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=−16x
2
+261x+130

Answer 1

SOLUTION

TO DETERMINE

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.

$\displaystyle \sf{y = - 16 {x}^{2} + 261x + 130 }$

Differentiating both sides with respect to x two times we get

$\displaystyle \sf{ \frac{dy}{dx} = -32x + 261 }$

$\displaystyle \sf{ \frac{ {d}^{2} y}{d {x}^{2} } = -32 }$

For extremum value of y we have

$\displaystyle \sf{ \frac{dy}{dx} = 0 }$

$\displaystyle \sf{ \implies \: -32x + 261 = 0 }$

$\displaystyle \sf{ \implies \: 32x = 261 }$

$\displaystyle \sf{ \implies \: x = 8.16 \: (approx) }$

Now

$\displaystyle \sf{ \frac{ {d}^{2} y}{d {x}^{2} } \bigg|_{x = 8.16} = -32 < 0 }$

So at x = 8.16 , maximum value of y occurs

Hence the required maximum value of y

$\displaystyle \sf{ = y \big|_{x = 8.16} }$

$\displaystyle \sf{ = - 16 \times {(8.16)}^{2} + (261 \times 8.16) + 332 }$

$\displaystyle \sf{ = 1194.39 }$

≈ 1190 foot ( nearest tenth)

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. a butterfly is moving in a straight line in the space.

let this path be denoted by a line l whose equation is x-2/2=2-y...

https://brainly.in/question/30868719

2. Find the distance of point p(3 4 5) from the yz plane

https://brainly.in/question/8355915