PRELIMINARY MATHEMATICS LECTURE 5 Optimization

Reading Chiang, A. C. and K. Wainwright (2005) Chapters 9 and 11. Thomas (1999) Chapters 10-12. Jacques (2006) Chapter 5.

Local maxima and local minima In this diagram, A and B represent turning points of the curve y = f(x). A is a maximum turning point and B is a minimum turning point. The value of y at point A is known as the local maxima (maximum) and the value of y at point B is called the local minima (minimum).

Local maxima and local minima Note that the slope of the curve y = f(x), measured by is 0 at both turning points.

Second order condition for local maxima and minima If we draw the derivative on a graph we can confirm that

Points of inflexion The condition however is a necessary condition for local maxima or minima but not a sufficient condition. In the function below the sign of at point A is changing from positive through 0 to positive again, so this is neither a maximum nor a minimum turning point, but a point of inflection.

Second order condition for local maxima and minima What distinguishes the two points is the slope/gradient of the derivative function

Second order condition for local maxima and minima At point A, which is the point of local maximum, the slope of the curve is negative, while at point B, which is the point of local minimum, the slope is positive.

Second order condition for local maxima and minima The slope of is measured by the derivative of known as the second derivative, denoted as

Second order condition for local maxima and minima we can see that at the local maximum on the original curve, and

Second order condition for local maxima and minima we can see that at the local maximum on the original curve, and and at the local minimum,

Second order condition for a point of inflection A point of inflexion can be characterised by

Second order condition for a point of inflection A point of inflexion can be characterised by However it is also possible to have a local maximum or a local minimum with

Second order condition for local maxima and minima Hence, the obtained second order condition the local maxima, and and local minima, is a second order sufficient condition but not necessary because it is also possible to have a local maximum or a local minimum with

n th order condition for local maxima and minima n th derivative test for relative extrema or inflection point. If at a stationary point, continue differentiating until a non-zero higher-order derivative is obtained.

n th order condition for local maxima and minima n th derivative test for relative extrema or inflection point. If at a stationary point, continue differentiating until a non-zero higher-order derivative is obtained. If when evaluated at the stationary point then if n is an odd number , we have a point of inflection if n is an even number AND → a local maximum if n is an even number AND → a local minimum

nth derivative test – an example. Let us consider y = x 6. The first order necessary condition is obtained as and x = 0, Next check the second order sufficient condition: Evaluating at x = 0 ,

nth derivative test – an example. Because we cannot exclude the possibility that the function is at a local maximum or local minimum, we carry out the nth derivative test to check if this is relative extrema or inflection points. Third order derivative: , at x = 0, Forth order derivative: , at x = 0, Fifth order derivative: , at x = 0,

nth derivative test – an example. Sixth order derivative: Because the sixth order derivative is non-zero, whereby the non-zero is obtained at an order which is an even number, and because the non-zero derivative is positive, we have a local minimum.

nth derivative test – an example. This is confirmed by drawing a diagram:

Recap: Differentiation of bivariate functions To recap: How to evaluate stationary points… Step 1. First order condition Equate the first derivative to zero and find all stationary points.

Recap: Differentiation of bivariate functions To recap: How to evaluate stationary points… Step 2. Second order sufficient conditions Evaluate the second order derivative at each stationary points. If we have a local maximum If we have a local minimum

Recap: Differentiation of bivariate functions To recap: How to evaluate stationary points… Step 3. If at any stationary point the second order derivative is zero, carry on differentiating until a non-zero higher order derivative is found. If n is an odd number, we have a point of inflection If n is an even number AND we have a local maximum if n is an even number AND we have a local minimum

Second order partial derivatives Given a function z = f (x, y), the second order (direct) partial derivative signifies that the function has been partially differentiated with respect to one of the independent variables twice while the other independent variable has been held constant The notation with a double subscript signifies that the primitive function f is differentiated partially with respect to x twice. The alternative notation resembles that of but with the partial derivatives symbol.

Second order partial derivatives The cross partial derivatives signifies that the function has been partially differentiated with respect to one of the independent variables and then in turn partially differentiated with respect to another independent variable. The two cross partial derivatives are identical with each other, as long as the two cross partial derivatives are both continuous (Young’s theorem).

Unconstrained optimisation Recall that for a function of a single variable such as y = f (x) any point for which and was a local minimum. Similarly, any point for which was a local maximum.

