A hiker starting at point P on a straight road wants to reach a forest cabin that is 2 km from a point Q, 3 km down the road from P . She can walk 8 km/hr along the road but only 3 km/hr through the forest. She wants to minimize the time required to reach the cabin. How far down the road should she walk before setting off through the forest straight for the cabin?

2.19 km

Step-by-step explanation:

If x is the distance she walks down the road before turning, then the total time is:

t = x/8 + √((3 − x)² + 2²) / 3

t = x/8 + √(9 − 6x + x² + 4) / 3

24t = 3x + 8√(13 − 6x + x²)

24t = 3x + 8(13 − 6x + x²)^½

Take derivative of both sides with respect to x.

24 dt/dx = 3 + 4(13 − 6x + x²)^-½ (-6 + 2x)

When t is a minimum, dt/dx = 0.

0 = 3 + 4(13 − 6x + x²)^-½ (-6 + 2x)

-3 = 4(13 − 6x + x²)^-½ (-6 + 2x)

3 / (6 − 2x) = 4(13 − 6x + x²)^-½

3 / (24 − 8x) = (13 − 6x + x²)^-½

(24 − 8x) / 3 = (13 − 6x + x²)^½

(24 − 8x)² / 9 = 13 − 6x + x²

576 − 384x + 64x² = 117 − 54x + 9x²

459 − 330x + 55x² = 0

Solve with quadratic formula.

x = [ 330 ± √((-330)² − 4(55)(459)) ] / 2(55)

x = (330 ± √7920) / 110

x = 2.19 or 3.81

Since 0 < x < 3, x = 2.19.

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