Optimization help

[jason0]

I can't figure out how to get a cost model, any help would be appreciated.

Manufacturing Co. will produce a box whose length is 3 times the base of the box. the material used to build the top and bottom of the box costs $10 per sq. ft and the material used to build the sides costs $6 per sq. ft. If the box must be 50 cubic feet find the dimensions that will minimize the cost.

Thanks

 

[soroban]

Hello, jason0!

Manufacturing Co. will produce a box whose length is 3 times the base of the box.
The material used to build the top and bottom of the box costs $10/sq.ft,
and the material used to build the sides costs $6/sq.ft.
If the box must be 50 cubic feet, find the dimensions that will minimize the cost.

         *- - - - - - -*
        /             /|
       /             / |y
      * - - - - - - *  |
      |             |  *
    y |             | /
      |             |/ x
      * - - - - - - *
            3x

\(\displaystyle \text{The width is }x\text{, the length is }3x\text{, the height is }y.\)

\(\displaystyle \text{The volume is 50: }\;(3x)(x)(y) \:=\:50 \quad\Rightarrow\quad y \:=\:\frac{50}{3x^2}\) .[1]


Now we want the Cost Function . . .

. . \(\displaystyle \begin{array}{cccc}& \text{Area} & \text{Price/ft}^2} & \text{Cost} \\ \hline \text{Top/bottom} & 2(3x)(x) & 10 & 60x^2 \\ \text{Left/right} & 2(x)(y) & 6 & 12xy \\ \text{Front/back} & 2(3x)(y) & 6 & 36xy \\ \hline & & \text{Total:} & 60x^2+48xy \end{array}\)

\(\displaystyle \text{We have: }\;C \;=\;60x^2 + 48xy\) .[2]


Substitute [1] into [2]:

. . \(\displaystyle C \;=\;60x^2 + 48x\left(\frac{50}{3x^2}\right) \quad\Rightarrow\quad C \;=\;60x^2 + 800x^{-1}\)

Go for it!