5. suppose you launch a potato straight up from the ground with a velocity of 96 ft/s. as the potato moves upward, gravity slows it and eventually the potato begins to fall back to the ground. the height of the potato after t seconds in the air is modeled by the equation ( h(t) = -16t^2 + 96t ).
a. find the rate of change of the potato in the first 3 seconds. include units.
b. what does the average velocity (application of rate of change) of the potato on the interval from ( t = 3 ) sec. to ( t = 4 ) sec. tell you about the behavior of the object. explain.
c. use a graphing utility to find where the velocity of the potato is stalled? how far off the ground is the potato at this time? (reference calculator skills sheet).
6. a local home and garden show planned for october depends on the ticket price for entry. the show sponsors used past receipts to find that the profit can be modeled by ( p(t) = -20t^2 + 800t + 100 ).
a. the goal of the sponsors is to have maximum profit. what would be the best price for the event tickets in order to maximize profits? explain how you know.
b. what is the maximum profit based on your ticket price from part a?
c. if sponsors decided to charge $25 per ticket, how much profit loss, on average, would be lost between ( t = 20 ) and ( t = 25 )?

Question

5. suppose you launch a potato straight up from the ground with a velocity of 96 ft/s. as the potato moves upward, gravity slows it and eventually the potato begins to fall back to the ground. the height of the potato after t seconds in the air is modeled by the equation ( h(t) = -16t^2 + 96t ).
   a. find the rate of change of the potato in the first 3 seconds. include units.
   b. what does the average velocity (application of rate of change) of the potato on the interval from ( t = 3 ) sec. to ( t = 4 ) sec. tell you about the behavior of the object. explain.
   c. use a graphing utility to find where the velocity of the potato is stalled? how far off the ground is the potato at this time? (reference calculator skills sheet).
6. a local home and garden show planned for october depends on the ticket price for entry. the show sponsors used past receipts to find that the profit can be modeled by ( p(t) = -20t^2 + 800t + 100 ).
   a. the goal of the sponsors is to have maximum profit. what would be the best price for the event tickets in order to maximize profits? explain how you know.
   b. what is the maximum profit based on your ticket price from part a?
   c. if sponsors decided to charge $25 per ticket, how much profit loss, on average, would be lost between ( t = 20 ) and ( t = 25 )?

Accepted Answer

### 5a Solution:
# Explanation:
## Step1: Recall average rate of change formula
The average rate of change of a function $ h(t) $ over the interval $[a, b]$ is $\frac{h(b) - h(a)}{b - a}$. Here, $ a = 0 $, $ b = 3 $, and $ h(t)=- 16t^{2}+96t$ (note: the original equation likely has a negative coefficient for $ t^{2} $ as gravity is a downward force, so we correct it to $ h(t)=-16t^{2}+96t $).
## Step2: Calculate $ h(0) $ and $ h(3) $
- $ h(0)=-16(0)^{2}+96(0) = 0 $
- $ h(3)=-16(3)^{2}+96(3)=-16\times9 + 288=-144 + 288 = 144 $
## Step3: Compute average rate of change
Using the formula: $\frac{h(3)-h(0)}{3 - 0}=\frac{144 - 0}{3}=48$ ft/s.
# Answer:
The rate of change in the first 3 seconds is $ 48 $ ft/s.

### 5b Explanation:
# Brief Explanations:
First, find $ h(3) = 144 $ ft (from 5a) and $ h(4)=-16(4)^{2}+96(4)=-256 + 384 = 128 $ ft. The average velocity (rate of change) is $\frac{h(4)-h(3)}{4 - 3}=\frac{128 - 144}{1}=-16$ ft/s. A negative average velocity means the potato's height is decreasing (falling) on $[3,4]$, so it's moving downward.
# Answer:
The average velocity is $-16$ ft/s, meaning the potato is falling (height decreasing) between $ t = 3 $ and $ t = 4 $ seconds.

### 5c Explanation:
# Brief Explanations:
The velocity function $ v(t) $ is the derivative of $ h(t) $. $ h(t)=-16t^{2}+96t $, so $ v(t)=h^\prime(t)=-32t + 96 $. Velocity is stalled (instantaneous velocity is 0) when $ v(t) = 0 $. Solve $-32t + 96 = 0\Rightarrow - 32t=-96\Rightarrow t = 3$ seconds. Then $ h(3)=-16(3)^{2}+96(3)=144 $ ft.
# Answer:
Velocity is stalled at $ t = 3 $ seconds, and the potato is $ 144 $ ft off the ground.

### 6a Solution:
# Explanation:
## Step1: Identify the profit function type
The profit function $ p(t)=-20t^{2}+800t + 100 $ (assuming a typo, should be $ -20t^{2} $) is a quadratic function with $ a=-20 $, $ b = 800 $, $ c = 100 $. For a quadratic $ ax^{2}+bx + c $, the vertex (maximum for $ a<0 $) occurs at $ t=-\frac{b}{2a} $.
## Step2: Calculate the ticket price (t) for maximum profit
$ t=-\frac{800}{2\times(-20)}=-\frac{800}{-40}=20 $. So the best ticket price is $ \$20 $ because the quadratic opens downward ( $ a=-20<0 $ ), so the vertex is the maximum point.
# Answer:
The best ticket price to maximize profit is $ \$20 $ (since the profit function is a downward - opening parabola, the vertex gives the maximum, and $ t =-\frac{b}{2a}=20 $).

### 6b Solution:
# Explanation:
## Step1: Substitute $ t = 20 $ into the profit function
$ p(t)=-20t^{2}+800t + 100 $, substitute $ t = 20 $:
$ p(20)=-20(20)^{2}+800(20)+100=-20\times400 + 16000+100=-8000 + 16000+100 = 8100 $
# Answer:
The maximum profit is $ \$8100 $.

### 6c Solution:
# Explanation:
## Step1: Find $ p(20) $ and $ p(25) $
- $ p(20)=8100 $ (from 6b)
- $ p(25)=-20(25)^{2}+800(25)+100=-20\times625+20000 + 100=-12500+20000 + 100 = 7600 $
## Step2: Calculate the average rate of change (profit loss per unit price change)
The average rate of change (profit loss on average) between $ t = 20 $ and $ t = 25 $ is $\frac{p(25)-p(20)}{25 - 20}=\frac{7600 - 8100}{5}=\frac{-500}{5}=-100$ dollars per ticket. The total profit loss is $ p(20)-p(25)=8100 - 7600 = 500 $ dollars, and the average loss per ticket (over the interval) is $ - 100 $ dollars (or a loss of $ \$100 $ per ticket on average in terms of rate, but the total loss between $ t = 20 $ and $ t = 25 $ is $ \$500 $).
# Answer:
The average profit loss between $ t = 20 $ and $ t = 25 $ is $ \$100 $ per ticket (or total loss of $ \$500 $ over the interval, with average loss of $ \$100 $ per ticket increase from $ \$20 $ to $ \$25 $).

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QUESTION IMAGE

Question

Explanation:

5a Solution:

Step1: Recall average rate of change formula

Step2: Calculate \( h(0) \) and \( h(3) \)

Step3: Compute average rate of change

Answer:

5b Explanation: