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Math Help - Business Calculus HW Minimizing and maximizing problems

  1. #1
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    Business Calculus HW Minimizing and maximizing problems

    1. Minimizing Construction Costs The management of the UNICO department store has decided to enclose an 800-ft squared area outside the building for displaying potted plants and flowers. One side will be formed by the external wall of the store (three sides), two sides will be constructed of pine boards, and the fourth side will be made of galvanized steel fencing material. If the pine board fencing costs $6/running foot and the steel fencing costs $3/running foot, determine the dimensions of the enclosure that can be erected at minimum cost.


    2. Minimizing Packaging Costs A rectangle box is to have a square base and a volume of 20-ft cubed. If the material for the base costs $0.30/square foot, the material for the sides costs $0.10/square foot, and the material for the top costs $0.20/square foot, determine the dimensions of the box that can be constructed at minimum costs.

    3. Parcel Post Regulations Postal regulations specify that a parcel sent by parcel post may have a combined length and girth of no more than 108 inches. Find the deminsions of a rectangular package that has a square cross section and the largest volume that my be sent through the mail. What is the volume of such a package?

    Hint: The length plus the girth is 4x + h

    Thank you
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  2. #2
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    1. Minimizing Construction Costs The management of the UNICO department store has decided to enclose an 800-ft squared area outside the building for displaying potted plants and flowers. One side will be formed by the external wall of the store (three sides), two sides will be constructed of pine boards, and the fourth side will be made of galvanized steel fencing material. If the pine board fencing costs $6/running foot and the steel fencing costs $3/running foot, determine the dimensions of the enclosure that can be erected at minimum cost.
    So we have the equations

    A=l \cdot w \iff 800= l \cdot w \iff w=\frac{800}{l}

    P=2l+w where l is the side made out of pine, and w steel.

    So the cost function is

    C= 6(2l)+3(w) using the last equation we get...

    C=12l +3 \cdot \frac{800}{l} taking the derivative with respect to l and setting equal to zero we get ...

    \frac{dC}{dl}=0=12-\frac{2400}{l^2} clearing the denominators

    0=12l^2-2400=12(l^2-200)

    This gives the solution

    l = \pm \sqrt{200} \approx \pm 14.14 using only the postivtive solution and the first equaiton

    w=\frac{800}{l}=\frac{800}{10 \sqrt{2}}=\frac{80}{\sqrt{2}}=40\sqrt{2} \approx 56.57
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  3. #3
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    Hello, stepho_145!

    Here's the first one . . .


    1. Minimizing Construction Costs
    The UNICO department store has decided to enclose an 800 ft² area
    outside the building. One side will be formed by the external wall of the store;
    two sides will be made of pine boards, and the fourth side will be galvanized steel fencing.
    If the pine boards cost $6/ft and the steel fencing costs $3/running foot,
    determine the dimensions of the enclosure that can be erected at minimum cost.
    Code:
        - * - - - - - - - * -
          |               |
          |               |
        y |               | y
          |               |
          |               |
          * - - - - - - - *
                  x

    The area will be 800 ft².
    We have: . xy \:=\:800\quad\Rightarrow\quad y \:=\:\frac{800}{x} .[1]

    The cost will be: . \begin{array}{ccc}x\text{ ft of steel fencing at \$3/ft} & = &3x\text{ dollars} \\<br />
2y\text{ ft of pine fencing at \$6/ft} &=& 12y\text{ dollars} \end{array}

    Hence, the total cost is: . C \;=\;3x + 12y .[2]

    Substitute [1] into [2]: . C \;=\;3x + 12\left(\frac{800}{x}\right)


    Therefore: . C \;=\;3x + 9600x^{-1}

    . . and that is the function we must minimize.


    Go for it!

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