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Math 133-1102a Unit 4 Group Project

MATH 133-1102A Unit 4 Group Project ABSTRACT Annual profit in thousands of dollars is given by the function, P(x) = 5000 – (1000/(x-1)), where x is the number of items sold in thousands, x > 1. 1. describe the meaning of the number 5000 in the formula 2. describe the meaning of the number 1 in the formula 3. find the profit for 5 different values of x 4. graph the profit function over its given domain; use the 5 values calculated in part 3 to construct the graph and connect these points with a smooth curve in Excel or another graphing utility. Insert the graph in a Word file and attach the graph in a Word file to the class DB thread. . will this profit function have a maximum, if so, what is it? 6. what steps should the company take to prepare for your answer to part 5? 1. Robin Lee 2. Derrick Roberts 3. Angela Johnson 4. June Stonehocker 5. Angela Johnson 6. Dedrana McCray #1The 5000 in the problem represents the limiting profit which is the maximum profit value, which can be the initial amount. www. algebra. com #2One is the limit of the annual profit equation where it determines the rise or increase in the annual profit or the lack their of the Elite Marketing Corporation. X=5 P=Profit P (5) =5000-(1000/ (5-1) I subtracted 1 from 5 and that leaves me with 4

P (5) =5000-(1000/ (5-1) P (5) =5000-(1000/(4) Next I begin to divide the 4 into 1000 and that gave me 250 P (5) =5000-(1000/4) P (5) =5000-250 Next step is to subtract 250 from 5000 and I will have my profit P (5) = 5000-250 P (5) =4750 Which means X is greater or higher than 1 because X which is 5 is greater than because X is what the profit will be at the end of the year and 1 is at the beginning of the year which means the company made a $4,750 profit. X>1 #3 Find the profit for 5 different values of x. ( both “x” and the profit, P(x), are listed in “thousands”) x = 3: P(3) = 5000 – [1000/(3 – 1)] = 5000 – (1000/2) = 4,500 = 5: P(5) = 5000 – [1000/(5 – 1)] = 5000 – (1000/4) = 4,750 x = 7: P(7) = 5000 – [1000/(7 – 1)] = 5000 – (1000/6) = 4,833. 33 x = 9: P(9) = 5000 – [1000/(9 – 1)] = 5000 – (1000/8) = 4,875 x = 11: P(11) = 5000 – [1000/(11 – 1)] = 5000 – (1000/10) = 4,900 #4 #5Will this profit function have a maximum, if so, what is it? Yes, there is a maximum profit; the maximum profit possible is 5000. The larger the value of x becomes the smaller the term 1000 / (x – 1) is, and the closer the function values are to 5,000. For very large values of x, the value of the function will be very, very close to 5,000. #6 REFERENCES: www. algebra. com