Question

1. during a football game, a punter punts the ball from a point 2 feet above the ground with an initial velocity of 60 feet per second. given the equation $h(t)= - 16t^{2}+60t + 2$. change the window: $xmin = 0$, $xmax = 4$, $xscl=0.25$, $ymin = 0$, $ymax = 80$, $yscl = 5$. $y_{1}=-16t^{2}+60t + 2$ and graph. 2nd graph (table). a: graph the function. find the maximum: 2nd trace (calc) 4:maximum. left bound? move cursor left of vertex and press enter. right bound? move cursor right of vertex and press enter twice. b: how long did it take for the ball to reach its maximum height? (x of vertex) c: identify the balls maximum height? (y of vertex) d: when does the ball hit the ground? (how long is the ball in the air?) $y_{1}=-16t^{2}+60t + 2$, $y_{2}=0$, 2nd trace (calc) 5:intersect. move cursor close to y = 0 (ground) and press enter 3 times. 2. the revenue from selling a certain product is modeled by the function $r(p)=-5p^{2}+250p$ where $r(p)$ is the revenue in dollars and $p$ is the price of the product in dollars. a. what price should the company charge to maximize its revenue? b. what is the maximum revenue the company can expect to earn?

Explanation:

Step1: Recall vertex - formula for a quadratic function

For a quadratic function \(y = ax^{2}+bx + c\), the \(x\) - coordinate of the vertex is given by \(x=-\frac{b}{2a}\). For the height - function of the ball \(h(t)=-16t^{2}+60t + 2\), where \(a=-16\), \(b = 60\), and \(c = 2\).

Step2: Find the time \(t\) when the ball reaches its maximum height

Using the formula \(t=-\frac{b}{2a}\), we substitute \(a=-16\) and \(b = 60\) into it.
\[t=-\frac{60}{2\times(-16)}=\frac{60}{32}=\frac{15}{8}=1.875\] seconds.

Step3: Find the maximum height of the ball

Substitute \(t = 1.875\) into the function \(h(t)=-16t^{2}+60t + 2\).
\[h(1.875)=-16\times(1.875)^{2}+60\times1.875 + 2\]
\[=-16\times3.515625+112.5 + 2\]
\[=-56.25+112.5 + 2=58.25\] feet.

Step4: Find when the ball hits the ground

Set \(h(t)=-16t^{2}+60t + 2 = 0\). Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), with \(a=-16\), \(b = 60\), and \(c = 2\).
\[t=\frac{-60\pm\sqrt{60^{2}-4\times(-16)\times2}}{2\times(-16)}=\frac{-60\pm\sqrt{3600 + 128}}{-32}=\frac{-60\pm\sqrt{3728}}{-32}=\frac{-60\pm61.06}{-32}\]
We take the positive root \(t=\frac{-60 + 61.06}{-32}\) (since time cannot be negative), \(t=\frac{60\pm\sqrt{3600+128}}{32}\approx\frac{60\pm61.06}{32}\). The positive solution is \(t=\frac{60 + 61.06}{32}\approx3.78\) seconds.

Step5: For the revenue function \(R(p)=-5p^{2}+250p\)

The price \(p\) that maximizes revenue: Using the vertex - formula \(p=-\frac{b}{2a}\), where \(a=-5\) and \(b = 250\).
\[p=-\frac{250}{2\times(-5)} = 25\] dollars.

Step6: Find the maximum revenue

Substitute \(p = 25\) into \(R(p)=-5p^{2}+250p\).
\[R(25)=-5\times(25)^{2}+250\times25=-5\times625+6250=-3125 + 6250=3125\] dollars.

Answer:

B. How long did it take for the ball to reach its maximum height? \(t = 1.875\) seconds
C. Identify the ball's maximum height? \(h = 58.25\) feet
D. When does the ball hit the ground? \(t\approx3.78\) seconds
a. What price should the company charge to maximize its revenue? \(p = 25\) dollars
b. What is the maximum revenue the company can expect to earn? \(R = 3125\) dollars