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Multivariable calculus

Course: Multivariable calculus > Unit 3

Lesson 4: Optimizing multivariable functions (articles)
Math
Multivariable calculus
Applications of multivariable derivatives
Optimizing multivariable functions (articles)
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Reasoning behind second partial derivative test

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For those of you who want to see why the second partial derivative works, I cover a sketch of a proof here.  

Background

In the last article, I gave the statement of the second partial derivative test, but I only gave a loose intuition for why it's true. This article is for those who want to dig a bit more into the math, but it is not strictly necessary if you just want to apply the second partial derivative test.

What we're building to

  • To test whether a stable point of a multivariable function is a local minimum/maximum, take a look at the quadratic approximation of the function at that point. It is easier to analyze whether this quadratic approximation has maximum/minimum.
  • For two-variable functions, this boils down to studying expression that look like this:
    start color #0c7f99, a, end color #0c7f99, x, squared, plus, 2, start color #0d923f, b, end color #0d923f, x, y, plus, start color #bc2612, c, end color #bc2612, y, squared
    These are known as quadratic forms. The rule for when a quadratic form is always positive or always negative translates directly to the second partial derivative test.

Single variable case via quadratic approximation

First, I'd like to walk through the formal reasoning behind why the single-variable second derivative test works. By formal, I mean capturing the idea of concavity into more of an airtight argument.
In single-variable calculus, when f, prime, left parenthesis, a, right parenthesis, equals, 0 for some function f and some input a, here's what the second derivative test looks like:
  • f has a local maximum at a if f, start superscript, prime, prime, end superscript, left parenthesis, a, right parenthesis, is less than, 0
  • f has a local minimum at a if f, start superscript, prime, prime, end superscript, left parenthesis, a, right parenthesis, is greater than, 0
  • If f, start superscript, prime, prime, end superscript, left parenthesis, a, right parenthesis, equals, 0, the second derivative alone cannot determine whether f has a maximum, minimum or inflection point at a.
To think about why this test works, start by approximating the function with a taylor polynomial out to the quadratic term, also known as a quadratic approximation.
f(x)≈f(a)+f′(a)(x−a)+12f′′(a)(x−a)2\begin{aligned} \quad f(x) \approx f(a) + f'(a)(x-a) + \dfrac{1}{2}f''(a)(x - a)^2 \end{aligned}
Since f, prime, left parenthesis, a, right parenthesis, equals, 0, this quadratic approximation simplifies like this:
f(a)+12f′′(a)(x−a)2\begin{aligned} \quad f(a) + \dfrac{1}{2}f''(a)(x - a)^2 \end{aligned}
The quadratic approximation at a local minimum.
The quadratic approximation at a local minimum.
Notice, left parenthesis, x, minus, a, right parenthesis, squared, is greater than or equal to, 0 for all possible x since squares are always positive or zero. That simple fact tells us everything we need to know! Why?
It means that when f, start superscript, prime, prime, end superscript, left parenthesis, a, right parenthesis, is greater than, 0, we can read our approximation like this:
f(a)+12f′′(a)(x−a)2⏟This is ≥0 for all values of x,and equals 0 only when x=a\begin{aligned} \quad f(a) + \underbrace{ \blueE{\dfrac{1}{2}f''(a)(x - a)^2} }_{\substack{ \text{This is $\ge 0$ for all values of $x$,} \\ \text{and equals $0$ only when $x=a$} }} \end{aligned}
Therefore a is a local minimum of our approximation. In fact, it is a global minimum, but we only care about the fact that it is a local minimum. When the quadratic approximation of a function has a local minimum at the point of approximation, the function itself must also have a local minimum there. I'll say more on this in the last section, but for now the intuition should be clear since the function and its approximation "hug" one another around the point of approximation a.
The quadratic approximation at a local maximum
The quadratic approximation at a local maximum
Similarly, if f, start superscript, prime, prime, end superscript, left parenthesis, a, right parenthesis, is less than, 0, we can read the approximation as
f(a)+12f′′(a)(x−a)2⏟This is ≤0 for all values of x,and equals 0 only when x=a\begin{aligned} \quad f(a) + \underbrace{ \redD{\dfrac{1}{2}f''(a)(x - a)^2} }_{\substack{ \text{This is $\le 0$ for all values of $x$,} \\ \text{and equals $0$ only when $x=a$} }} \end{aligned}
In this case, the approximation has a local maximum at x, equals, a, indicating that the function itself also has a local maximum there.
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
The quadratic approximation at the inflection point is flat.
The quadratic approximation at the inflection point is flat.
When f, start superscript, prime, prime, end superscript, left parenthesis, a, right parenthesis, equals, 0, our quadratic approximation always equals the constant f, left parenthesis, a, right parenthesis, meaning our function is in some sense too flat to be analyzed by the second derivative alone.
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
What to take away from this:
When f, prime, left parenthesis, a, right parenthesis, equals, 0, studying whether f has a local maximum or minimum at a comes down to whether the quadratic term of the Taylor approximation start fraction, 1, divided by, 2, end fraction, f, start superscript, prime, prime, end superscript, left parenthesis, a, right parenthesis, left parenthesis, x, minus, a, right parenthesis, squared is always positive or always negative.

