**Unformatted text preview: **t congruent squares from each corner of the cardboard and then folding the resulting
tabs. Let x denote the length of the side of the square which is removed from each corner.
x x x x 12 in
height
x x
x depth
width x 10 in 1. Find the volume V of the box as a function of x. Include an appropriate applied domain.
2. Use a graphing calculator to graph y = V (x) on the domain you found in part 1 and approximate the dimensions of the box with maximum volume to two decimal places. What is the
maximum volume?
Solution.
1. From Geometry, we know Volume = width × height × depth. The key is to now ﬁnd each of
these quantities in terms of x. From the ﬁgure, we see the height of the box is x itself. The
cardboard piece is initially 10 inches wide. Removing squares with a side length of x inches
from each corner leaves 10 − 2x inches for the width.5 As for the depth, the cardboard is
initially 12 inches long, so after cutting out x inches from each side, we would have 12 − 2x
inches remaining. As a function6 of x, the volume is
V (x) = x(10 − 2x)(1...

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