Folding Up the Sides for an Open Box

The Problem A 9-by-12-inch piece of cardboard is to be made into an open box by cutting slits into square-shaped corners, folding up the sides, and gluing the tabs to the ends. What is the volume of the open box if the corner squares are 2 inches on a side  

The volume of the right rectangular prism that s formed is found with the formula V = Iwh. The height, h, of the prism is 2 inches — the size of the squares at the corners. The length of the prism is 12 inches minus the two squares in the corners 12-4 = 8 inches for the length. The width of the prism is 

In the calculus problem involving the open box, you get to determine the size of squares in the corners that gives you the largest possible volume. If the squares are small, then the box isn t very deep, but the area of the base is big. If the squares are large, then the box is deep, but the area of the base is small. Calculus allows you to balance the depth and the base area to find the best dimensions. In the next problem, you get to do somewhat the same process with a table of the possible values. 

The Problem A sandbox is to be formed from a 9 foot square piece of metal by cutting equal squares from the comers and folding up the sides. The edges are then welded together to form the box. What size squares should be cut from the corners to form a box that will hold the greatest amount of sand possible (have the greatest volume)  

determine the dimensions of the box in terms of the length in feet of the sides of the squares, x The height is, of course, x. The length and width are both 9 minus the two squares (one on either end) or 9 - 2x feet. The volume of the sandbox is V = Iwh = (9 - 2x)(9 - 2x)x. In Table 20-1, you see measures for x in half-foot increments and the resulting volumes. 

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