For each of the following quadratic functions, determine whether there is a minimum. If so, find

For each of the following quadratic functions, determine...

Key Concepts

Quadratic Functions

A quadratic function in several variables is a polynomial of degree two that can be expressed as a sum of squared terms, cross-product terms, linear terms, and a constant. These functions naturally arise in many areas of mathematics and engineering, particularly in optimization, as their curvature properties make them well-suited for analysis using differential calculus and linear algebra.

Matrix Representation of Quadratic Forms

Many quadratic functions of multiple variables can be written in the compact form Q(x) = x?Ax + b?x + c, where A is a symmetric matrix, b is a vector, and c is a constant. This representation facilitates the use of linear algebra tools, enabling techniques such as finding stationary points by solving linear systems, and analyzing the function’s curvature by examining the matrix A.

Positive Definiteness

A key concept in the study of quadratic functions is positive definiteness. A symmetric matrix A is positive definite if for every nonzero vector x, the quadratic form x?Ax is positive. This property guarantees that the quadratic function is convex and thus has a unique global minimum. Determining whether A is positive definite is essential in optimization and is typically done by checking the eigenvalues or leading principal minors of A.

Completing the Square

Completing the square is an algebraic technique used to reframe a quadratic function as a sum of squared terms plus a constant. This method can simplify the process of finding the minimizer and the minimum value of the function by isolating the squared portion that reaches its minimum when equal to zero. It is particularly useful when the quadratic function is not immediately expressed in its matrix form.

Stationary Points and Gradient Methods

In optimization, the minimization of quadratic functions is often performed by finding stationary points where the gradient is zero. For a function expressed in the form Q(x) = x?Ax + b?x + c, setting the gradient equal to zero leads to a system of linear equations whose solution gives the candidate minimizer. If the Hessian matrix (which, in this case, is 2A) is positive definite, then the stationary point is indeed a global minimum.

Transcript

00:01 We're given a function and we're asked to find this function's minimum value subject to two constraints.

00:10 The function is f of x, y, z equals x squared plus y squared plus c squared, and our constraints are the planes x plus 2y plus z equals 3 and x minus y equals 4.

00:30 So we'll call these g of x, y, z equals x plus 2y plus z equals 3, and h of x, y, z equals x minus y equals 4.

01:06 So first let's write out the lagrange equations.

01:11 We have the gradient of f is 2x, 2y, 2 z, the gradient of g, is 1, 2, and 2, and the gradient of to 1, and the gradient of h is 1 negative 1 ,0.

01:30 Therefore, the grange condition says that the gradient of f is equal to lambda times the gradient of g plus mu times the gradient of h.

01:37 So we have 2x, 2y, 2y, 2z equals lambda times 1 to 1, 2, plus mu times 1 negative 1 ,0.

01:54 And so we get the 3 ,000.

01:56 Or range equations, 2x equals lambda plus mu, 2y equals 2 lambda minus mu, and 2z equals lambda i want to solve for lambda and mu.

02:21 So our first equation gives us that lambda is equal to 2x minus mu.

02:30 Combining this with our third equation, we get that 2 z is equal to 2x x minus mu.

02:45 Our second equation, solving it for mu, we get that mu is equal to 2 lambda minus 2 y.

03:03 And combining this with our third equation, we get that mu is equal to 4 z minus 2y.

03:19 So we have these two equations in mu x and z.

03:25 And so we have 2 z is equal to 2.

03:33 2x minus and then u is 4 z minus 2y which is the same as 2x minus 4 z plus 2y and therefore we have that 6 z equals 2x plus 2y or z is equal to x plus y over 3 this is the line along which we're looking now now i want to solve for x y and z using our constraints so we have have that x minus y equals 4.

04:31 Well, this tells us that y is equal to x minus 4.

04:40 Likewise, x plus 2y plus z equals 3...

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