The answers to all the parts of the question are:
- (A) q(p) = -15p + 300
- (B) R(p) = -15p² + 300p
- (C) C(p) = -30p + 1600
- (D.1) P(p) = -15p²+ 330p - 1600
- (D.2) p = $11
(A) Before we begin, consider the cartesian plane, where the x-axis represents the cover charge and the y-axis represents the number of guests per night.
According to the previous information, we will find the following coordinates there: (9,165) (10,150).
Remember that the linear equation's slope-intercept form is y=mx+b, where m is the slope and b is the intercept.
The slope can be calculated using the following formula:
In this case, the first coordinate will be (9,165) and the second will be (10,150).
The order is irrelevant.
Finally, we get the same value for m.
So,
- m = (150 - 165)/(10 - 9) = -15
We now have the slope and two points. Find a linear demand equation using this formula and remember that the point-slope form of a line's equation is y - y1 = m(x - x1).
Any of two coordinates can be selected.
In this case, we will select the second one.
We can choose between two coordinates.
In this case, we'll go with the second option.
- y-150=-15(x-10) ....... (1)
Solve equation (1) for y:
- y = -15x + 150 + 150
- y = -15x + 300
We defined x as the cover charge (p) and y as the number of guests above (q).
As a result, rewrite equation 1.
Its formula depicts the number of guests per night as a function of the cover charge.
(B) The nightly revenue can be calculated by multiplying the cover charge price by the number of guests.
So we simply multiply the equation 1 by p.
- p × q(p) = p × (-15p +300)
- R(p) = -15p² + 300p ......(2)
(C) To calculate the nightclub's cost function, multiply the cost of two non-alcoholic drinks by the number of guests (equation 1) and add nightly overheads.
- C(p) = 2 × (-15p+300) + 1000
- C(p) = -30p + 600 + 1000
- C(p) = -30p + 1600
(D) To calculate the profit in terms of the cover charge, subtract the nightly costs (equation 3) from the nightly revenue (equation 2).
- P(p) = R(p)-C(p)
- P(p) = -15p² + 300p - (-30p + 1600)
- P(p) = -15p² + 300p+30p - 1600
- P(p) = -15p² + 330p - 1600
We will use the first derivative test (y'(x)=0) to find an extremum point to determine the entrance fee we should charge for maximum profit.
P's first derivative should be computed (p)
P'(p) = -30p + 330 = 0
30p = 330
p = 330/30
p = $11
For maximum profit, the entrance fee should be p=$11 per guest.
Therefore, the answers to all the parts of the question are:
- (A) q(p) = -15p + 300
- (B) R(p) = -15p² + 300p
- (C) C(p) = -30p + 1600
- (D.1) P(p) = -15p²+ 330p - 1600
- (D.2) p = $11
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The complete question is given below:
You have just opened a new nightclub, Russ' Techno Pitstop, but are unsure of how high to set the cover charge (entrance fee). One week you charged $9 per guest and averaged 165 guests per night. The next week you charged $10 per guest and averaged 150 guests per night.
(a) Find a linear demand equation showing the number of guests q per night as a function of the cover charge p. q(p) =
(b) Find the nightly revenue R as a function of the cover charge p. R(p) =
(c) The club will provide two free non-alcoholic drinks for each guest, costing the club $2 per head. In addition, the nightly overheads (rent, salaries, dancers, DJ, etc.) amount to $1,000. Find the cost C as a function of the cover charge p. C(p) =
(d) Now find the profit in terms of the cover charge p. P(p) =
Determine the entrance fee you should charge for a maximum profit. p = $ per guest.