(1 point) During the summer months Terry makes and sells necklaces on the beach.

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 30 24 15 12 15 (a) Find a linear function of the form n = Co + cip that best fits these data, using least squares. n = n(p) = 1761/37-798/37p (b) Find the revenue (number of items sold times the price of each item) as a function of price p R = R(p) = 1761/37p-798/37 p^2 (c) If the material for each necklace costs Terry 6 dollars, find the profit (revenue minus cost of the material) as a function of price p. P = P(p) = 1 1 7p-798/37p12-10565 (d) Finally, find the price that will maximize the profit. p= 2.21

Explanation / Answer

Ans(a):
type given data in excel.
Insert scatter plot from insert menu
right click on any point in the graph and select Add trend line (linear model) which gives equation:
y = -2.108x + 47.59
or we get n=-2.108p+ 47.59

Ans(b):
revenue is given by product times price
so that will be
R=R(p)=(-2.108p+ 47.59)*p
or R=R(p)=-2.108p^2+ 47.59p

Ans(c):
given that 1 necklace costs $6 then cost of n neckace 6n
Hence profit= revenue -cost
P(p)=-2.108p^2+ 47.59p-6p
or P(p)=-2.108p^2+ 41.59p

Ans(d):
above profit equation is quadratic and we know that for quadratic equation profit occurs at it's vertex.
which is given by
p=-b/(2a)=-41.59/(2*-2.108)=-41.59/(-4.216)=9.86480075901
Hence required value of p is approx 9.86

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