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1. A second hand car showroom has a space of 420 m^2 for displaying two different models of cars: a family car and a sport car. A family car needs 8 m^2 of space and a sports car needs 10 m^2 of space for display. The cost for displaying a family car and a sport car is $300 and $600 respectively. The total cost for... 顯示更多 1. A second hand car showroom has a space of 420 m^2 for displaying two different models of cars: a family car and a sport car. A family car needs 8 m^2 of space and a sports car needs 10 m^2 of space for display. The cost for displaying a family car and a sport car is $300 and $600 respectively. The total cost for the display should not exceed $18000. Asume that there are x family cars and y sports cars displayed. The constraints are: 8x+10y(<=)420...(1) 300x+600y<=18000...(2) x and y are non negative integers...(3) The net profit on selling a family car and a sport car is $6000 and $10000 respectively. Asume that all the cars displayed will be sold. 40 family cars should be displayed to get the maximum profit $340000. (If the number of family cars sold is at least five times the number of sports cars sold, find the maximum profit and the corresponding number of family cars sold)

最佳解答:

There is one more constraint : x > 5y. Max. profit happens at the intersection of x = 5y and 8x + 10y < 420. That is x = 42 and y = 8.4. Since x and y must be integer, so x = 42 and y = 8. Max. profit = 42 x 6000 + 8 x 10000 = $332,000.

其他解答:31C9A75CB3B14398
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