The park had initially planned to charge $8 for admission and expected to have 2400 visitors a day. Allison and Juan were assigned the task of analyzing the park's admission revenues.
1.a)How much revenue would the park have for one day at the current price?
b)They have been provided with some market research that shows for every $0.50 the admission price is raised, the park will have 80 fewer visitors. How much would the park revenue be if park raised their admission price by $1?
c)After a few calculations, Allison and Hannah realize the park will make more money if they raise the price of admission. However, they also understand that there must be a limit to how much the park can charge. As a result, they model the situation with the equation, R = (2400 - 80x)(8 + 0.5x), where R represents the revenue from sales and x represents the number of price increases.
i)Write this equation in standard form, R = ax2 + bx + c. Show all of the steps leading to the final answer.
ii)What price should the park charge to maximize its revenue?
iii)What price range would produce a revenue over $20,000?
a) To calculate the revenue for one day at the current price, we need to multiply the admission price by the number of visitors. In this case, the admission price is $8, and the number of visitors is 2400. So, the revenue for one day would be:
Revenue = Admission Price * Number of Visitors
= $8 * 2400
= $19,200
Therefore, the park would have $19,200 in revenue for one day at the current price.
b) To calculate the revenue if the park raised the admission price by $1, we need to determine the new number of visitors and then multiply it by the new admission price. We know that for every $0.50 increase in admission price, there will be 80 fewer visitors.
So, if the admission price is increased by $1, it means that the price has been raised by $1 / $0.50 = 2 increments of $0.50. Therefore, the number of visitors will decrease by 2 increments * 80 visitors/increment = 160 visitors.
The new number of visitors would be 2400 - 160 = 2240 visitors. The new admission price would be $8 + $1 = $9.
Revenue = Admission Price * Number of Visitors
= $9 * 2240
= $20,160
Therefore, the park revenue would be $20,160 if the admission price is raised by $1.
c) i) To write the equation in standard form, we need to expand and arrange it accordingly:
R = (2400 - 80x)(8 + 0.5x)
First, distribute the 2400 to both terms inside the parentheses:
R = 2400 * 8 + 2400 * 0.5x - 80x * 8 - 80x * 0.5x
Simplify each term:
R = 19200 + 1200x - 640x - 40x^2
Rearrange the terms in descending order of exponents:
R = -40x^2 + 560x + 19200
The equation R = -40x^2 + 560x + 19200 is now in standard form.
ii) To find the price at which the park should charge to maximize its revenue, we need to find the vertex of the quadratic equation. The x-coordinate of the vertex can be obtained using the formula: x = -b / 2a, where a, b, and c are the coefficients of the quadratic equation in standard form.
In this case, a = -40, b = 560, and c = 19200. Plugging these values into the formula:
x = -560 / (2 * -40)
= -560 / -80
= 7
So, the park should charge $7 to maximize its revenue.
iii) To find the price range that would produce a revenue over $20,000, we need to solve the quadratic equation for R > 20,000.
-40x^2 + 560x + 19200 > 20,000
Rearrange the equation to have zero on one side:
-40x^2 + 560x + 19200 - 20,000 > 0
-40x^2 + 560x - 8000 > 0
To find the price range, we need to solve this quadratic inequality. Due to the complexity of solving quadratic inequalities, it would be best to use graphing or software to determine the range.
a) To calculate the revenue for one day at the current price, we multiply the admission price by the expected number of visitors:
Revenue for one day = Admission price * Number of visitors
= $8 * 2400
= $19,200
b) If the park raises the admission price by $1, the new admission price would be $8 + $1 = $9. According to the market research provided, for every $0.50 increase in price, there would be 80 fewer visitors. Thus, for a $1 increase in price, the park would have 2 * 80 = 160 fewer visitors.
Revenue for one day with increased price = (Admission price + Increase in price) * (Number of visitors - Decrease in visitors)
= $9 * (2400 - 160)
= $20,160
c) We model the situation with the equation, R = (2400 - 80x)(8 + 0.5x), where R represents the revenue from sales and x represents the number of price increases.
i) To write this equation in standard form, we need to simplify the expression:
R = (2400 - 80x)(8 + 0.5x)
= 2400(8 + 0.5x) - 80x(8 + 0.5x)
= 19200 + 1200x + 16x + 0.5x^2 - 640x - 40x^2
= 0.5x^2 - 40x^2 + 1200x - 640x + 19200
So, the equation in standard form is:
R = -39.5x^2 + 560x + 19200
ii) To find the price that maximizes revenue, we need to find the vertex of the parabola represented by the equation in standard form. The formula to find the x-coordinate of the vertex is x = -b/(2a), where a and b are coefficients in the quadratic equation ax^2 + bx + c.
Using the formula, we have:
x = -560/(2*(-39.5))
= -560/(-79)
= 7
So, the park should charge 7 increases in price to maximize its revenue.
iii) To find the price range that produces a revenue over $20,000, we substitute R = 20000 into the equation:
20000 = -39.5x^2 + 560x + 19200
Next, we rearrange the equation to standard form, so it becomes:
0.5x^2 - 560x + 800 = 0
Now, we can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/2a
Using the coefficients of the equation, we can plug in the values:
x = (-(-560) ± √((-560)^2 - 4*0.5*800))/2*0.5
= (560 ± √(313600 - 1600))/1
= (560 ± √312000)/1
= (560 ± 558.80)/1
This gives us two solutions for x:
x = 560 - 558.80 = 1.20
x = 560 + 558.80 = 1118.80
Since x represents the number of price increases, we can't have a fractional or decimal number of price increases. Therefore, the price range would be between 1 and 1118.
In summary:
a) The park would have $19,200 for one day at the current price.
b) The park revenue would be $20,160 with a $1 price increase.
c)
i) The equation in standard form is R = -39.5x^2 + 560x + 19200.
ii) The park should charge 7 increases in price to maximize its revenue.
iii) The price range that would produce a revenue over $20,000 is between 1 and 1118.
just remember that revenue = price * no.tickets.sold
You can use that to answer all these questions.
Give it a try, and come back if you get stuck.
also, recall that the vertex of a parabola ax^2+bx+c is at x = -b/2a. For R(x), that is where the maximum revenue occurs.