Respuesta :
The function that gives the amount of profit at a given price and number
given are obtained from the relationship between the variables.
a) The profit in terms of the number of tickets sold, P = -0.015·n² + 250·n
(b) The profit in terms of ticket price, is P = -66.[tex]\overline 6[/tex]·p² + 3000·p + 2,333,333.[tex]\overline 3[/tex]
Reasons:
Number of tickets sold at x = $200, is y = 10,000 tickets
Number of extra tickets sold for every $15 price reduction = 1,000 tickets
Number of tickets sold at $185 = 11,000
The cost to fly a person = $100
(a) The rate of change, [tex]\frac{\Delta y}{\Delta x}[/tex], of the number of ticket sold wit price is given as
follows;
[tex]\dfrac{\Delta y}{\Delta x} = \dfrac{11,000 - 10,000}{185 - 200 } = -66. \overline 6[/tex]
The equation in slope and intercept form is therefore;
n - 11,000 = -66.[tex]\overline 6[/tex]·(p - 185)
n = -66.[tex]\overline 6[/tex]·p + 12,333.[tex]\overline 3[/tex] + 11,000 = -66.[tex]\overline 6[/tex]·p + 23,333.[tex]\overline 3[/tex]
Number of tickets sold, n = -66.[tex]\overline 6[/tex]·p + 23,333.[tex]\overline 3[/tex]
[tex]Price \ per \ ticket, \ p = \dfrac{23,333.\overline {3} - n}{66. \overline{6}}[/tex]
The revenue, R = n·x
Therefore;
[tex]R = \dfrac{23,333.\overline {3} \cdot n- n^2}{66. \overline{6}}[/tex]
The profit = Revenue - Cost
The cost = 100·n
[tex]Profit, P = \dfrac{23,333.\overline {3} \cdot n- n^2}{66. \overline{6}} - 100 \cdot n = -0.015 \cdot n^2 + 250 \cdot n[/tex]
- Total profit, P = -0.015·n² + 250·n
(b) The total profit, P, in terms of the price p, for one ticket
Number of tickets sold, n = -66.[tex]\overline 6[/tex]·p + 23,333.[tex]\overline 3[/tex]
The revenue, R = n·p
∴ R = -66.[tex]\overline 6[/tex]·p² + 23,333.[tex]\overline 3[/tex]·p
Cost = 100·n
∴ Cost = 100·(-66.[tex]\overline 6[/tex]·p + 23,333.[tex]\overline 3[/tex])
Therefore, the profit is given as follows;
Profit = -66.[tex]\overline 6[/tex]·p² + 23,333.[tex]\overline 3[/tex]·p - (100·(-66.[tex]\overline 6[/tex]·p + 23,333.[tex]\overline 3[/tex]))
Therefore;
- Total profit, P = -66.[tex]\overline 6[/tex]·p² + 3000·p + 2,333333.[tex]\overline 3[/tex]
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We want to express the total profit in two different ways.
The answers are:
a) P(n) = n*[n - 23,333.3]*(-15/1,000) - n*100
b) P(p) = n(p)*p - n(p)*100 = [ (-1,000/15)*p + 23,333.3]*(p - 100)
The given information is:
- If the ticket price is $200, they sell 10,000 tickets.
- If the ticket price decreases $15, they sell 1,000 tickets more.
- It costs $100 to fly a person, then the cost of each ticket is $100.
a) We want to get the profit in terms of the number n of tickets sold.
The number of tickets sold depends on the price, we know that it is a linear relation like:
n(p) = a*p + b
where p is the price.
we know that:
n(200) = 10,000
n(185) = 11,000
If this is a linear relation, the slope is known:
a = -1,000/15$
Then the line is something like:
n(p) = (-1,000/15)*p + b
We know that if k = 200, then n = 10,000
Replacing that we have:
n(200) = 10,000 = (-1,000/15)*200 + b
10,000 + (1,000/15)*200 = b = 23,333.3
Then the number of tickets sold as a function of the price is:
n(p) = (-1,000/15)*p + 23,333.3
Then the revenue is given by the number of tickets sold times the price, this is:
r(n) = n*p
We can write k in terms of n as:
k = [n - 23,333.3]*(-15/1,000)
Then the revenue as a function of the number of tickets sold is:
r(n) = n*[n - 23,333.3]*(-15/1,000)
The profit is the difference between the revenue and the cost, then the profit function is:
P(n) = n*[n - 23,333.3]*(-15/1,000) - n*100
Where n*100 is the cost.
b) now we want to write the same equation, but this time as a function of the price.
now the revenue is:
r(p) = n*p = [ (-1,000/15)*p + 23,333.3]*p
Then the profit is:
P(p) = n(p)*p - n(p)*100 = [ (-1,000/15)*p + 23,333.3]*(p - 100)
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