Questions: An open-top box is constructed by starting with a square section of material 594 mm by 594 mm, removing a small square of material from each corner, and folding up each edge. See the diagram below. Determine the dimensions of such a box such that the volume is maximal. Determine the maximum volume of such a box. Volume = mm^3

Solution Steps

Let $w$ be the side length of the initial square sheet, which is 594 mm. Let $x$ be the side length of the small squares removed from each corner. The dimensions of the resulting open-top box are length $l = w - 2x$, width $b = w - 2x$, and height $h = x$. The volume of the box is given by $V = l \times b \times h = (w - 2x)^2 x$.

To maximize the volume, we need to find the critical points of the volume function. First, substitute $w = 594$ into the volume formula: $V(x) = (594 - 2x)^2 x = 4x^3 - 2376x^2 + 352836x$ Now, take the derivative of the volume function with respect to $x$: $V'(x) = 12x^2 - 4752x + 352836$ Set the derivative equal to zero to find the critical points: $12x^2 - 4752x + 352836 = 0$ Divide the equation by 12: $x^2 - 396x + 29403 = 0$ Solve the quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $x = \frac{396 \pm \sqrt{(-396)^2 - 4(1)(29403)}}{2(1)}$ $x = \frac{396 \pm \sqrt{156816 - 117612}}{2}$ $x = \frac{396 \pm \sqrt{39204}}{2}$ $x = \frac{396 \pm 198}{2}$ We have two possible solutions for $x$: $x_1 = \frac{396 + 198}{2} = \frac{594}{2} = 297$ and $x_2 = \frac{396 - 198}{2} = \frac{198}{2} = 99$. Since the side length of the initial square is 594, cutting out squares of side length 297 would mean cutting out the whole sheet, so we choose $x = 99$.

The dimensions of the box are: Length = $594 - 2(99) = 594 - 198 = 396$ mm Width = $594 - 2(99) = 594 - 198 = 396$ mm Height = $99$ mm The maximum volume is: Volume = $(396)(396)(99) = 15488816$ mm$^3$

Final Answer: