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 Question from Dina, a student: A meat market purchases steak from a local meat packinghouse. The meat is purchased on Monday at a price of $2 per pound, and the meat market sells the steak for$3 per pound. Any steak left over at the end of the week is sold to a local Zoo for $0.50 per pound. The demand for steak and the probabilities of occurrence are as follows: Demand Probability 20 10% 21 10% 22 15% 23 20% 24 20% 25 15% 26 10% Determine the amount of stock to maximize the profit. Draw the graph and explain. Hi Dina, I can get you started. You need to order between 20 and 26 pounds of meat and in each case calculate the expected profit and from this determine which size order has the largest expected profit. I will do two of the calculations. Suppose you order 20 pounds of meat. This will cost you$2 × 20 = $40. The demand table then says that you expect to sell all the meat so your revenue will be$3 × 20 = $60. Thus in this case expected profit = expected revenue - cost =$60 - $40 =$20.

Suppose you order 22 pounds of meat. This will cost you $2 × 22 =$44. From the demand table you expect to sell all this meat with probability (0.15 + 0.20 + 0.20 + 0.15 + 0.10) = 0.80. If you sell it all your revenue will be $3 × 22 =$66. There is a probability of 10% that your demand will only be for 21 pounds. In this case your revenue will be 0$3 × 21 +$0.50 × 1 = $63 +$0.50 = $63.50. There is also a probability of 10% that the demand will only be for 20 pounds and this this case your revenue will be$3 × 20 + $0.50 × 2 =$60 + $1 =$61. Hence if you order 22 pounds

expected profit = expected revenue - cost = 0.80 × $66 + 0.10 ×$63.50 + 0.10 × $61 -$44 = \$21.25

I hope this helps,
Harley

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.