Page 135 - How To Solve Word Problems In Calculus

2
2
d(x) = x + (x − 1) 2
√
4
2
2
= x + x − 2x + 1
√
4
2
= x − x + 1
2
Rather than minimize d(x), we consider D = d , observing that the
minimum values of both d and D occur at the same value of x.
2
4
D(x) = x − x + 1
Since we are restricting our attention to the first quadrant,
0 ≤ x < ∞.
3
D (x) = 4x − 2x

3
0 = 4x − 2x
2
0 = 2x(2x − 1)
√
2
Either x = 0or2x − 1 = 0, in which case x = 1/ 2.
√
(x =−1/ 2 can be ignored.)
2

D (x) = 12x − 2

D (0) =−2 < 0 ⇒ relative maximum at 0

1 1
D √ = 4 > 0 ⇒ relative minimum at √
2 2

1 3
Since d(0) = 1 and d √ = , we conclude that the point
2 4

1 1
closest to (0, 1) in the first quadrant is √ , . The
2 2

1 1
corresponding point in the second quadrant is −√ , .
2 2
14.
x
1"
x − 3
y 2" y − 2 1"

1"
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