Question

(1 point) During the summer months Terry makes and sells necklaces on the beach. Terry notices that if he lowers the price, he can sell more necklaces, and if he raises the price than he sells fewer necklaces. The table below shows how the number n of necklaces sold in one day depends on the price p (in dollars) Price |Number of necklaces sold її 24 16 15 (a) Using least squares find a linear function (i.e., a line) of the form that best fts this data C 51 D-1-(27/11) (b) Find the revenue (number n of items sold times the price p of each item) as a function of the price p. (In the box to the right of the equal sign enter an equation that contains only the variable p and the values for C' and D you entered in part (a) above. Note: This equation in p may not be a line.) Revenue- (С the material for each necklace costs er y 5 dollars, r nd the prot revenue minus cost of the material as a function of the price p p the values for C and D you entered in part (a) and the cost 5. Note: As in part (b), your equation in p may not be a line) As in part b above enter an equation that contains only the variable Profit- (d) Finally, find the price p that will maximize the profit. (Use what you learned in calculus about finding the maximum of a function of single variable p.) P 31.67-0.49p

Answer 1

Given that: - (a) p^bar = 2P/3 Sigma P = 8 + 11 + 16 = 35 implies p^bar = 35/3 n^bar = sigma n/3 pi = 72/3 Slope of the line, m = Sigma (Pi - p) (n_i - n^bar)/(p_i - p^bar)^2 \r\n m = (8 - 35/3) (33 - 72/3) + (11 - 35/3) (24 - 72/3) + (16 - 35/3) (15 - 72/3)/(8 - 35/3)^2 + (11 - 35/3)^2 + (16 - 35/3)^2 m = -33 + 0 - 39/13.44 + 0.44 + 18.77 m = -2.2052 Slope of the line, m = -2.2052 b^bar = n^bar - m p^bar n^bar = 72/3, m = -2.2052, p^bar = 35/3 \r\n b^bar = 72/3 - (-2.2052 times 35/3) b^bar = 49.727 implies n = 49.727 - 2.2052 p Therefore C = 49.727 D = -2.2052 \r\n (b) R = n times P where n = number of products p = price implies R = 49.7271 - 2.2052 p^2 Therefore Revenue = 49.727 P - 2.2052 p^2