Dive into the world of knowledge and get your queries resolved at IDNLearn.com. Ask any question and receive accurate, in-depth responses from our dedicated team of experts.
Sagot :
Let's address each part of the problem step-by-step.
### Part (a): Writing Functions for Each Option
Option A: Charge a \[tex]$10 entrance fee and sell game tickets for \$[/tex]0.25 each.
- Let [tex]\( x \)[/tex] be the number of game tickets sold.
- The total profit [tex]\( P_A \)[/tex] can be modeled as:
[tex]\[ P_A(x) = 10 + 0.25x \][/tex]
Option B: Charge \[tex]$0.50 per game ticket without an entrance fee. - Let \( x \) be the number of game tickets sold. - The total profit \( P_B \) can be modeled as: \[ P_B(x) = 0.50x \] ### Part (b): Finding the Intersection Points Algebraically We are given two equations: 1. \( y = x^2 + 3x - 4 \) 2. \( y = -x - 7 \) To find the intersection, set the equations equal to each other: \[ x^2 + 3x - 4 = -x - 7 \] Step 1: Move all terms to one side: \[ x^2 + 3x - 4 + x + 7 = 0 \] \[ x^2 + 4x + 3 = 0 \] Step 2: Factor the quadratic equation: \[ (x + 1)(x + 3) = 0 \] Step 3: Solve for \( x \): \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \] \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] Step 4: Find the corresponding \( y \) values by substituting \( x \) back into either equation (we'll use \( y = -x - 7 \)): For \( x = -1 \): \[ y = -(-1) - 7 = 1 - 7 = -6 \] For \( x = -3 \): \[ y = -(-3) - 7 = 3 - 7 = -4 \] Thus, the intersection points are: \[ (-1, -6) \quad \text{and} \quad (-3, -4) \] Next, we solve the second equation: \[ 3|x + 4| - 2 = 4 \] Step 1: Isolate the absolute value term: \[ 3|x + 4| - 2 = 4 \] \[ 3|x + 4| = 6 \] \[ |x + 4| = 2 \] Step 2: Solve the absolute value equation: \[ x + 4 = 2 \quad \Rightarrow \quad x = -2 \] \[ x + 4 = -2 \quad \Rightarrow \quad x = -6 \] ### Intersection Points from Part (b) - The intersections of \( y = x^2 + 3x - 4 \) and \( y = -x - 7 \) are \( (-1, -6) \) and \( (-3, -4) \). - The solutions for \( 3|x + 4| - 2 = 4 \) are \( x = -2 \) and \( x = -6 \). ### Part (c): Advice Based on the Intersection Points Based on the intersection points, we can determine that \( x = -2 \) and \( x = -6 \) are significant values where the functions of ticket sales equate to each other. However, since ticket numbers cannot be negative in a realistic scenario, these solutions do not directly help in deciding which option to choose. In terms of maximizing profits: - For Option A: Profit will always include the fixed \$[/tex]10 plus [tex]\( 0.25x \)[/tex] per ticket.
- For Option B: Profit will be [tex]\( 0.50x \)[/tex] per ticket.
To truly maximize profits, consider the potential number of tickets sold:
- If fewer tickets are expected to be sold, Option A might be preferable due to the fixed entrance fee, ensuring at least \$10.
- If a high number of tickets are expected to be sold, Option B might yield higher profits due to the higher per-ticket charge.
By factoring any assumptions about the number of tickets, your friend can choose the best option for maximizing the fundraiser's profits.
### Part (a): Writing Functions for Each Option
Option A: Charge a \[tex]$10 entrance fee and sell game tickets for \$[/tex]0.25 each.
- Let [tex]\( x \)[/tex] be the number of game tickets sold.
- The total profit [tex]\( P_A \)[/tex] can be modeled as:
[tex]\[ P_A(x) = 10 + 0.25x \][/tex]
Option B: Charge \[tex]$0.50 per game ticket without an entrance fee. - Let \( x \) be the number of game tickets sold. - The total profit \( P_B \) can be modeled as: \[ P_B(x) = 0.50x \] ### Part (b): Finding the Intersection Points Algebraically We are given two equations: 1. \( y = x^2 + 3x - 4 \) 2. \( y = -x - 7 \) To find the intersection, set the equations equal to each other: \[ x^2 + 3x - 4 = -x - 7 \] Step 1: Move all terms to one side: \[ x^2 + 3x - 4 + x + 7 = 0 \] \[ x^2 + 4x + 3 = 0 \] Step 2: Factor the quadratic equation: \[ (x + 1)(x + 3) = 0 \] Step 3: Solve for \( x \): \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \] \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] Step 4: Find the corresponding \( y \) values by substituting \( x \) back into either equation (we'll use \( y = -x - 7 \)): For \( x = -1 \): \[ y = -(-1) - 7 = 1 - 7 = -6 \] For \( x = -3 \): \[ y = -(-3) - 7 = 3 - 7 = -4 \] Thus, the intersection points are: \[ (-1, -6) \quad \text{and} \quad (-3, -4) \] Next, we solve the second equation: \[ 3|x + 4| - 2 = 4 \] Step 1: Isolate the absolute value term: \[ 3|x + 4| - 2 = 4 \] \[ 3|x + 4| = 6 \] \[ |x + 4| = 2 \] Step 2: Solve the absolute value equation: \[ x + 4 = 2 \quad \Rightarrow \quad x = -2 \] \[ x + 4 = -2 \quad \Rightarrow \quad x = -6 \] ### Intersection Points from Part (b) - The intersections of \( y = x^2 + 3x - 4 \) and \( y = -x - 7 \) are \( (-1, -6) \) and \( (-3, -4) \). - The solutions for \( 3|x + 4| - 2 = 4 \) are \( x = -2 \) and \( x = -6 \). ### Part (c): Advice Based on the Intersection Points Based on the intersection points, we can determine that \( x = -2 \) and \( x = -6 \) are significant values where the functions of ticket sales equate to each other. However, since ticket numbers cannot be negative in a realistic scenario, these solutions do not directly help in deciding which option to choose. In terms of maximizing profits: - For Option A: Profit will always include the fixed \$[/tex]10 plus [tex]\( 0.25x \)[/tex] per ticket.
- For Option B: Profit will be [tex]\( 0.50x \)[/tex] per ticket.
To truly maximize profits, consider the potential number of tickets sold:
- If fewer tickets are expected to be sold, Option A might be preferable due to the fixed entrance fee, ensuring at least \$10.
- If a high number of tickets are expected to be sold, Option B might yield higher profits due to the higher per-ticket charge.
By factoring any assumptions about the number of tickets, your friend can choose the best option for maximizing the fundraiser's profits.
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is committed to providing accurate answers. Thanks for stopping by, and see you next time for more solutions.
