Question 1202278
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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 
<font color=red>P =  -60t^2 + 720t - 162 </font> is the final answer.


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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
{{{t = (-b +- sqrt(b^2 - 4ac))/(2a)}}}


{{{t = (-(-120) +- sqrt((-120)^2 - 4(10)(27)))/(2(10))}}}


{{{t = (120 +- sqrt(13320))/(20)}}}


{{{t = (120 + sqrt(13320))/(20)}}} or {{{t = (120 - sqrt(13320))/(20)}}}


{{{t = 11.77}}} or {{{t = 0.23}}}
Each decimal value is approximate.



<font color=red>Answer:</font> 
Break-even point happens when t = 0.23 and when t = 11.77


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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


<font color=red>Answer:</font>
The max profit is $1,998. It occurs when the ticket price is $6.
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