Two variable case, visual warmup

Now suppose you have a function f, left parenthesis, x, comma, y, right parenthesis with two inputs and one output, and you find a stable point. That is, a point where both its partial derivatives are 0,
fx(x0,y0)=0fy(x0,y0)=0\begin{aligned} \quad f_x(x_0, y_0) = 0 \\ f_y(x_0, y_0) = 0 \\ \end{aligned}
which is more succinctly written as
del, f, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, equals, start bold text, 0, end bold text, left arrow, start color gray, start text, Z, e, r, o, space, v, e, c, t, o, r, end text, end color gray
Tangent plane
del, f, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, equals, start bold text, 0, end bold text indicates that the tangent plane at left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis is flat.
In order to determine whether this is a local maximum, local minimum, or neither, we look to it's quadratic approximation. Let's start with a visual preview of what we want to do:
  • f will have a local minimum at a stable point left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis if the quadratic approximation at that point is a concave-up paraboloid.
    Local min
  • f will have local maximum there if the quadratic approximation is a concave down paraboloid:
    Local max
  • If the quadratic approximation is saddle-shaped, f has neither a maximum nor a minimum, but a saddle point.
    Saddle point
  • If the quadratic approximation is flat in one or all directions, we do not have enough information to make conclusions about f.
    Quadratic approximation is flat in one direction.
    Quadratic approximation is constant.

