|
||||||||||||
|
||||||||||||
| ||||||||||||
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
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
I hope this helps, | ||||||||||||
|
||||||||||||
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. |