Studio Funzione

SOLUTION EXERCISES OPTIMIZATION

140 120 100 80 f(x) 60 40 20 0 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 x 3 4 5 6 7 8 9 10 11 12 13

Exercises

Optimization

Lesson 6

1)

Find and classify the stationary points of the following functions; then draw their graphs. a) f(x) = x2 - 4x + 5

f ’ (x) = 2x -4 we impose 0 = f ’ (x)

or

0= 2x -4 f ’’(x)= 2 x

the solution of this equation is x = 2

(for any x)

then in the point (2, 1) there is a minimum b) 2x3 + 3 x2 -12x + 4 = y y’ = 6x2 + 6x -12 I impose 0 = f ’ (x) = 6x2 + 6x -12 -2 -6 +/- (36+288) 12

-1/2

= ( -6+/- 18) /12 = +1

y’’ = 12x + 6

50 40 30 20 f(x) 10 0

9 3 -1 -0 .7 -0 .4 -0 .1 0. 2 0. 5 0. 8 1. 1 1. 4 1. 7 5 2 7 4 1 8 6 3 2 2. -1 .

f ’’ (-2) = -18 in (-2, 24 ) there is a maximum f ’’ (1) = 18 in (1, -3 ) there is a minimum Both are local.

-1 . -2 . -2 . -1 . -3 . -3 . -2 . 2. 6

-10 -20 x

-3 .

SOLUTION EXERCISES OPTIMIZATION

2)

A firm’s short-run production function is given by Q = 6L2 - 0.2L3

a) find the size of the workforce that maximizes output and hence sketch a graph of this production function Q’ = 12L -0,6L2 = 6L (2 -0,01L) = Solutions L =0 and L=20 Q’(0) = 0 = Q’ ( 20) 900

800 700 600 500 400 300 200 100 0

14 16 18 22 24 26 28 12 10 20 30 0 2 4 6 8

Q’’ = 12-1,2L the second derivative is nil if L=10, in that point there is an inflexion point, but if L =20 it is negative, Q’’(20) = -12, (there is a maximum) and for L = 0, Q’’(0) = +12, it is positive, there is a minimum. For LQ(20),but a

negative number of workers....... For L >0 Q(20)is a maximum. In this plant the maximum which I can produce is with 20 workers. Find the size of the workforce that maximizes the average product of labour. Q = 6L2 - 0.2L3 Q/ L = average product of labour = ALP ALP= 6L-0.2L2 ALP’ = 6-0.4L and ALP’’= -0,4 the second derivative is 0, x with Q ≠ 0 4 7 2 AC ’ = 0 se +10 = 10 /Q Then: then 103 = Q2 then Q = +31,62 1= 103/Q2 Even if this number is closer to 32 than to 31,...