Arock formation is 3 miles due north of its closest point along a straight shoreline. a visitor is staying in a tent that is 4 miles west of that point. the visitor is planning to go from the tent to the rock formation. suppose the visitor runs at a rate of 6 mph and swims at a rate of 3 mph. how far should the visitor run to minimize the time it takes to reach the rock formation?

  about 2.268 miles

Step-by-step explanation:

Let x represent the distance the visitor should run. Then the distance he swims will be ...

  d = √((4 -x)² +3²) = √(x² -8x +25)

The total travel time is given by ...

  time = distance / speed

  time = x/6 +d/3

  time = x/6 +1/3√(x² -8x +25)

The time will be minimized at the value of x that makes the derivative zero.

  d(time)/dx = 0 = 1/6 +1/6(2x -8)/√(x² -8x +25)

  0 = √(x² -8x +25) +2x -8

  (8 -2x)² = x² -8x +25 . . . . . . subtract (2x-8), square both sides

  3x^2 -24x +39 = 0 . . . . . . subtract the right side

  x² -8x +13 = 0 . . . . . . . . . divide by 3

  (x -4)² = 3 . . . . . . . . . . . complete the square

  x = 4 -√3 ≈ 2.268 . . . . . . . . . . square root, add 4

The visitor should run about 2.268 miles to minimize the time.

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Additional comment

The ratio of swim rate in water to running rate is 3/6 = 1/2. The angle with respect to straight out from the shoreline will be arcsin(1/2) = 30°, so the distance from the point closest to the rock will be (3 mi)tan(30°) = 1.732 mi. The distance to run is 4 mi - 1.732 mi = 2.268 mi. The angle relation is the generic solution to a problem like this.

y = 0.77(x)

step-by-step explanation:

the jacket retains this value after 23% cut or retains 77% of original value. plug original 87 jacket price for x and the remaining value should be y.

3< x< 5

step-by-step explanation:

|x - 4| < 1

there are two solutions, one positive and one negative.   remember to flip the inequality when taking the negative solution

x-4 < 1       x-4 > -1

add 4 to each side

x-4+4 < 1+4       x-4+4 > -1+4

x< 5                 x> 3

3< x< 5