Analyzing the quadratic approximation

The formula for the quadratic approximation of f, in vector form, looks like this:
Q, start subscript, f, end subscript, left parenthesis, start bold text, x, end bold text, right parenthesis, equals, start underbrace, f, left parenthesis, start bold text, x, end bold text, start subscript, 0, end subscript, right parenthesis, end underbrace, start subscript, start text, C, o, n, s, t, a, n, t, end text, end subscript, plus, start underbrace, del, f, left parenthesis, start bold text, x, end bold text, start subscript, 0, end subscript, right parenthesis, dot, left parenthesis, start bold text, x, end bold text, minus, start bold text, x, end bold text, start subscript, 0, end subscript, right parenthesis, end underbrace, start subscript, start text, L, i, n, e, a, r, space, t, e, r, m, end text, end subscript, plus, start underbrace, start fraction, 1, divided by, 2, end fraction, left parenthesis, start bold text, x, end bold text, minus, start bold text, x, end bold text, start subscript, 0, end subscript, right parenthesis, start superscript, T, end superscript, start bold text, H, end bold text, f, left parenthesis, start bold text, x, end bold text, start subscript, 0, end subscript, right parenthesis, left parenthesis, start bold text, x, end bold text, minus, start bold text, x, end bold text, start subscript, 0, end subscript, right parenthesis, end underbrace, start subscript, start text, Q, u, a, d, r, a, t, i, c, space, t, e, r, m, end text, end subscript
Since we care about points where the gradient is zero, we can get rid of that gradient term
Q, start subscript, f, end subscript, left parenthesis, start bold text, x, end bold text, right parenthesis, equals, f, left parenthesis, start bold text, x, end bold text, start subscript, 0, end subscript, right parenthesis, plus, start fraction, 1, divided by, 2, end fraction, left parenthesis, start bold text, x, end bold text, minus, start bold text, x, end bold text, start subscript, 0, end subscript, right parenthesis, start superscript, T, end superscript, start bold text, H, end bold text, f, left parenthesis, start bold text, x, end bold text, start subscript, 0, end subscript, right parenthesis, left parenthesis, start bold text, x, end bold text, minus, start bold text, x, end bold text, start subscript, 0, end subscript, right parenthesis
To see this spelled out for the two-variable case, let's expand out the Hessian term,
Qf(x,y)=f(x0,y0)+12fxx(x0,y0)(x−x0)2+fxy(x0,y0)(x−x0)(y−y0)+12fyy(x0,y0)(y−y0)2\begin{aligned} \quad Q_f(x, y) &= f(x_0, y_0) + \\ \\ &\quad {\dfrac{1}{2}f_{xx}(x_0, y_0)}(x-x_0)^2 + \\ \\ &\quad {f_{xy}(x_0, y_0)}(x-x_0)(y-y_0) + \\ \\ &\quad {\dfrac{1}{2}f_{yy}(x_0, y_0)}(y-y_0)^2 \end{aligned}
(Note, if this approximation or any of the notation feels shaky or unfamiliar, consider reviewing the article on quadratic approximations).
As I showed with the single variable case, the strategy is to study if the quadratic term of this approximation is always positive or always negative.
Qf(x,y)=f(x0,y0)+12fxx(x0,y0)(x−x0)2+fxy(x0,y0)(x−x0)(y−y0)+12fyy(x0,y0)(y−y0)2}Is this always ≥0?Is it always ≤0?Can it be either?\begin{aligned} \quad Q_f(x, y) &= f(x_0, y_0) + \\ &\left. \begin{array}{c} \blueE{\dfrac{1}{2}f_{xx}(x_0, y_0)(x-x_0)^2 +} \\ \\ \qquad \greenE{f_{xy}(x_0, y_0)(x-x_0)(y-y_0) +} \\ \\ \redE{\dfrac{1}{2}f_{yy}(x_0, y_0)(y-y_0)^2} \quad \end{array} \right\} \substack{ \text{Is this always $\ge 0$?} \\ \text{Is it always $\le 0$?} \\ \text{Can it be either?} } \end{aligned}
Right now, this term is a lot to write down, but we can distill its essence by studying expressions of the following form:
ax2+2bxy+cy2\begin{aligned} \quad \blueE{a}x^2 + 2\greenE{b}xy + \redE{c}y^2 \end{aligned}
Such expressions are often fancifully called "quadratic forms".
  • The word "quadratic" indicates that the terms are of order two, meaning they involve the product of two variables.
  • The word "form" always threw me off here, and it makes the idea of a quadratic form sound more complicated than it really is. Mathematicians say "quadratic form" instead of "quadratic expression" to emphasize that all terms are of order 2, and there are no linear or constant terms mucking up the expression. A phrase like "purely quadratic expression" would have been much too reasonable and understandable to adopt.
To make the notation for quadratic forms easier to generalize into higher dimensions, they are often written with respect to a symmetric matrix M
x⊺Mx=[xy][abbc][xy]\begin{aligned} \quad \textbf{x}^{\intercal} M \textbf{x} = \left[x \quad y\right] \left[ \begin{array}{cc} \blueD{a} & \greenD{b} \\ \greenD{b} & \redD{c} \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] \end{aligned}
Here is the crucial question:
  • How can we tell whether the expression start color #11accd, a, end color #11accd, x, squared, plus, 2, start color #1fab54, b, end color #1fab54, x, y, plus, start color #e84d39, c, end color #e84d39, y, squared is always positive, always negative, or neither, just by analyzing the constants start color #11accd, a, end color #11accd, start color #1fab54, b, end color #1fab54 and start color #e84d39, c, end color #e84d39?

