8. -/1 points details my notes tanapcalc10 4.4.045.mi. flight of a rocket the altitude in feet attained by a…

8. -/1 points details my notes tanapcalc10 4.4.045.mi. flight of a rocket the altitude in feet attained by a model rocket t seconds into flight is given by the function h(t). find the maximum altitude (in ft) attained by the rocket. (round your answer to the nearest foot.) h(t)= - 1/3t³ + 2t² + 12t + 21 (t ≥ 0) ft need help? read it master it submit answer

8. -/1 points details my notes tanapcalc10 4.4.045.mi. flight of a rocket the altitude in feet attained by a model rocket t seconds into flight is given by the function h(t). find the maximum altitude (in ft) attained by the rocket. (round your answer to the nearest foot.) h(t)= - 1/3t³ + 2t² + 12t + 21 (t ≥ 0) ft need help? read it master it submit answer

Answer

Explanation:

Step1: Find the derivative of $h(t)$

Using the power - rule $\frac{d}{dt}(t^n)=nt^{n - 1}$, we have $h'(t)=-t^{2}+4t + 12$.

Step2: Set the derivative equal to zero

$-t^{2}+4t + 12 = 0$. Multiply through by - 1 to get $t^{2}-4t - 12=0$. Factor the quadratic equation: $(t - 6)(t+2)=0$. So $t = 6$ or $t=-2$. Since $t\geq0$, we discard $t=-2$.

Step3: Determine if $t = 6$ is a maximum

Find the second - derivative $h''(t)=-2t + 4$. Evaluate $h''(6)=-2\times6 + 4=-8<0$. Since $h''(6)<0$, $t = 6$ is a point of maximum.

Step4: Find the maximum altitude

Substitute $t = 6$ into the original function $h(t)=-\frac{1}{3}t^{3}+2t^{2}+12t + 21$. $h(6)=-\frac{1}{3}(6)^{3}+2(6)^{2}+12\times6 + 21$. $h(6)=-\frac{1}{3}\times216+2\times36 + 72+21$. $h(6)=-72 + 72+72+21$. $h(6)=93$.

Answer:

93