Question

1. a quarterback releases the football from a point 6 feet above the ground with an initial velocity of 30 feet per second. the height of the ball above the field is modeled by the equation $h(t)=-16t^{2}+30t + 6$, where $h$ is the height in feet and $t$ is the time in seconds.

part a: what is the maximum height the ball will reach?
part b: how long will it take the ball to reach its maximum height?
part c: assuming no one catches the ball, when will the ball hit the ground?

1. a local theater currently charges $25 per ticket and sells 600 tickets for each performance. market research indicates that for every $1 increase in the ticket price, the number of tickets sold will decrease by 12. the model is $r(x)=-12x^{2}+300x + 15000$

a. how many $1 increases will maximize the revenue?
b. what is the maximum possible revenue?

Explanation:

Step1: Identify the quadratic - function form

The height function of the ball is $h(t)=-16t^{2}+30t + 6$, which is in the form of a quadratic function $y = ax^{2}+bx + c$ where $a=-16$, $b = 30$, and $c = 6$. The revenue function $R(x)=-12x^{2}+300x + 15000$ is also a quadratic function with $a=-12$, $b = 300$, and $c = 15000$. For a quadratic function $y = ax^{2}+bx + c$, the vertex of the parabola (which gives the maximum or minimum value depending on the sign of $a$) has its $x$ - coordinate (or $t$ - coordinate in the height - time case) at $x=-\frac{b}{2a}$.

Step2: Solve Part A of the ball - height problem

For the ball - height function $h(t)=-16t^{2}+30t + 6$, first find the time $t$ at which the ball reaches its maximum height. Using the formula $t=-\frac{b}{2a}$, we have $t=-\frac{30}{2\times(-16)}=\frac{30}{32}=\frac{15}{16}$ seconds. Then substitute $t = \frac{15}{16}$ into the height function:
\[

$$\begin{align*} h(\frac{15}{16})&=-16\times(\frac{15}{16})^{2}+30\times\frac{15}{16}+6\\ &=-16\times\frac{225}{256}+ \frac{450}{16}+6\\ &=-\frac{225}{16}+\frac{450}{16}+6\\ &=\frac{-225 + 450}{16}+6\\ &=\frac{225}{16}+6\\ &=\frac{225+96}{16}\\ &=\frac{321}{16}=20.0625 \end{align*}$$

\]
So the maximum height is $\frac{321}{16}=20.0625$ feet.

Step3: Solve Part B of the ball - height problem

We already found the time $t$ at which the ball reaches its maximum height in Step 2. Using the formula $t =-\frac{b}{2a}$ for $h(t)=-16t^{2}+30t + 6$, $t=\frac{15}{16}=0.9375$ seconds.

Step4: Solve Part C of the ball - height problem

When the ball hits the ground, $h(t)=0$. So we solve the quadratic equation $-16t^{2}+30t + 6 = 0$. Divide through by - 2 to get $8t^{2}-15t - 3=0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$, here $a = 8$, $b=-15$, and $c=-3$.
\[

$$\begin{align*} t&=\frac{15\pm\sqrt{(-15)^{2}-4\times8\times(-3)}}{2\times8}\\ &=\frac{15\pm\sqrt{225 + 96}}{16}\\ &=\frac{15\pm\sqrt{321}}{16} \end{align*}$$

\]
We take the positive root since time cannot be negative. $t=\frac{15+\sqrt{321}}{16}\approx\frac{15 + 17.92}{16}=\frac{32.92}{16}\approx2.06$ seconds.

Step5: Solve Part a of the revenue problem

For the revenue function $R(x)=-12x^{2}+300x + 15000$, using the formula $x=-\frac{b}{2a}$ to find the number of $\$1$ increases that maximizes the revenue. Here $a=-12$ and $b = 300$. So $x=-\frac{300}{2\times(-12)}=\frac{300}{24}=12.5$.

Step6: Solve Part b of the revenue problem

Substitute $x = 12.5$ into the revenue function $R(x)=-12x^{2}+300x + 15000$.
\[

$$\begin{align*} R(12.5)&=-12\times(12.5)^{2}+300\times12.5+15000\\ &=-12\times156.25+3750+15000\\ &=-1875+3750+15000\\ &=16875 \end{align*}$$

\]

Answer:

PART A: $\frac{321}{16}=20.0625$ feet
PART B: $\frac{15}{16}=0.9375$ seconds
PART C: $\frac{15 + \sqrt{321}}{16}\approx2.06$ seconds
a. $12.5$
b. $\$16875$