Page 35 - How To Solve Word Problems In Calculus

(x, y)
10
y
x x

√
2 2 2
A = 2xy. Since x + y = 100, y = 100 − x . By substitution,
√
2
A = 2x 100 − x . Since the point on the circle (x, y) was selected
in the first quadrant, 0 ≤ x ≤ 10. The area function is

A(x) = 2x 100 − x 2 0 ≤ x ≤ 10

5. Let l , w, and h represent the length, width, and height, respectively,
of the resulting open box. V = lw h.

x x
x x
8 w

x x
x x
l
12

Since the length and width of the boxwill be the corresponding
dimensions of the sheet metal diminished by 2x, l = 12 − 2x,
w = 8 − 2x, and the height of the boxwill be just x itself. By
substitution,

V = (12 − 2x)(8 − 2x)x

2
V (x) = 96x − 40x + 4x 3
The values of x are restricted by the size of the smaller dimension
of the rectangle: 0 ≤ x ≤ 4.

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