The Asterix theme park can serve at most 1,000 people per day. The park charges

Question

The Asterix theme park can serve at most 1,000 people per day. The park charges an admission fee, and all rides are free. If p is the price of admission on day i, the demand will be Di(p). The table below displays the demand curves for each day.

Suppose the theme park charges the same admission price on all days. Use the table above to identify a revenue maximizing price.

Suppose the park were free to charge a different price on each day and the demands on each day were independent of each other. What price on each day should be charged to maximize revenue?

Table 1: Demand Functions for the Asterisk Theme Park

Now suppose the demands on each day are not independent. A customer planning to attend on day i may choose to attend on day j instead if the price difference is large enough. In particular, for every $1 difference in price, 8 customers will shift days. For example, suppose the base prices for Monday, Tuesday and Wednesday are $20, $22 and $24 respectively. First compute D1(20), D2(22) and D3(24). The actual demand on Monday will be D1(20) + 2 × 8+4×8 and the actual demand on Tuesday will be D2(22)2×8+2×8. In other words, the $2 difference in price between Monday and Tuesday, means that 2 × 8 people will shift from Tuesday to Monday. The $2 difference in price between Tuesday and Wednesday, means that 2 × 8 people will shift from Wednesday to Tuesday. Under this assumption find revenue maximizing daily prices.

Index Day Di (p) 1 Sunday 3100 - 62p 2 Monday 1500 - 50p 3 Tuesday 1400 - 40p 4 Wednesday 1510 - 42p 5 Thursday 2000 - 52.6p 6 Friday 2500 - 55.6p 7 Saturday 3300 - 60p

Table 1: Demand Functions for the Asterisk Theme Park

Explanation / Answer

Index Day Di (p) Demand (People) Possible servings Price Revenue 1 Sunday 3100 - 62p 3100 1000 $    62.00 $           62,000 2 Monday 1500 - 50p 1500 1000 $    50.00 $           50,000 3 Tuesday 1400 - 40p 1400 1000 $    40.00 $           40,000 4 Wednesday 1510 - 42p 1510 1000 $    42.00 $           42,000 5 Thursday 2000 - 52.6p 2000 1000 $    52.60 $           52,600 6 Friday 2500 - 55.6p 2500 1000 $    55.60 $           55,600 7 Saturday 3300 - 60p 3300 1000 $    60.00 $           60,000 $       3,62,200 As max people served are 1000, The maximum price is on Sunday i.e. 62 So revenue maximising price is Di 1000 x 62 (p) = $ 62,000 If the demand is independent each day, park is free to charge the max price which is 62 which will result in $ 362, 200 per week for the park As the lowest price is on Tuesday which is $ 40 (considered as base price) Daily prices Difference from base price 8 customers per $ change Daily available customers (1000 less) Max revenue Sunday $              62.00 $              22.00 176                 824 $           51,088 Monday $              50.00 $              10.00 80                 920 $           46,000 Tuesday $              40.00 $                     -   0              1,000 $           40,000 Wednesday $              42.00 $                2.00 16                 984 $           41,328 Thursday $              52.60 $              12.60 101                 899 $           47,298 Friday $              55.60 $              15.60 125                 875 $           48,661 Saturday $              60.00 $              20.00 160                 840 $           50,400 $       3,24,775

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.