SOLUTION: Dances at the community centre produce revenue R=-60t^2 +600t, where R is the revenue and t the ticket price in dollars. Francine, the manager, found that the expenses, C, for the
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Question 1202278: Dances at the community centre produce revenue R=-60t^2 +600t, where R is the revenue and t the ticket price in dollars. Francine, the manager, found that the expenses, C, for the dances is modelled by C=162 - 120t. Note: Profit P = R- C.
a) Determine the equation to represent the profit.
b) Determine the break-even point (zero profit).
c) Find the maximum profit and the ticket price that yields this profit.
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!
Part (a)
Revenue = money coming in
Cost = money going out
Profit = Revenue - Cost
P = R - C
P = ( R ) - ( C )
P = ( -60t^2 +600t ) - ( 162 - 120t )
P = -60t^2 +600t - 162 + 120t
P = -60t^2 + 720t - 162 is the final answer.
===================================
Part (b)
The break-even point is when the company neither gains money nor loses money.
Set profit equal to zero to determine t.
P = 0
-60t^2 + 720t - 162 = 0
-6(10t^2 - 120t + 27) = 0
10t^2 - 120t + 27 = 0/(-6)
10t^2 - 120t + 27 = 0
Let's use the quadratic formula.
Plugging in a = 10, b = -120, c = 27
or
or
Each decimal value is approximate.
Answer:
Break-even point happens when t = 0.23 and when t = 11.77
===================================
Part (c)
The vertex in this case is the highest point. It corresponds to the max profit.
The x coordinate of the vertex is the midpoint of the roots.
Each root is a break-even point.
Average the two break-even points
(0.23+11.77)/2 = 6
The max profit happens when t = 6 is the ticket price.
Plug this into the profit function.
P = -60t^2 + 720t - 162
P = -60*6^2 + 720*6 - 162
P = 1998
Answer:
The max profit is $1,998. It occurs when the ticket price is $6.
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