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Math  /  Calculus

Question8. [-/1 Points] DETAILS MY NOTES TANAPCALC10 4.4.045.MI. ASK YOUR TEACHER PRACTICE ANOTHER
Flight of a Rocket The altitude in feet attained by a model rocket tt seconds into flight is given by the function h(t)h(t). Find the maximum altitude (in ft) attained by the rocket. (Round your answer to the nearest foot.) h(t)=−13t3+4t2+20t+20(t≥0)h(t)=-\frac{1}{3} t^{3}+4 t^{2}+20 t+20 \quad(t \geq 0) \qquad ft Need Help? Read It Master It Submit Answer View Previous Question Question 8 of 10 VieynNext Question Home My Assignments Request Extension

Studdy Solution

STEP 1

1. The function h(t)=−13t3+4t2+20t+20 h(t) = -\frac{1}{3} t^3 + 4t^2 + 20t + 20 represents the altitude of the rocket in feet as a function of time t t in seconds.
2. We are looking for the maximum value of this function for t≥0 t \geq 0 .
3. The maximum value of a continuous function can be found by identifying critical points and evaluating the function at these points and endpoints.

STEP 2

1. Find the derivative of h(t) h(t) .
2. Determine the critical points by setting the derivative equal to zero.
3. Evaluate the function at the critical points and endpoints to find the maximum altitude.
4. Round the maximum altitude to the nearest foot.

STEP 3

Find the derivative of h(t) h(t) :
h(t)=−13t3+4t2+20t+20 h(t) = -\frac{1}{3} t^3 + 4t^2 + 20t + 20
The derivative h′(t) h'(t) is found using the power rule:
h′(t)=−t2+8t+20 h'(t) = -t^2 + 8t + 20

STEP 4

Determine the critical points by setting the derivative equal to zero:
−t2+8t+20=0 -t^2 + 8t + 20 = 0
This is a quadratic equation. We will solve for t t using the quadratic formula:
t=−b±b2−4ac2a t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=−1 a = -1 , b=8 b = 8 , and c=20 c = 20 .

STEP 5

Calculate the discriminant and solve the quadratic equation:
b2−4ac=82−4(−1)(20)=64+80=144 b^2 - 4ac = 8^2 - 4(-1)(20) = 64 + 80 = 144
t=−8±1442(−1) t = \frac{-8 \pm \sqrt{144}}{2(-1)}
t=−8±12−2 t = \frac{-8 \pm 12}{-2}
This gives us two solutions:
t1=−8+12−2=−2 t_1 = \frac{-8 + 12}{-2} = -2
t2=−8−12−2=10 t_2 = \frac{-8 - 12}{-2} = 10
Since t≥0 t \geq 0 , we discard t1=−2 t_1 = -2 .
So, t=10 t = 10 is a critical point.

STEP 6

Evaluate the function at the critical point t=10 t = 10 and at the endpoint t=0 t = 0 :
h(10)=−13(10)3+4(10)2+20(10)+20 h(10) = -\frac{1}{3}(10)^3 + 4(10)^2 + 20(10) + 20
h(10)=−13(1000)+400+200+20 h(10) = -\frac{1}{3}(1000) + 400 + 200 + 20
h(10)=−10003+620 h(10) = -\frac{1000}{3} + 620
h(10)=−333.33+620 h(10) = -333.33 + 620
h(10)=286.67 h(10) = 286.67
Rounding to the nearest foot gives 287 287 feet.
Now evaluate h(0) h(0) :
h(0)=−13(0)3+4(0)2+20(0)+20 h(0) = -\frac{1}{3}(0)^3 + 4(0)^2 + 20(0) + 20
h(0)=20 h(0) = 20

STEP 7

Compare the values at the critical point and the endpoint:
- h(10)=287 h(10) = 287 feet - h(0)=20 h(0) = 20 feet
The maximum altitude is 287 287 feet.
The maximum altitude attained by the rocket is 287 \boxed{287} feet.

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