Respuesta :
Answer:
[tex]h'(t) = -12.6 t +53[/tex]
Now we can set up the derivate equal to 0 and we have:
[tex] -12.6 t +53 = 0[/tex]
And solving for t we got:
[tex] t = \frac{53}{12.6}= 4.206[/tex]
For the second derivate respect the time we got:
[tex] h''(t) = -12.6<0[/tex]
So then we can conclude that t = 4.206 is a maximum for the function.
And the corresponding height would be:
[tex] h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft[/tex]
So the maximum occurs at t = 4.206 s and with a height of 135.468 m
Step-by-step explanation:
For this case we have the following function:
[tex] h(t) = -6.3t^2 +53 t+24[/tex]
In order to maximize this function we need to take the first derivate respect the time and we have:
[tex]h'(t) = -12.6 t +53[/tex]
Now we can set up the derivate equal to 0 and we have:
[tex] -12.6 t +53 = 0[/tex]
And solving for t we got:
[tex] t = \frac{53}{12.6}= 4.206[/tex]
For the second derivate respect the time we got:
[tex] h''(t) = -12.6<0[/tex]
So then we can conclude that t = 4.206 is a maximum for the function.
And the corresponding height would be:
[tex] h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft[/tex]
So the maximum occurs at t = 4.206 s and with a height of 135.468 m