Analyzing quadratic forms

If we plug in a constant value y, start subscript, 0, end subscript for y, we get some single variable quadratic function:
start color #0c7f99, a, end color #0c7f99, x, squared, plus, 2, start color #0d923f, b, end color #0d923f, x, y, start subscript, 0, end subscript, plus, start color #bc2612, c, end color #bc2612, left parenthesis, y, start subscript, 0, end subscript, right parenthesis, squared
The graph of this function is a parabola, and it will only cross the x-axis if this quadratic function has real roots.
A quadratic with two real roots can be both positive and negative.
A quadratic with two real roots can be both positive and negative.
Otherwise, it either stays entirely positive or entirely negative, depending on the sign of start color #0c7f99, a, end color #0c7f99.
A quadratic with no real roots can either be entirely positive or entirely negative.
A quadratic with no real roots can either be entirely positive or entirely negative.
We can apply the quadratic formula to this expression to see whether it's roots are real or complex.
start color #0c7f99, a, end color #0c7f99, x, squared, plus, 2, start color #0d923f, b, end color #0d923f, x, y, start subscript, 0, end subscript, plus, start color #bc2612, c, end color #bc2612, left parenthesis, y, start subscript, 0, end subscript, right parenthesis, squared
  • The leading term is start color #0c7f99, a, end color #0c7f99.
  • The linear term is 2, start color #0d923f, b, end color #0d923f, y, start subscript, 0, end subscript.
  • The constant term is start color #bc2612, c, end color #bc2612, y, start subscript, 0, end subscript, squared
Applying the quadratic formula looks like this:
−2by0±(−2by0)2−4acy022a⇓−2by0±2y0b2−ac2a⇓y0(−b±b2−aca)\quad \dfrac{-2\greenE{b}y_0 \pm \sqrt{(-2\greenE{b}y_0)^2 - 4\blueE{a}\redE{c}y_0^2}}{2\blueE{a}} \\ \qquad \qquad \qquad \Downarrow \\ \qquad \dfrac{-2\greenE{b}y_0 \pm 2y_0 \sqrt{\greenE{b}^2 - \blueE{a}\redE{c}}}{2\blueE{a}} \\ \qquad \qquad \qquad \Downarrow \\ \qquad \boxed{y_0\left( \dfrac{-\greenE{b} \pm \sqrt{\greenE{b}^2 - \blueE{a}\redE{c}}}{\blueE{a}} \right)}
If y, start subscript, 0, end subscript, equals, 0, the quadratic has a double root at x, equals, 0, meaning the parabola barely kisses the x-axis at that point. Otherwise, whether or not these roots are real depends only on the sign of the expression start color #0d923f, b, end color #0d923f, squared, minus, start color #0c7f99, a, end color #0c7f99, start color #bc2612, c, end color #bc2612.
  • If start color #0d923f, b, end color #0d923f, squared, minus, start color #0c7f99, a, end color #0c7f99, start color #bc2612, c, end color #bc2612, is greater than or equal to, 0, there are real roots, so the graph of start color #0c7f99, a, end color #0c7f99, x, squared, plus, 2, start color #0d923f, b, end color #0d923f, x, y, start subscript, 0, end subscript, plus, start color #bc2612, c, end color #bc2612, left parenthesis, y, start subscript, 0, end subscript, right parenthesis, squared crosses the x-axis.
  • Otherwise, if start color #0d923f, b, end color #0d923f, squared, minus, start color #0c7f99, a, end color #0c7f99, start color #bc2612, c, end color #bc2612, is less than, 0, there are no real roots, so the graph of start color #0c7f99, a, end color #0c7f99, x, squared, plus, 2, start color #0d923f, b, end color #0d923f, x, y, start subscript, 0, end subscript, plus, start color #bc2612, c, end color #bc2612, left parenthesis, y, start subscript, 0, end subscript, right parenthesis, squared either stays entirely positive or entirely negative.
For example, consider the case
  • start color #0c7f99, a, end color #0c7f99, equals, 1
  • start color #0d923f, b, end color #0d923f, equals, 3
  • start color #bc2612, c, end color #bc2612, equals, 5
In this case, start color #0d923f, b, end color #0d923f, squared, minus, start color #0c7f99, a, end color #0c7f99, start color #bc2612, c, end color #bc2612, equals, start color #0d923f, 3, end color #0d923f, squared, minus, left parenthesis, start color #0c7f99, 1, end color #0c7f99, right parenthesis, left parenthesis, start color #bc2612, 5, end color #bc2612, right parenthesis, equals, 4, is greater than, 0, so the graph of f, left parenthesis, x, right parenthesis, equals, x, squared, plus, 6, x, y, start subscript, 0, end subscript, plus, 5, y, start subscript, 0, end subscript, squared always crosses the x-axis. Here is a video showing how that graph moves around as we let the value of y, start subscript, 0, end subscript slowly change.
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This corresponds with the fact that the graph of f, left parenthesis, x, comma, y, right parenthesis, equals, x, squared, plus, 6, x, y, plus, 5, y, squared can be both positive and negative.
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In contrast, consider the case
  • start color #11accd, a, end color #11accd, equals, 2
  • start color #1fab54, b, end color #1fab54, equals, 2
  • start color #e84d39, c, end color #e84d39, equals, 3
Now, start color #1fab54, b, end color #1fab54, squared, minus, start color #11accd, a, end color #11accd, start color #e84d39, c, end color #e84d39, equals, start color #1fab54, 2, end color #1fab54, squared, minus, left parenthesis, start color #11accd, 2, end color #11accd, right parenthesis, left parenthesis, start color #e84d39, 3, end color #e84d39, right parenthesis, equals, minus, 2, is less than, 0. This means the graph of f, left parenthesis, x, right parenthesis, equals, 2, x, squared, plus, 4, x, y, start subscript, 0, end subscript, plus, 3, y, start subscript, 0, end subscript, squared never crosses the x-axis, although it kisses it if the constant y, start subscript, 0, end subscript is zero. Here is a video showing how that graph changes as we let the constant y, start subscript, 0, end subscript vary:
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This corresponds with the fact that the multivariable function f, left parenthesis, x, comma, y, right parenthesis, equals, 2, x, squared, plus, 4, x, y, plus, 3, y, squared is always positive.
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Rule for the sign of quadratic forms

