5*x)-(200-5*x)*6

    p(x)=-2.5x^2+80x+800

    1. The graph of this function is a parabola where the vertex is at the maximum value. If you tranform this equation into the vertex form (?) of the parabola you'll get:
    p(x)=-2.5(x-16)^2+1440. That means: If the price is increased 16 times by 0.5 $ then the profit will be 1440$ instead of the initial 800 $.

    2. You'll get the maximum of p if the 1rst derivative equals zero:
    {dp(x) \over dx}=-5x+80 . Let the RHS be zero, then x = 16. Plug in this value for x into p(x) and you'll get 1440 $.

    Bon appetit!

    EB
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