Calculus Optimization Methods

A key application of calculus is in optimization: finding maximum and minimum values of a function, and which points realize these extrema.

Context

Formally, the field of mathematical optimization is called mathematical programming, and calculus methods of optimization are basic forms of nonlinear programming. We will primarily discuss finite-dimensional optimization, illustrating with functions in 1 or 2 variables, and algebraically discussing n variables. We will also indicate some extensions to infinite-dimensional optimization, such as calculus of variations, which is a primary application of these methods in physics.

Techniques

Basic techniques include the first and second derivative test, and their higher-dimensional generalizations.

A more advanced technique is Lagrange multipliers, and generalizations as Karush–Kuhn–Tucker conditions and Lagrange multipliers on Banach spaces.

Applications

Optimization, particularly via Lagrange multipliers, is particularly used in the following fields:

Further, several areas of mathematics can be understood as generalizations of these methods, notably Morse theory and calculus of variations.

Terminology

• Input points, output values
• Maxima, minima, extrema, optima
• Stationary point, critical point; stationary value, critical value
• Objective function
• Constraints – equality and inequality
• Especially sublevel sets
• Feasible region, whose points are candidate solutions

Statement

This tutorial presents an introduction to optimization problems that involve finding a maximum or a minimum value of an objective function $f(x_{1},x_{2},\ldots ,x_{n})$  subject to a constraint of the form $g(x_{1},x_{2},\ldots ,x_{n})=k$ .

Maximum and minimum

Finding optimum values of the function $f(x_{1},x_{2},\ldots ,x_{n})$  without a constraint is a well known problem dealt with in calculus courses. One would normally use the gradient to find stationary points. Then check all stationary and boundary points to find optimum values.

Example

• $f(x,y)=2x^{2}+y^{2}$
• $f_{x}(x,y)=4x=0$
• $f_{y}(x,y)=2y=0$

$f(x,y)$  has one stationary point at (0,0).

The Hessian

A common method of determining whether or not a function has an extreme value at a stationary point is to evaluate the hessian of the function at that point. where the hessian is defined as

$H(f)={\begin{bmatrix}{\frac {{\partial }^{2}f}{\partial x_{1}^{2}}}&{\frac {{\partial }^{2}f}{\partial x_{1}\partial x_{2}}}&\dots &{\frac {{\partial }^{2}f}{\partial x_{1}\partial x_{n}}}\\{\frac {{\partial }^{2}f}{\partial x_{2}\partial x_{1}}}&{\frac {{\partial }^{2}f}{\partial x_{2}^{2}}}&\dots &{\frac {{\partial }^{2}f}{\partial x_{2}\partial x_{n}}}\\\vdots &\vdots &\ddots &\vdots \\{\frac {{\partial }^{2}f}{\partial x_{n}\partial x_{1}}}&{\frac {{\partial }^{2}f}{\partial x_{n}\partial x_{2}}}&\dots &{\frac {{\partial }^{2}f}{\partial x_{n}^{2}}}\\\end{bmatrix}}.$

Second derivative test

The Second derivative test determines the optimality of stationary point $x$  according to the following rules :

• If $H(f)>0$  at point x then $f$  has a local minimum at x
• If $H(f)<0$  at point x then $f$  has a local maximum at x
• If $H(f)$  has negative and positive eigenvalues then x is a saddle point
• Otherwise the test is inconclusive

In the above example.

$H(f)={\begin{bmatrix}4&0\\0&2\end{bmatrix}}.$

Therefore $f(x,y)$  has a minimum at (0,0).