Answer:
[tex]h_{max} = 26.114\,ft[/tex]
Step-by-step explanation:
The maximum altitude of the rocket can be found with the help of First Derivative and Second Derivative Tests:
[tex]\frac{dh}{dt} = -39\cdot t^{2}+4\cdot t + 12[/tex]
[tex]\frac{d^{2}h}{dt^{2}}=-78\cdot t + 4[/tex]
Critical values are:
[tex]t_{1}\approx 0.608\,s[/tex], [tex]t_{2}\approx -0.506\,s[/tex]
The second derivative values related to critical values are:
[tex]t_{1}:[/tex]
[tex]\frac{d^{2}h}{dt^{2}}= -43.424[/tex] (Maximum)
[tex]t_{2}:[/tex]
[tex]\frac{d^{2}h}{dt^{2}}= 43.468[/tex] (Minimum)
Only the first solution offers a physically reasonable solution, as time is a positive variable. Then, maximum height is:
[tex]h_{max} = 26.114\,ft[/tex]