As if to confuse students who are familiar with the quadratic formula, rules regarding quadratic forms are often phrased with respect to start color #0c7f99, a, end color #0c7f99, start color #bc2612, c, end color #bc2612, minus, start color #0d923f, b, end color #0d923f, squared instead of start color #0d923f, b, end color #0d923f, squared, minus, start color #0c7f99, a, end color #0c7f99, start color #bc2612, c, end color #bc2612. Since one is the negative of the other, this requires switching when you say is greater than or equal to, 0 and when you say is less than or equal to, 0. The reason mathematicians prefer start color #0c7f99, a, end color #0c7f99, start color #bc2612, c, end color #bc2612, minus, start color #0d923f, b, end color #0d923f, squared is because this is the determinant of the matrix describing the quadratic form:
det⁡([abbc])=ac−b2 \det\left( \left[ \begin{array}{cc} \blueE{a} & \greenE{b} \\ \greenE{b} & \redE{c} \end{array} \right] \right) = \blueE{a}\redE{c} - \greenE{b}^2
As a reminder, this is how the quadratic form looks using the matrix.
ax2+2bxy+cy2=[xy][abbc][xy] \blueE{a}x^2 + 2\greenE{b}xy + \redE{c}y^2 = \left[x \quad y\right] \left[ \begin{array}{cc} \blueE{a} & \greenE{b} \\ \greenE{b} & \redE{c} \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right]
Tying this convention together with what we found in the previous section, we write the rule for the sign of a quadratic form as follows:
  • If start color #0c7f99, a, end color #0c7f99, start color #bc2612, c, end color #bc2612, minus, start color #0d923f, b, end color #0d923f, squared, is less than, 0, the quadratic form can attain both positive and negative values, and it's possible for it to equal 0 at values other than left parenthesis, x, comma, y, right parenthesis, equals, left parenthesis, 0, comma, 0, right parenthesis.
  • If start color #0c7f99, a, end color #0c7f99, start color #bc2612, c, end color #bc2612, minus, start color #0d923f, b, end color #0d923f, squared, is greater than, 0 the form is either always positive or always negative depending on the sign of start color #0c7f99, a, end color #0c7f99, but in either case it only equals 0 at left parenthesis, x, comma, y, right parenthesis, equals, left parenthesis, 0, comma, 0, right parenthesis.
    • If start color #0c7f99, a, end color #0c7f99, is greater than, 0, the form is always positive, so left parenthesis, 0, comma, 0, right parenthesis is a global minimum point of the form.
    • If start color #0c7f99, a, end color #0c7f99, is less than, 0, the form is always negative, so left parenthesis, 0, comma, 0, right parenthesis is a global maximum point of the form.
  • If start color #0c7f99, a, end color #0c7f99, start color #bc2612, c, end color #bc2612, minus, start color #0d923f, b, end color #0d923f, squared, equals, 0, the form will again either be always positive or always negative, but now it's possible for it to equal 0 at values other than left parenthesis, x, comma, y, right parenthesis, equals, left parenthesis, 0, comma, 0, right parenthesis

