Answer:
- r = 12.5p(32 -p)
- $16 per ticket
- $3200 maximum revenue
Step-by-step explanation:
The number of tickets sold (q) at some price p is apparently ...
q = 150 + 25(20 -p)/2 = 150 +250 -12.5p
q = 12.5(32 -p)
The revenue is the product of the price and the number of tickets sold:
r = pq
r = 12.5p(32 -p) . . . . revenue equation
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The maximum of revenue will be on the line of symmetry of this quadratic function, which is halfway between the zeros at p=0 and p=32. Revenue will be maximized when ...
p = (0 +32)/2 = 16
The theater should charge $16 per ticket.
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Maximum revenue will be found by using the above revenue function with p=16.
r = 12.5(16)(32 -16) = $3200 . . . . maximum revenue
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<em>Additional comment</em>
The number of tickets sold at $16 will be ...
q = 12.5(32 -16) = 200
It might also be noted that if there are variable costs involved, maximum revenue may not correspond to maximum profit.
Answer:
A: there is not enough information because there are no numbers to incorporate the sizes on it so that would mean that there is <u><em>not enough information!!!</em></u>
Answer:
37.6%
Step-by-step explanation:
P(A|B) = P(A and B) / P(B)
P(red | car) = P(red and car) / P(car)
P(red | car) = 0.32 / 0.85
P(red | car) = 0.376
3-5=-2. 2×4=8. 8×-2=-16 the answer is -16
The maximum profit is the y-coordinate of the vertex of the parabola represented by the equation.
The maximum value is 6400, but the profit is given in hundreds of dollars, so multiply the value by 100.
The maximum profit the company can make is $640,000.
