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Find the largest box that can be made from a fixed amount of material

Find the largest open box (i.e., no material covering the top face) that can be made from a rectangular piece of material by removing squares at each corner and turning up the sides in the following cases:

  1. the rectangular material has edges 10 and 10;
  2. the rectangular material has edges 12 and 18.

  1. We begin with a rectangle with sides each of length 10. The picture is:

    Rendered by QuickLaTeX.com

    So, the edges of the base of the box are each of length l = w = 10 - 2x, and the height is x; therefore, the volume is

        \[ V = x(10-2x)^2 = 4x^3 - 40x^2 + 100x. \]

    Taking the derivative,

        \[ \frac{dV}{dx} = 12x^2 - 80x + 100. \]

    Setting this equal to zero and solving,

        \[ 12x^2 - 80x + 100 = 0 \quad \implies \quad x = 5 \text{ or } \frac{5}{3}. \]

    Then looking at the sign of the derivative around these critical points we have

        \begin{align*}  \frac{dV}{dx} &> 0 & \text{when} &&  0 &< x < \frac{5}{3} \\  \frac{dV}{dx} &< 0 & \text{when} && \frac{5}{3} &< x < 5. \end{align*}

    Therefore, V is increasing on 0 < x < \frac{5}{3} and decreasing on \frac{5}{3} < x < 5. Hence, V has a maximum at x = \frac{5}{3}. Solving for the length and width of the base of the box we then have

        \[ l = w = 10 - 2x = 10 - 2 \left( \frac{5}{3} \right) = \frac{20}{3}. \]

  2. This time we begin with a rectangle with length 18 and width 12. The picture is as follows:

    Rendered by QuickLaTeX.com

    So, the edges of the base of the box have lengths w = 12-2x and l = 18-2x, and the height of the box is x. Thus, the volume is

        \[ V = x (12-2x)(18-2x) = 4x^3 - 60x^2 + 216x. \]

    Taking the derivative,

        \[ \frac{dV}{dx} = 12x^2 - 120x + 216. \]

    Setting this equal to 0 we obtain the critical points,

        \[ 12x^2 - 120x + 216 = 0 \quad \implies \quad x = 5 - \sqrt{7} \text{ or } 5 + \sqrt{7}. \]

    Then,

        \begin{align*}  \frac{dV}{dx} &>0 & \text{when} && 0 &< x < 5 - \sqrt{7} \\  \frac{dV}{dx} &<0 & \text{when} && 5-\sqrt{7} & < x < 5 + \sqrt{7}. \end{align*}

    Hence, V has a maximum at x = 5 - \sqrt{7}. Solving for l and w we then have

        \[ l = 18-2(5 - \sqrt{7}) = 8 + 2 \sqrt{7}, \qquad w = 12 - 2(5-\sqrt{7}) = 2 + 2 \sqrt{7}. \]

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