Some terminology:

When start color #0c7f99, a, end color #0c7f99, x, squared, plus, 2, start color #0d923f, b, end color #0d923f, x, y, plus, start color #bc2612, c, end color #bc2612, y, squared, is greater than, 0 for all left parenthesis, x, comma, y, right parenthesis other than left parenthesis, x, comma, y, right parenthesis, equals, left parenthesis, 0, comma, 0, right parenthesis, the quadratic form and the matrix associated with it are both called positive definite.
When start color #0c7f99, a, end color #0c7f99, x, squared, plus, 2, start color #0d923f, b, end color #0d923f, x, y, plus, start color #bc2612, c, end color #bc2612, y, squared, is less than, 0 for all left parenthesis, x, comma, y, right parenthesis other than left parenthesis, x, comma, y, right parenthesis, equals, left parenthesis, 0, comma, 0, right parenthesis, they are both negative definite.
If you replace the is greater than and is less than with is greater than or equal to and is less than or equal to, the corresponding properties are positive semi-definite and negative semi-definite.

Applying this to Q, start subscript, f, end subscript

Okay zooming back out to where we started, let's write down our quadratic approximation again:
Qf(x,y)=f(x0,y0)+12fxx(x0,y0)(x−x0)2+fxy(x0,y0)(x−x0)(y−y0)+12fyy(x0,y0)(y−y0)2\begin{aligned} \quad Q_f(x, y) &= f(x_0, y_0) + \\ \\ &\quad \blueE{\dfrac{1}{2}f_{xx}(x_0, y_0)}(x-x_0)^2 + \\ \\ &\quad \greenE{f_{xy}(x_0, y_0)}(x-x_0)(y-y_0) + \\ \\ &\quad \redE{\dfrac{1}{2}f_{yy}(x_0, y_0)}(y-y_0)^2 \end{aligned}
The quadratic portion of Q, start subscript, f, end subscript is written with respect to left parenthesis, x, minus, x, start subscript, 0, end subscript, right parenthesis and left parenthesis, y, minus, y, start subscript, 0, end subscript, right parenthesis instead of simply x and y, so everywhere where the rule for the sign of quadratic forms references the point left parenthesis, 0, comma, 0, right parenthesis, we apply it instead to the point left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis.
As with the single-variable case, when the quadratic approximation Q, start subscript, f, end subscript has a local maximum (or minimum) at left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, it means f has a local maximum (or minimum) at that point. This means we can translate the rule for the sign of a quadratic form directly to get the second derivative test:
Suppose del, f, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, equals, 0, then
  • If start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99, start color #bc2612, f, start subscript, y, y, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #bc2612, minus, left parenthesis, start color #0d923f, f, start subscript, x, y, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0d923f, right parenthesis, squared, is less than, 0, f has a neither minimum nor maximum at left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, but instead has a saddle point.
    Saddle point
  • If start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99, start color #bc2612, f, start subscript, y, y, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #bc2612, minus, left parenthesis, start color #0d923f, f, start subscript, x, y, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0d923f, right parenthesis, squared, is greater than, 0, f definitely has either a maximum or minimum at left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, and we must look at the sign of start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99 to figure out which one it is.
    • If start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99, is greater than, 0, f has a local minimum.
      Local min
    • If start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99, is less than, 0, f has a local maximum.
      Local max
  • If start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99, start color #bc2612, f, start subscript, y, y, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #bc2612, minus, left parenthesis, start color #0d923f, f, start subscript, x, y, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0d923f, right parenthesis, squared, equals, 0, the second derivatives alone cannot tell us whether f has a local minimum or maximum.

