Ok, here is the question…. A museum has an admission fee of $14 and averages 300 visitors per day. The museum board decides to raise the fee. Research indicates that for every $1 increase, there would be 10 fewer visitors per day. What admission fee would maximize the revenue?
What im looking for is a complete answer with work written out, i am also wondering how make an equation for a porabola from the information given.
x = amount to raise admission fee
Revenue = (14+x)(300 – 10x) [That's fee times # of visitors]
R = -10x^2 +160x +4200
dR/dx = -20x +160 = 0 [Set dR/dx to 0 to find maximum
-20x = -160
x = 8
So 14+8 = $22 admissin fee maximizes revenue
Revenue is the number of tickets sold, 300-10x, times the price per ticket, 14+x.
So R(x) = (300 -10x) (14 + x)
References :
x = amount to raise admission fee
Revenue = (14+x)(300 – 10x) [That's fee times # of visitors]
R = -10x^2 +160x +4200
dR/dx = -20x +160 = 0 [Set dR/dx to 0 to find maximum
-20x = -160
x = 8
So 14+8 = $22 admissin fee maximizes revenue
References :
Column 1 = dollars per ticket
Column 2 = people per day
Column 3 = total dollars taken in
14 * 300 = 4200 = Starting Point
15 * 290 = 4350
16 * 280 = 4480
17 * 270 = 4590
Keep going till the total decreases.
To see the parabola plot dollars per ticket on x-axis
and total dollars taken in on y-axis.
This is a problem from calculus called max-min.
You can solve it without calculs using trial and error.
References :