Exercises 1-3, tell which equation you would choose to solve for one of the variables. Explain.
2 answers:
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Answer:
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.
