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.

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

Step-by-step explanation:

(a).Before getting started, we are going to consider  the cartesian plane, where x-axis corresponds to the cover charge and y-axis corresponds to number of guests per night . There, we will locate the following coordinates according to the previous information:

(9,165)  (10,150).

Remember that  the slope-intercept form of the  linear equation is y=mx+b where m is the slope and b  is te intercept.

The slope can be calculated  by the next formula:

In this case we will assign  (9,165) as the first coordinate and (10,150) as the second one. The order does not matter. At the end we get the same value for m.

Therefore:

Now, we have the slope and two poitns. According to this, and taking into account that  point-slope form of the equation of a line is y-y1=m(x-x1),   find a linear demand equation.

Any of two coordintates can be selected. In this case we will select the second one.

y-150=-15(x-10)   equation 1

Solve equation 1 for y

y=-15x+150+150

y=-15x+300

Above we had defined x as cover charge (p) and y as number of guests (q). Thus, rewrite equation 1.

q(p)=-15p+ +300.  

it equation shows the number of guests per night as a function of the cover charge,

(b). The nightly revenue can be calculated multplying the price of the cover charge by  the number of guests. So, we just have to multiply the equation 1 by p.

p*q(p)=p*(-15p +300)

R(p)=-15p²+300p   equation 2

(c) To get the cost function of the nightclub  we just have to multiply the cost of the two non-alcoholic drinks   by the number of the guests ( equation 1) and add to it  nightly overheads.

C(p)=2*(-15p+300)+1000

C(p)= -30p+600+1000

C(p)=-30p+1600  equation 3

(d) In order to find the profit in terms of the cover charge we will  subtract nightly costs ( equation 3) from 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

To determine that  the entrance fee we should charge for a maximum profit we are going to use the first derivative test (y'(x)=0) in order to find an extremum point.

Compute the first derivative of P(p)

P'(p)=-30p+330=0

30p=330

p=330/30

p= $11

The entrance fee that we should charge for a maximum profit is p=$11 per guest.

a 270° rotation about point p says on e d g e n u it y (had to space it out because it says it's a misspelling so i couldn't submit it