(a) The number of guest per night=p,
and the cover charge for one night per person=q.
Let the liner demand equation be
,
where m and c are constants.
Given that for the first week: q=345 and p=$7 and
for the second night: q=300 and p=$10.
Put these values in equation (i), we have
345=7m+c
![\Rigntarrow c=345-7m\cdots(ii)](https://img.qammunity.org/2021/formulas/mathematics/college/biz2464kq5zqcro4ff3bujik97k1qbhdcq.png)
and
300=10m+c
[from equation (ii)]
![\Rightarrow 3m=-45](https://img.qammunity.org/2021/formulas/mathematics/college/mqi5iivcwohr0i8gw08pxgzdynx778j4f8.png)
![\Rightarrow m=-15](https://img.qammunity.org/2021/formulas/mathematics/college/q01irynz84xcu2pbmetdcgvj9indgu1r90.png)
and from equation (ii),
.
Hence, the required demand equation is:
![q(p)=-15p+450\cdots(iii)](https://img.qammunity.org/2021/formulas/mathematics/college/kktge9801txw7qvhj2l3zljr4dhglkw0fv.png)
(b) Revenue = (Number of guests) x (Cover charge)
R=qp
[from equation (iii)]
![\Rightarrow R(p)=-15p^2+450p\;\cdots(iv)](https://img.qammunity.org/2021/formulas/mathematics/college/1f2r45xxc7jtr71xrhdjf3jd7wxoda76oe.png)
(c) The costs per night for drinks for the number of guest p is $2p and the nightly overheads is $2000.
So, the required cost function is
![C(p)=2p+2000\;\cdots(v)](https://img.qammunity.org/2021/formulas/mathematics/college/2029f1z6848fdmzvfmmlafwu3nv26rhha6.png)
(d) Profit=(Revenue)-(Total costs)
i.e
![P(p)=R(p)-C(p)](https://img.qammunity.org/2021/formulas/mathematics/college/mrn3lak98xgv0iv300y0sejx3ku1f18ax6.png)
Using equations (iv) and (v), we have
![\Rightarrow P(p)=(-15p^2+450p)-(2p+2000)](https://img.qammunity.org/2021/formulas/mathematics/college/q34dhxvlhyqj00y930mx3he7m9bf3zgqba.png)
![\Rightarrow P(p)=-15p^2+448p-2000.](https://img.qammunity.org/2021/formulas/mathematics/college/ov1tdmkve0xzskp5jgj6alwlucp6fsb0xu.png)
Now we have the profit function as function of cover charge (entry fee) per person.
So, to get the maximum profit, P, differentiate the profit function with respect to the cover charge, q and equate it to zero to get the eqtremum point.
![\frac {dP}{dq}=0](https://img.qammunity.org/2021/formulas/mathematics/college/ekyosg24uoi6dpzk4ybuz1fiq4lebsoz5w.png)
![\Rightarrow -30p+448=0](https://img.qammunity.org/2021/formulas/mathematics/college/yt67blq7ct39pm2r3h37fsgr0k8y4otise.png)
![\Rightarrow p=14.93](https://img.qammunity.org/2021/formulas/mathematics/college/ofogzv9065viwl9afkx0wjc9d2rxe4ybba.png)
This point is either maxima or minima, so differentiate the profit function twice to to ensure the value of p corrosponding to what.
![\frac {d^2P}{dq^2}=-30](https://img.qammunity.org/2021/formulas/mathematics/college/gkoqk4i4wsdfnl9cw2rcc6ooicxfyqyjqv.png)
This the second deravitive is negative, so p=14.93 is the value for maximum profit.
Hence, for the maximum profit, the cover charge per guest is $14.93.