Question

what is the maximum of this equation???
h(t) = -16t^2 + 60t + 82

Answer 1

take the derivative and then put it equal to zero then find thae values of t...you will gt the result..

Answer 2

that didn't make since..

Answer 3

take its derivative w.r.t. "t"

wht will youget..??

Answer 4

If you don't know calculus, an easy way to do this is to find the parabola into the vertex form. So going from the form \(ax^2+bx+c\) into \(a(x-h)^2+k\). Here, they are using \(t\) instead of \(x\) though as the variable.
\[\large \begin{align} -16t^2 + 60t+82 &= -16\left[ t^2+\left( \frac{60}{-16}\right)t+\left( \frac{82}{-60}\right)\right] \\&=-16\left[ t^2 -\frac{15}{4}t-\frac{41}{8}\right] \\ &=-16\left[\underbrace{t^2-\frac{15}{4}t\color{red}{+\frac{225}{64}}}\color{red}{-\frac{225}{64}}-\frac{41}{8} \right] \\ &\text{completed the square doing}\pm\left( -\frac{15}{4}\right)^2 \\&=-16\left[\underbrace{\left(t-\frac{15}{8}\right)^2 }-\frac{553}{64}\right]\\ \, \\ &=-16\left( t-\frac{15}{8}^2\right)-16\left(-\frac{553}{64} \right)\\ &=-16\left( t-\frac{15}{8}^2\right)+\frac{553}{4}\\ &=a(t-h)^2+k\end{align}
\]

Since the vertex of the parabola is at the point \((h, k)\), then the maximum is \(\large \frac{553}{4}\) and it occurs at \(\large t=\frac{15}{8}\)

Answer 5

Typo sorry, the last term on the first line should be \(\large \frac{82}{-16}\)

Answer 6

Also, for completing the square... it's \(\Large \left(-\frac{15}{4}\div 2\right)^2\)