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Answer :
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### Problem 1: Arrow Shot Vertically
1. Height Formula: The height of the arrow as a function of time [tex]\( t \)[/tex] is given by the equation:
[tex]\[ h(t) = -5t^2 + 60t \][/tex]
2. Maximum Height:
The maximum height is reached when the velocity is 0. The velocity function is obtained by differentiating the height function with respect to [tex]\( t \)[/tex]:
[tex]\[ v(t) = \frac{d}{dt}(-5t^2 + 60t) = -10t + 60 \][/tex]
Set [tex]\( v(t) = 0 \)[/tex] to find the time when the maximum height is reached:
[tex]\[ -10t + 60 = 0 \][/tex]
[tex]\[ t = \frac{60}{10} = 6 \, \text{seconds} \][/tex]
3. Calculate Maximum Height:
Substitute [tex]\( t = 6 \)[/tex] back into the height equation:
[tex]\[ h(6) = -5(6)^2 + 60(6) \][/tex]
[tex]\[ h(6) = -180 + 360 = 180 \, \text{meters} \][/tex]
4. Total Flight Time:
The arrow takes the same amount of time to go up as it takes to come down. Therefore, total flight time is twice the time to reach maximum height:
[tex]\[ \text{Total flight time} = 2 \times 6 = 12 \, \text{seconds} \][/tex]
### Problem 2: Rocket Flight
1. Height Formula: The height of the rocket as a function of time [tex]\( t \)[/tex] is:
[tex]\[ h(t) = -5t^2 + 40t + 45 \][/tex]
2. Maximum Height:
To find the maximum height, first find the time when the velocity is 0 (similar process as above):
[tex]\[ v(t) = -10t + 40 \][/tex]
Set [tex]\( v(t) = 0 \)[/tex]:
[tex]\[ -10t + 40 = 0 \][/tex]
[tex]\[ t = \frac{40}{10} = 4 \, \text{seconds} \][/tex]
3. Calculate Maximum Height:
Substitute [tex]\( t = 4 \)[/tex] into the height equation:
[tex]\[ h(4) = -5(4)^2 + 40(4) + 45 \][/tex]
[tex]\[ h(4) = -80 + 160 + 45 = 125 \, \text{meters} \][/tex]
4. Total Flight Time:
Use the quadratic formula to find when the rocket returns to the initial height (height = 0):
[tex]\[ -5t^2 + 40t + 45 = 0 \][/tex]
Solving the quadratic equation:
[tex]\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\( a = -5 \)[/tex], [tex]\( b = 40 \)[/tex], [tex]\( c = 45 \)[/tex].
Calculate:
[tex]\[
t = \frac{-40 \pm \sqrt{40^2 - 4 \times (-5) \times 45}}{2 \times (-5)}
\][/tex]
This calculation gives two possible times, but the positive one is the total flight time of about 9 seconds.
### Problem 3: Business Profit Calculation
1. Price and Customers: Initially, you sell a product for [tex]$10 and have 1000 customers.
2. Change in Revenue: For every $[/tex]1 increase in price, you lose 100 customers.
Let [tex]\( x \)[/tex] be the number of [tex]$1 increases in price. Then the price becomes \( 10 + x \), and the customers become \( 1000 - 100x \).
3. Revenue Function:
\[ R(x) = (10 + x)(1000 - 100x) \]
4. Maximizing Profit:
Simplify this to find the value of \( x \) that maximizes revenue and solve using calculus:
You'll find that increasing the price by 0 dollars maximizes revenue, meaning keeping the price at $[/tex]10 will give maximum profit. This means it's best not to raise the price at all.
In summary, the maximum height of the arrow is 180 meters and it stays in the air for 12 seconds. The rocket reaches a maximum height of 125 meters and is in the air for 9 seconds. For the business product, maximum profit is achieved by not increasing the price, keeping it at $10.
### Problem 1: Arrow Shot Vertically
1. Height Formula: The height of the arrow as a function of time [tex]\( t \)[/tex] is given by the equation:
[tex]\[ h(t) = -5t^2 + 60t \][/tex]
2. Maximum Height:
The maximum height is reached when the velocity is 0. The velocity function is obtained by differentiating the height function with respect to [tex]\( t \)[/tex]:
[tex]\[ v(t) = \frac{d}{dt}(-5t^2 + 60t) = -10t + 60 \][/tex]
Set [tex]\( v(t) = 0 \)[/tex] to find the time when the maximum height is reached:
[tex]\[ -10t + 60 = 0 \][/tex]
[tex]\[ t = \frac{60}{10} = 6 \, \text{seconds} \][/tex]
3. Calculate Maximum Height:
Substitute [tex]\( t = 6 \)[/tex] back into the height equation:
[tex]\[ h(6) = -5(6)^2 + 60(6) \][/tex]
[tex]\[ h(6) = -180 + 360 = 180 \, \text{meters} \][/tex]
4. Total Flight Time:
The arrow takes the same amount of time to go up as it takes to come down. Therefore, total flight time is twice the time to reach maximum height:
[tex]\[ \text{Total flight time} = 2 \times 6 = 12 \, \text{seconds} \][/tex]
### Problem 2: Rocket Flight
1. Height Formula: The height of the rocket as a function of time [tex]\( t \)[/tex] is:
[tex]\[ h(t) = -5t^2 + 40t + 45 \][/tex]
2. Maximum Height:
To find the maximum height, first find the time when the velocity is 0 (similar process as above):
[tex]\[ v(t) = -10t + 40 \][/tex]
Set [tex]\( v(t) = 0 \)[/tex]:
[tex]\[ -10t + 40 = 0 \][/tex]
[tex]\[ t = \frac{40}{10} = 4 \, \text{seconds} \][/tex]
3. Calculate Maximum Height:
Substitute [tex]\( t = 4 \)[/tex] into the height equation:
[tex]\[ h(4) = -5(4)^2 + 40(4) + 45 \][/tex]
[tex]\[ h(4) = -80 + 160 + 45 = 125 \, \text{meters} \][/tex]
4. Total Flight Time:
Use the quadratic formula to find when the rocket returns to the initial height (height = 0):
[tex]\[ -5t^2 + 40t + 45 = 0 \][/tex]
Solving the quadratic equation:
[tex]\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\( a = -5 \)[/tex], [tex]\( b = 40 \)[/tex], [tex]\( c = 45 \)[/tex].
Calculate:
[tex]\[
t = \frac{-40 \pm \sqrt{40^2 - 4 \times (-5) \times 45}}{2 \times (-5)}
\][/tex]
This calculation gives two possible times, but the positive one is the total flight time of about 9 seconds.
### Problem 3: Business Profit Calculation
1. Price and Customers: Initially, you sell a product for [tex]$10 and have 1000 customers.
2. Change in Revenue: For every $[/tex]1 increase in price, you lose 100 customers.
Let [tex]\( x \)[/tex] be the number of [tex]$1 increases in price. Then the price becomes \( 10 + x \), and the customers become \( 1000 - 100x \).
3. Revenue Function:
\[ R(x) = (10 + x)(1000 - 100x) \]
4. Maximizing Profit:
Simplify this to find the value of \( x \) that maximizes revenue and solve using calculus:
You'll find that increasing the price by 0 dollars maximizes revenue, meaning keeping the price at $[/tex]10 will give maximum profit. This means it's best not to raise the price at all.
In summary, the maximum height of the arrow is 180 meters and it stays in the air for 12 seconds. The rocket reaches a maximum height of 125 meters and is in the air for 9 seconds. For the business product, maximum profit is achieved by not increasing the price, keeping it at $10.
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