SOLUTION: One hundred students will attend a concert if tickets cost $30 each. For each $5 raise in the price of the ticket, 10 fewer students will attend. What price will deliver the maximu
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Question 1189393: One hundred students will attend a concert if tickets cost $30 each. For each $5 raise in the price of the ticket, 10 fewer students will attend. What price will deliver the maximum dollar sales?
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
100 students buy a ticket at $30 each.
This draws in a revenue (aka sales) of 100*30 = 3,000 dollars. This is before any price increase.
After 1 price increase of $5, we have- 100 students drop to 100-10 = 90 students
- old price = $30, new price = $35
- revenue = (number of students)*(price) = 90*35 = 3,150 dollars
This shows that despite 10 fewer students showing up, the revenue has gone up overall.
After 2 price increases of $5 each, aka 2*5 = 10 dollar increase overall, we have:- 100 students drop to 100-2*10 = 100-20 = 80 students
- old price = $30, new price = 30+2*5 = 30+10 = $40
- revenue = (number of students)*(price) = 80*40 = 3,200 dollars
So far, the revenue keeps going up.
After 3 price increases of $5 each, aka 3*5 = 15 dollar increase overall, we have:- 100 students drop to 100-3*10 = 100-30 = 70 students
- old price = $30, new price = 30+3*5 = 30+15 = $45
- revenue = (number of students)*(price) = 70*45 = 3,150 dollars
The revenue has gone down.
Therefore, the max revenue occurs when the price is increased twice ($5 increments each time). The max revenue itself is $3200.
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A slightly different approach using algebra.
x = number of times the price is raised $5
x is some positive whole number.
After x price increases, where x is a positive whole number, then:- 100 students drop to 100-10x students
- old price = $30, new price = 30+5x dollars
- revenue = (number of students)*(price) = (100-10x)*(30+5x) dollars
Let's expand out that expression to get some quadratic.
(100-10x)*(30+5x)
y*(30+5x) ..... let y = 100-10x
30y+5xy
30( y ) + 5x( y )
30(100-10x) + 5x(100-10x) .... plug in y = 100-10x
3000 - 300x + 500x - 50x^2
3000 + 200x - 50x^2
-50x^2 + 200x + 3000
That expression is in the form ax^2+bx+c with
a = -50
b = 200
c = 3000
This will plot out a parabola that opens downward to form a "frowny" face. At the peak of this parabola will represent when the revenue is maxed out.
We know that the parabola opens downward since a = -50 is negative.
(h,k) is the vertex of the parabola
h = x coordinate of vertex
h = -b/(2a)
h = -200/(2*(-50))
h = -200/(-100)
h = 2
The max revenue occurs when the x coordinate of the vertex is x = 2. In other words, when exactly 2 sales increases of $5 each happens. This confirms our previous result above.
Plug this x value into the expression found to get the revenue
So,
(100-10*2)*(30+5*2)
(100-20)*(30+10)
(80)*(40)
3200
To wrap things up, if the company increases the price by $5 a total of two times, then we have- attendance = 80
- price per ticket = $40
- revenue maxes out at 80*40 = $3200
Despite losing 20 people, it's in the company's best interest to raise the price this amount.
Answer: 3200 dollars
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