Unconstrained optimisation Now consider a function of two variables z = f (x, y) When does z have a local minimum? Extrapolating our arguments of local minima of a single variable function, we may expect z to have a local minimum when it is not possible to obtain a smaller value of z by making slight changes in the values of x and y. There are three ways in which we can make ‘slight changes’ in x and y: (a) we can keep x constant and make y vary; (b) we can keep y constant and make x vary; (c) we can make both x and y vary at the same time. If z has a local minimum at point x = a and y = b, then we should be able to change the values of x and y by any of the methods (a), (b), or (c), and yet not make the value of z smaller.

Unconstrained optimisation Let us first consider case (a). In this case, the value of x is held constant. At a point where and we cannot obtain a smaller z by varying y alone. Similarly, if we consider case (b), when and we cannot obtain a smaller z by varying x alone.

Unconstrained optimisation Case (c) is slightly complicated. The condition that must be satisfied here is or

Unconstrained optimisation Hence, the second order conditions are as follows: A point at which the first partial derivatives are zero is: a local maximum if a local minimum if Note that if then we have neither a maxima nor a minima but what is called a saddle point

In the two-variable case, the total differential is Determinantal test In the two-variable case, the total differential is We can also derive the second-order total differential as

In the two-variable case, the total differential is Determinantal test In the two-variable case, the total differential is We can also derive the second-order total differential as

If d 2 z > 0 (positive definite), local minima Determinantal test If d 2 z > 0 (positive definite), local minima If d 2 z < 0 (negative definite), local maxima

The quadratic form can be expressed in matrix form as: Determinantal test Since the variables x and y appear only in squares, this can be seen as a quadratic form. As discussed in lecture 1, the condition for d 2 z to be positive/ negative definite may be stated by use of determinants. The quadratic form can be expressed in matrix form as:

Determinantal test The determinant with the second order partial derivatives as its elements is called a Hessian determinant.

The principal minors in this case are: Determinantal test The principal minors in this case are:

Hence the second order conditions are: Determinantal test Hence the second order conditions are: d 2 z > 0 (positive definite), iff and , then local minima. d 2 z < 0 (negative definite), iff and , then local maxima.

If we consider a function of three choice variables, Determinantal test If we consider a function of three choice variables, The Hessian determinant is

And the principal minors are Determinantal test And the principal minors are

d 2 z > 0 (positive definite), iff Determinantal test d 2 z > 0 (positive definite), iff and , then local minima. d 2 z < 0 (negative definite), and , then local maxima.

Unconstrained optimisation: application Application: profit maximisation Let us consider a firm that is a price taker, produces and sells two goods, goods 1 and goods 2 in perfectly competitive markets, and has a profit function: What would be the level of output for goods 1 and 2 satisfying the first order condition for maximum profit?

Unconstrained optimisation: application Application: profit maximisation Let us consider a firm that is a price taker, produces and sells two goods, goods 1 and goods 2 in perfectly competitive markets, and has a profit function: What would be the level of output for goods 1 and 2 satisfying the first order condition for maximum profit?

Unconstrained optimisation: application Application: profit maximisation Setting both condition equal to zero,

Unconstrained optimisation: application Application: profit maximisation Setting both condition equal to zero, Rewrite as

Unconstrained optimisation: application Application: profit maximisation Setting both condition equal to zero, Rewrite as Express this in terms of matrices

Unconstrained optimisation: application Application: profit maximisation Use Cramer’s rule

Unconstrained optimisation: application Application: profit maximisation Let us consider a firm that is a price taker, produces and sells two goods, goods 1 and goods 2 in perfectly competitive markets, and has a profit function: What would be the level of output for goods 1 and 2 satisfying the first order condition for maximum profit? The level of output/supply for good 1 must be 40 and for good 2 be 24 in order to satisfy the first order conditions.

Unconstrained optimisation: application Application: profit maximisation Let us consider a firm that is a price taker, produces and sells two goods, goods 1 and goods 2 in perfectly competitive markets, and has a profit function: What would be the level of output for goods 1 and 2 satisfying the first order condition for maximum profit? The level of output/supply for good 1 must be 40 and for good 2 be 24 in order to satisfy the first order conditions. Next, check the second order conditions to verify if we have local maxima.

Unconstrained optimisation: application Application: profit maximisation Let us consider a firm that is a price taker, produces and sells two goods, goods 1 and goods 2 in perfectly competitive markets, and has a profit function: Let us consider whether the stationary point we found at Q1 = 40 and Q2 = 24 is a local maxima or a local minima.

Unconstrained optimisation: application Application: profit maximisation The Hessian matrix can be written as Hence the profit is at local maximum at Q1 = 40 and Q2 = 24