Page 127 - How To Solve Word Problems In Calculus

The volume of the boxis V = lwh where l = 12 − 2x,
w = 12 − 2x, and h = x. Substituting,
2
V (x) = (12 − 2x) x
2
= (144 − 48x + 4x )x
2
= 144x − 48x + 4x 3
Since the size of the cutout cannot be negative and cannot exceed
6 inches (otherwise we would cut away more than we have),
0 ≤ x ≤ 6. We next compute the critical values.

V (x) = 144 − 96x + 12x 2

0 = 144 − 96x + 12x 2

0 = 12 − 8x + x 2

0 = (x − 2)(x − 6)
x = 6 x = 2

If x = 0or x = 6, V (x) = 0. The maximum volume occurs,
therefore, when x = 2. The length and width of the boxare each
12 − 2x = 8 inches and the height is 2 inches. The maximum
3
volume is 128 in .
7.

x h x

x
2
y y

x
The area of the window is the sum of areas of the rectangular
bottom and the triangular top. The area of the rectangle is xy and
√
3 2
the area of the triangle is x . The combined area of the
4
rectangle and triangle is then
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