Our current tools are lacking

Everything presented here almost constitutes a full proof, except for one final step.
Intuitively, it might make sense that when a quadratic approximation bends and curves in a certain way, the function should bend and curve in that same way near the point of approximation. But how do we formalize this beyond intuition?
Unfortunately, we will not do that here. Making arguments about derivatives fully rigorous requires using real analysis, the theoretical backbone of calculus.
Furthermore, you might be wondering how this generalizes to functions with more than two inputs. There is a notion of quadratic forms with multiple variables, but phrasing the rule for when such forms are always positive or always negative uses various ideas from linear algebra.

Summary

  • To test whether a stable point of a multivariable function is a local minimum/maximum, take a look at the quadratic approximation of the function at that point. It is easier to analyze whether this quadratic approximation has maximum/minimum.
  • For two-variable functions, this boils down to studying expression that look like this:
    start color #0c7f99, a, end color #0c7f99, x, squared, plus, 2, start color #0d923f, b, end color #0d923f, x, y, plus, start color #bc2612, c, end color #bc2612, y, squared
    These are known as quadratic forms. The rule for when a quadratic form is always positive or always negative translates directly to the second partial derivative test.

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  • mr pants teal style avatar for user sauj123
    Posted 7 years ago. Direct link to sauj123's post “Can khan academy add vide...”
    Can khan academy add videos/articles on real analysis? Perhaps add a subtopic to the mathematics topic called "Analysis", with sub-subtopics on real, complex and functional analysis.
    (83 votes)
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  • leaf yellow style avatar for user Jose Molina
    Posted 6 years ago. Direct link to Jose Molina's post “When he says "If ac-b2<0,...”
    When he says "If ac-b2<0, then the quadratic form can attain both positive and negative values, and is zero only at
    (x,y)=(0,0)" I am not sure I understand, because this implies that the form has real solutions for any y value. So there should be other points for which the form = 0.
    (8 votes)
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  • leaf orange style avatar for user Sam
    Posted 7 years ago. Direct link to Sam's post “Why is that when fxx*fyy ...”
    Why is that when fxx*fyy - fxy^2 = 0 then the second derivatives alone cannot tell us whether f has a local minimum or maximum?
    How does this relate to the aforementioned statement that if ac- b^2 = 0, the form will again either be always positive or always negative, but now it's possible for it to equal 0 at values other than (x, y) = (0, 0).
    I am having trouble figuring this out. My strategy is to figure out what the parabola looks like when ac-b^2 = 0 (or f'xx*f'yy - f'xy^2). But I don't understand intuitively or analytically the implications that "now it's possible for it to equal 0 at values other than (x, y) = (0, 0)."
    Any help will be appreciated.
    (4 votes)
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    • leaf green style avatar for user AbhiSharm
      Posted 6 years ago. Direct link to AbhiSharm's post “Hey Sam, You have to loo...”
      Hey Sam,
      You have to look at the whole function to understand this. We are saying that for a particular y, the zeros of the approximating function at (x, y) [where x is a variable] will occur at x = y*((-b - sqrt(b^2 - ac))/2a). If ac-b^2 = b^2 - ac = 0, then the only zeros of the function will occur at x = 0.
      For that to be the case, the function can NEVER cross the z axis. Why is that? It's hard to prove without using real analysis, but you can definitely picture it. We know for a fact that every zero of this function has to be on a line at x = 0. Now imagine that the function crosses the x axis. Then there's some point where the function is zero. By the thing we proved earier , it has to cross at x = 0. Now, we have to use the fact that the function that we're talking about is quadratic, and is kind of shaped like a sombrero. So if it crosses the z axis, the set of points where z=0 is a circular shape in the x-y plane. However, there's no circle that can fit on the line x = 0.
      (2 votes)
  • piceratops ultimate style avatar for user Robert
    Posted 4 years ago. Direct link to Robert's post “Would it then be right to...”
    Would it then be right to conclude that for f(x,y,z) to have a local min/max at (x0,y0,z0) you need

