what is the minimum value of X

Question

in golf a typical hole is about 400-yards-long, while the cup measures a 4.25 inches in diameter. Suppose that, with every swing, you hit the ball X percent closer to the center of the hole.

For example, if X were 75 percent, then with every swing the ball would be four times closer to the hole than it was previously.

For a 400-yard hole, assuming there are no hazards (water, sand or otherwise) in the way, what is the minimum value of X so that you’ll shoot par, meaning you’ll hit the ball into the cup in exactly four strokes?

Answer 1

Hi Emily:

Since we get 75% closer to the hole on each stroke, the distance to the hole falls geometrically.

Also, to reach the hole, the ball must ≤ the radius of the hole from the pin ⇒ it must be within 4.25/2 = 2.125 inches from the pin.

We can now write the inequality: X⁴ (400) ≤ 2.125 / 36 . . . the 36 is to convert inches to yards

⇒ X⁴ ≤ 2.125 / (36 x 400)

⇒ X⁴ ≤ 0.00015

Taking logs on both sides: 4 log X ≤ log (0.00015)

So, log X ≤ log (0.00015) / 4 = -0.956

∴ X ≤ 0.11 = 11%

So X must be ≤ l11% for this to work.

Answer 2

If D is the distance in yards from the hole, then D - aD is the distance the first swing is from the hole where a is the value desired. The next swing gets the ball (D-aD) - a(D-aD) yards from the hole. Repeating this process and additional two times yields the formula D x (1 -a)⁴ as the final distance from the hole after 4 swings. The calculation using the formula must be ≤ 4.25 inches. This looks like this:

(1 - a) = (4.25/14400)^1/4 where 400 yds x 36" = 14400

solve for a to get a = 0.869

Put this number into the formula to show that the ball is within 4.25 " of the hole.