1. Maximization without Calculus Not all economic maximization problems can be solved using calculus –If a manager does not know the profit function, but can approximate parts of it by straight lines Quantity  ** q*  = f(q) d  /dq does not exist at q*

2. Maximization without Calculus Calculus also cannot be used in the case where a firm cannot produce fractional values of output d  /dq does not exist at q*

3. Second Order Conditions - Functions of One Variable Let y = f(x) A necessary condition for a maximum is that dy/dx = f ’(x) = 0 To ensure that the point is a maximum, y must be decreasing for movements away from it

4. Second Order Conditions - Functions of One Variable The total differential measures the change in y dy = f ‘(x) dx To be at a maximum, dy must be decreasing for small increases in x To see the changes in dy, we must use the second derivative of y

5. Second Order Conditions - Functions of One Variable Note that d 2 y < 0 implies that f ’’(x)dx 2 < 0 Since dx 2 must be positive, f ’’(x) < 0 This means that the function f must have a concave shape at the critical point

6. Second Order Conditions - Functions of Two Variables Suppose that y = f(x 1, x 2 ) First order conditions for a maximum are  y/  x 1 = f 1 = 0  y/  x 2 = f 2 = 0 To ensure that the point is a maximum, y must diminish for movements in any direction away from the critical point

7. Second Order Conditions - Functions of Two Variables The slope in the x 1 direction (f 1 ) must be diminishing at the critical point The slope in the x 2 direction (f 2 ) must be diminishing at the critical point But, conditions must also be placed on the cross-partial derivative (f 12 = f 21 ) to ensure that dy is decreasing for all movements through the critical point

8. Second Order Conditions - Functions of Two Variables The total differential of y is given by dy = f 1 dx 1 + f 2 dx 2 The differential of that function is d 2 y = (f 11 dx 1 + f 12 dx 2 )dx 1 + (f 21 dx 1 + f 22 dx 2 )dx 2 d 2 y = f 11 dx 1 2 + f 12 dx 2 dx 1 + f 21 dx 1 dx 2 + f 22 dx 2 2 By Young’s theorem, f 12 = f 21 and d 2 y = f 11 dx 1 2 + 2f 12 dx 2 dx 1 + f 22 dx 2 2

9. Second Order Conditions - Functions of Two Variables d 2 y = f 11 dx 1 2 + 2f 12 dx 2 dx 1 + f 22 dx 2 2 For this equation to be unambiguously negative for any change in the x’s, f 11 and f 22 must be negative If dx 2 = 0, then d 2 y = f 11 dx 1 2 –For d 2 y < 0, f 11 < 0 If dx 1 = 0, then d 2 y = f 22 dx 2 2 –For d 2 y < 0, f 22 < 0

10. Second Order Conditions - Functions of Two Variables d 2 y = f 11 dx 1 2 + 2f 12 dx 2 dx 1 + f 22 dx 2 2 If neither dx 1 nor dx 2 is zero, then d 2 y will be unambiguously negative only if f 11 f 22 - f 12 2 > 0 –The second partial derivatives (f 11 and f 22 ) must be sufficiently large that they outweigh any possible perverse effects from the cross- partial derivatives (f 12 = f 21 )

11. Constrained Maximization Suppose we want to choose x 1 and x 2 to maximize y = f(x 1, x 2 ) subject to the linear constraint c - b 1 x 1 - b 2 x 2 = 0 We can set up the Lagrangian L = f(x 1, x 2 ) - (c - b 1 x 1 - b 2 x 2 )

12. Constrained Maximization The first-order conditions are f 1 - b 1 = 0 f 2 - b 2 = 0 c - b 1 x 1 - b 2 x 2 = 0 To ensure we have a maximum, we must use the “second” total differential d 2 y = f 11 dx 1 2 + 2f 12 dx 2 dx 1 + f 22 dx 2 2

13. Constrained Maximization Only the values of x 1 and x 2 that satisfy the constraint can be considered valid alternatives to the critical point Thus, we must calculate the total differential of the constraint -b 1 dx 1 - b 2 dx 2 = 0 dx 2 = -(b 1 /b 2 )dx 1 These are the allowable relative changes in x 1 and x 2

14. Constrained Maximization Because the first-order conditions imply that f 1 /f 2 = b 1 /b 2, we can substitute and get dx 2 = -(f 1 /f 2 ) dx 1 Since d 2 y = f 11 dx 1 2 + 2f 12 dx 2 dx 1 + f 22 dx 2 2 we can substitute for dx 2 and get d 2 y = f 11 dx 1 2 - 2f 12 (f 1 /f 2 )dx 1 + f 22 (f 1 2 /f 2 2 )dx 1 2

15. Constrained Maximization Combining terms and rearranging d 2 y = f 11 f 2 2 - 2f 12 f 1 f 2 + f 22 f 1 2 [dx 1 2 / f 2 2 ] Therefore, for d 2 y < 0, it must be true that f 11 f 2 2 - 2f 12 f 1 f 2 + f 22 f 1 2 < 0 This equation characterizes a set of functions termed quasi-concave functions –Any two points within the set can be joined by a line contained completely in the set

16. Constrained Maximization Recall the fence problem: Maximize A = f(x,y) = xy subject to the constraint P - 2x - 2y = 0 Setting up the Lagrangian [L = x·y + (P - 2x - 2y)] yields the following first-order conditions:  L/  x = y - 2 = 0  L/  y = x - 2 = 0  L/  = P - 2x - 2y = 0

17. Constrained Maximization Solving for the optimal values of x, y, and yields x = y = P/4 and = P/8 To examine the second-order conditions, we compute f 1 = f x = yf 2 = f y = x f 11 = f xx = 0f 12 = f xy = 1 f 22 = f yy = 0

18. Constrained Maximization Substituting into f 11 f 2 2 - 2f 12 f 1 f 2 + f 22 f 1 2 we get 0 ·x 2 - 2 ·1 ·y ·x + 0 ·y 2 = -2xy Since x and y are both positive in this problem, the second-order conditions are satisfied