    fxxfyy + fxxfzz + fyyfzz - fxy² - fyz² - fxz²

    to be greater than 0, since for f to have an min/max you need a min/max over the xy-plane (which would be accounted for by z=0, y=y0, x=variable, then the same reasoning as in the article to get fxxfyy - fxy² > 0), as well as over the yz-plane (fyyfzz - fyz² > 0) and the xz-plane (fxxfzz - fxz² > 0), and adding all of those together gives us

    fxxfyy + fyyfzz + fxxfzz - fxy² - fyz² - fxz² > 0

    , or would that only be analogous to having fxx and fyy be greater/smaller than 0 and there is still information missing?
    (2 votes)
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  • blobby purple style avatar for user John Smith
    Posted 2 years ago. Direct link to John Smith's post “I cannot understand why, ...”
    I cannot understand why, when the quadratic form can be both negative and positive means that the function must have a saddle.

    "This corresponds with the fact that the graph of
    f(x,y) = x^2 + 6xy + 5y^2" No matter how many times I read this section, I don't really get how this makes the graph a saddle at the point?
    (2 votes)
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  • primosaur ultimate style avatar for user Hemen Taleb
    Posted 7 years ago. Direct link to Hemen Taleb's post “does there exist videos/a...”
    does there exist videos/articles on limits and continuity in 3d?
    (2 votes)
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  • blobby green style avatar for user Kalev Maricq
    Posted 3 years ago. Direct link to Kalev Maricq's post “Why does "the quadratic ...”
    Why does
    "the quadratic form can attain both positive and negative values, and it's possible for it to equal 0 at values other than (x, y) = (0, 0)(x,y)=(0,0)"
    =
    "f has a neither minimum nor maximum at (x_0, y_0)(x_0 ,y_0), but instead has a saddle point"?

    I'm missing the implicit connection here.
    (2 votes)
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  • leaf green style avatar for user Will Simon
    Posted 5 years ago. Direct link to Will Simon's post “This article raised a few...”
    This article raised a few questions for me:
    1) If a point p in R^k is not a critical point of a scalar-(or vector-)valued function f (i.e., the gradient of f at p is not 0), yet the quadratic approximation does have a local extremum at p, do we still have that f has a local extremum at p?
    2) Does a scalar- or vector-valued function f defined on R^k having a local extremum at a point p in R^k always (or at least sometimes) imply that its quadratic approximation has a local minimum there?
    3) In the context of this article, if b^2-ac > 0, how do we know the only zero of the quadratic form is at (0, 0)?
    4) Again in the context of this article, why does b^2-4ac=0 mean both that the quadratic form will always be positive or always be negative and that it can have zeros at points other than (0, 0)?
    (1 vote)
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  • blobby green style avatar for user Miss H
    Posted 2 months ago. Direct link to Miss H's post “At the beginning of the s...”
    At the beginning of the section titled "Analyzing Quadratic Forms," why can we plug in a constant value y0 for y?
    (1 vote)
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  • area 52 yellow style avatar for user Surya Raju
    Posted 2 years ago. Direct link to Surya Raju's post “Can this test be arrived ...”
    Can this test be arrived at by interpreting the Hessian matrix as the Jacobian matrix of the gradient field and then finding an interpretation for the determinant? Maybe the eigen-values of the Hessian could be used somehow.
    (1 vote)
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