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bearhunter [10]
3 years ago
5

A downhill ski area is experiencing a decline in the number of lift tickets sold, falling revenues, and inadequate profits. The

average price of a lift ticket is $20 and there are 2,500 tickets sold daily on average. The estimated price elasticity of demand is 1.5 and the lifts are currently operating at an average of 75 percent of capacity. Which of the following methods is most likely to increase the ski area's revenues and profits.
A. a 10 percent increase in the average price of a lift ticket.B. an aggressive advertising campaign.C. a 10 percent increase in the average price of a lift ticket combined with an aggressive advertising campaign.D. a 10 percent decrease in the average price of a lift ticket.
Business
1 answer:
sukhopar [10]3 years ago
5 0

Answer:

D. a 10 percent decrease in the average price of a lift ticket.

Explanation:

When Price elasticity is greater than 1, that suggests that the demand for that particular good or service is highly responsive to price or is price-sensitive . Furthermore, If price elasticity is greater than 1 then an increase in price will cause revenue to decrease.

Applying the above-stated principle to the given scenario, it has been stated that 'The estimated price elasticity of demand is 1.5.' implying that the demand for downhill ski is highly sensitive and responsive to changes in price.

Therefore, the only logical economic strategy to improve revenues will be to decrease price so that revenue can increase.

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elena-s [515]

Answer:

The minimum cost will be "$214085".

Explanation:

D = 1700 units \\\\S =  \$ 50 \\\\H=  20%\\

i) When quantity = 1-1500,  price = $ 12.50 , and holding price is $12.50 * 20 %= $2.50.

ii) When quantity = 1501 -10,000,  price = $ 12.45 , and holding price is $12.45 * 20 %= $2.49.

iii) When quantity = 10,0001- and more,  price = $ 12.40 , and holding price is $12.40 * 20 %= $2.48.

EOQ= \sqrt{\frac{2DS}{H}} \\\\EOQ1= \sqrt{\frac{2\times 17000\times 50}{2.50}} \\\\EOQ1=824.62 \ \ \ or \ \ \ 825\\

EOQ2= \sqrt{\frac{2\times 17000\times 50}{2.49}} \\\\EOQ1=826.2T \ \ \ or \ \ \ 826\\

EOQ3= \sqrt{\frac{2\times 17000\times 50}{2.48}} \\\\EOQ3=827.93 \ \ \ or \ \ \ 828\\

know we should calculate the total cost of EOQ1 and break ever points (1501 to 10,000)units

total \ cost = odering \ cost + holding \ cost + \ Annual \ product \ cost\\\\total_c  = \frac{D}{Q} \times S +  \frac{Q}{2} \times H + (p \times D) \\\\T_c  = \frac{17000}{825} \times 50+  \frac{825}{2} \times 2.50 + (12.50 \times 17000)\\\\T_c = 1030 .30 +1031.25+212500\\\\T_c =$ 214561.55\\\\

T_c  = \frac{17000}{1501} \times 50+  \frac{1501}{2} \times 2.49 + (12.45 \times 17000)\\\\T_c = 566.28 +1868.74+211650\\\\T_c =$ 214085.02 \ \ \ or \ \ \  $ 214085\\\\

T_c  = \frac{17000}{10001} \times 50+  \frac{10001}{2} \times 2.48 + (12.40 \times 17000)\\\\T_c = 84.99+ 12401.24+210800\\\\T_c =$ 223286.23 \\

The total cost is less then 15001. So, optimal order quantity is 1501, that's why cost is = $214085.

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• eqm Q = 175

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