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 31 23 16 15 (a) Using least squares find a linear function (i.e., a line) of the form n = C + DP that best fits this data. C = D = (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 = (c) If the material for each necklace costs Terry 6 dollars, find the profit (revenue minus cost of the material) as a function of the price p. (As in part (b) above enter an equation that contains only the variable p, the values for C and D you entered in part (a) and the cost 6. Note: As in part (b), your equation in p may not be a line.) 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 =

Answer 1

2 n=c+ DP +(C+16D-15)? of p=7 then n=C+7D, necklaces sold = 31 p=11 then n=C+IID, necklaces sold = 23 þ= 16 then n=C+I6D, neck laces sold = 15 Squaled ellos E = (C+70-31)+(c +11D-23) this errol is leart luhen E=0, 2 = 0 3 =») 2 (c+70-31)+ 2 (C+1D-23) +2 (C+16D-15) = 0 30+34D=69 0 20+0 = 14(C+70-31)+22 (C+11D-23) + 32 (C+16 D-15) = 0. 34C+ 426 D = 710 - © solving ① and ② we get C = 43.06557 ID=-1.77049 n = 43.06557-1.770495

(6 Revenue = number of items sold & plice = np - (43.06557-1.77049 p) Revenue = 143.065575-1.770495? © Cost of each necklace = $6 Cost of n necklaces = 60 (dollars) Profit = Revenue - Cost = np - 6n = 43.065575-1.7704956% 43.06557 + 6x 1.77049 p plofit = -1.77049 p +5368855-258.3434) o when petit us maximum & poefit )-0, (pofit) <0 (peofit)=0 -2x1.77044f+53.6885 = 0 53.6885 P 3.54098 -= 15.1620 I (Peofit) <o therefore != $15.1620