a. Find the demand function, assuming that it is linear. b. What should the selling price be to maximize his profit?

Question

During the summer months, Terry makes and sells necklaces on the beach. Last summer, he sold the necklaces for $10 each and his sales averaged 20 per day. When he increased his price by $1, he found that the average decreased by two sales per day.

a) Find the demand function (price p as a function of units sold, x), assuming that it is linear

p(x)=?

b) If the material for each necklace costs Terry $8, what should the selling price be to maximize his profit?

$? per necklace

Answer 1

20 for 10

18 for 11

0 for $20,priced out of the market

40 for free, market saturation, can't give more than 40 away

x(P)= 40- 2P is the linear demand function, giving quantity demanded as a function of price, but you apparently want the inverse function, price as a function of quantity demanded: so, switch x and P and solve for the new x. P=40-2x, 2x =40-P x = 20-P/2, but switch back, P=20-x/2. 10=20-20/2 = 20-10=10

so your linear demand function is

P(x)=20-x/2

R=xP= 40P-2P^2 = Revenue as a function of Price

cost=C=8x=8(40-2P)= 320-16P

profit= R-C=-2P^2+56P-320

take the derivative and set equal to zero, solve for P

(R-C)'=-4P+56=0

P= 56/4 = $14 = profit maximizing price

max profit= -2(14^2)+56(14)-320= 28(14)-320=$72

10 sells 20, making (10-8)(20)=$40

14 sells 12 making (14-8)(12)= 6(12)=$72

if you don't know calculus, here's an alternative algebraic or graphical method to find Max Profit

the profit function is a downward opening parabola with vertex = (14, 72)

-2P^2+56P -320

rewrite in vertex form

-2(P^2-28P +14^2) -320+(2)14^2=-320+392 = 72

= -2(P-14)^2 + 72 = a(x-h)^2+k where (h,k) = vertex, = maximum = (14,72)

although apparently not the P(x) demand function your problem seems to want