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    RARORA's Avatar
    RARORA Posts: 6, Reputation: 1
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    #1

    Oct 7, 2006, 10:12 AM
    Algebra-maxima minima
    A man sells biscuits in boxes of different sizes. The biscuits are priced at Rs 2 per biscuit up to 200 biscuits. For every additional 20 biscuits in a box, the price of the whole lot goes down by 10 paise per biscuit. What should be the size of the box, that would maximize the revenue?
    kyop's Avatar
    kyop Posts: 48, Reputation: 11
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    #2

    Nov 15, 2006, 11:37 AM
    A man sells biscuits in boxes of different sizes. The biscuits are priced at Rs 2 per biscuit up to 200 biscuits. For every additional 20 biscuits in a box, the price of the whole lot goes down by 10 paise per biscuit. What should be the size of the box, that would maximize the revenue?

    The price of a box of biscuits is the price times the number in the box. It's safe to assume that the linear value associated with the number times the price will not be at a maximum when we have less than 200 biscuits. So, since the price decreases $0.10 for every 20 new bisquits, we can make two different expressions:

    (2 - .1x) for the price and (200 + 20x) for the number. (If you play with different values of x, you will see that this works)

    So, your function is:

    f(x) = (2 - .1x)(200 + 20x)

    this gives:

    f(x) = 400 + 20x - 2x^2

    Now, the maximum value will happen at -b/(2a), so -20/2(-2) or x=5.

    Plug in 5 for x and you get 400 + 20(5) - 2(5)^2 or 400 + 100 - 50 or 450.

    So, the maximum revenue would be 450 and the size of the box would be 300 biscuits.

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