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Thread: Word problems #2

  1. #1
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    Word problems #2

    I am aware it is late so i will keep this short as possible:
    1.According to postal regulations, the girth plus the length of a parcel sent by fourth-class mail may not exceed 108 in. What is the largest possible volume of a rectangular parcel with two square sides that can be sen by fourth-class mail? The answer says 11664in^3 = 6.75 ft^3.

    2.A poster is to contain 108 cm^2 of printed matter with margins of 6cm each at top and bottom and 2 cm on the sides. What is the minimum cost of the of the poster if it is to be made of material costing 20 cents/cm^2? ans. is $60
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  2. #2
    Senior Member tukeywilliams's Avatar
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    1. So $\displaystyle V = x^{2}y $ which we want to maximize. Then $\displaystyle 4x+y \leq 108 $. Note that girth is $\displaystyle 2(h+w) $ or in this case $\displaystyle 2(x+x) = 4x $. To get the maximum volume we set $\displaystyle 4x+y = 108 $. Then $\displaystyle y = 108 - 4x $ and $\displaystyle 0 \leq x \leq 27 $.

    Then $\displaystyle V = x^2(108-4x) $.
    $\displaystyle V'(x) = 216x-12x^2 $.

    $\displaystyle 0 = 216x-12x^2 = 12x(18-x) $.

    Critical values are $\displaystyle x = 0, \ x = 18 $. Checking the endpoints $\displaystyle 0,27 $ (volume is 0) we conclude that the maximum volume occurs when $\displaystyle x = 18 $.

    So $\displaystyle V_{\text{max}} = 11,644 \ \text{in}^3 $.

    2. So we want to minimize $\displaystyle C = 20(x-4)(y-12) $. Since $\displaystyle xy = 108 $, $\displaystyle y = \frac{108}{x} $. Then $\displaystyle C(x) = 20(x-4)\left(\frac{108}{x}-12 \right) $. So $\displaystyle C'(x) = \frac{2160}{x} - 240 - \frac{2160(x-4)}{x^2} $. Critical values are $\displaystyle 6,-6 $. We get $60 when we plug in $\displaystyle C(-6) $.
    Last edited by tukeywilliams; Jul 30th 2007 at 11:51 PM.
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