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Annual profit in thousands of dollars is given by the function, P(x) = -.1×2 + 50x – 300, where x is the number of items sold, x ? 0.Describe the meaning of the number -.1 in the formula, in terms of its meaning in relation to the profit.
Describe the meaning of the number -300 in the formula, in terms of its meaning in relation to the profit.
Find the profit for 5 different values of x
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?
What steps should the company take to prepare for your answer to part 5?
Post your final draft as a response to this post; use the small group area for collaboration.See the attached file.-.1 in the equation is the coefficient that is multiplied to the square of the number of items sold. This term will reduce the profit by a factor of .1 for the square of every item sold, and can be attributed as a factor that deducts a small portion of profit for selling each item.-300 in the formula is a constant, which is a representation of a fixed reduction of 300 regardless of the number of items sold. So for every every year, 300 thousand dollars is deducted from profit independent on the number of items sold on every year.We select x to have values of 100, 200, 300, 400 and 500. We have:
P(100) = -.1(100)^2 + 50(100) – 300 = \$3,700 thousand
P(200) = -.1(200)^2 + 50(200) – 300 = \$5,700 thousand
P(300) = -.1(300)^2 + 50(300) – 300 = \$5,700 thousand
P(400) = -.1(400)^2 + 50(400) – 300 = \$3,700 thousand
P(500) = -.1(500)^2 + 50(500) – 300 = \$-300 thousandGraph is attached on the word document.From the graph, it has a maximum point. The maximum profit can be computed by getting the number of sold items that will result in maximum (vertex of the parabolic curve). We have:h = -b/2a
h = -50/(2(-.1))
h = 250The maximum profit is:k = Pmax = -.1(250)^2 + 50(250) – 300
Pmax = \$5,950 thousandThe step to get the maximum profit, as stated in the previous problem, is to get the vertex of the parabolic curve. Another method is to use differential calculus and get the maximum, which will also result on the same answer.

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