Question

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the

time after launch, x in seconds, by the given equation. Using this equation, find the
maximum height reached by the rocket, to the nearest tenth of a foot.
y=-16x2 + 200x + 82

Answer 1

Explanation:

dy/dx = -32x + 200

maximum height when dy/dx = 0

-32x + 200 = 0

x = 6.25 seconds

y = -16·6.25² + 200·6.25 + 82 = 707 feet