Example 42 When a company launches a new product its share of the market after x months is calculated by the formula S(x) = x x 2 So after 5 months the share is S(5) = 2 / 5 – 4 / 25 = 6 / 25 Find the maximum share of the market that the company can achieve. (x  2)

End points S(2) = 1 – 1 = 0 There is no upper limit but as x   S(x)  0. SPs S(x) = x x 2 = 2x -1 – 4x -2 So S (x) = -2x x -3 = x 2 x 3 = x 3 x 2 SPs occur where S (x) = x 3 x 2 = 0 ctd

or 8 x 3 = 2 x 2 ( cross mult!) 8x 2 = 2x 3 8x 2 - 2x 3 = 0 2x 2 (4 – x) = 0 x = 0 or x = 4 Outwith interval In interval We now check the gradients either side of 4

x  4  S (x) S (3.9 ) = … S (4.1) = … Hence max TP at x = 4 And max share of market = S(4)= 2 / 4 – 4 / 16 = 1 / 2 – 1 / 4 = 1 / 4