With functions of one variable we were interested in places where the derivative is zero, since they made candidate points for the maximum or minimum of a function. If the derivative is not zero, we have a direction that is downhill and moving a little in that direction gives a lower value of the function. Similarly, with functions of two variables we can only find a minimum or maximum for a function if both partial derivatives are 0 at the same time. Such points are called critical points.

The point \((a,b)\) is a critical point for the multivariable function \(f(x,y)\text{,}\) if both partial derivatives are 0 at the same time.

We need to find the places where both partial derivatives are 0. With this simple system, I can solve this system algebraically and find the only critical point is \((9/4, -1/4)\text{.}\)

If the partials are more complicated, I will want to find the critical points another way. I can find the point with Solver.

To get solver to set both partials to 0 at the same time, I ask it to solve for \(f_y=0\text{,}\) while setting \(f_x=0\) as a constraint. Make sure to uncheck the box that makes unconstrained variables non-negative.

This finds our critical point within our error tolerance.

Critical Point by CAS

We can also use Wolfram|Alpha to find the solution to our system of equations.

Determining the Critical Point is a Minimum

We thus get a critical point at \((9/4,-1/4)\) with any of the three methods of solving for both partial derivatives being zero at the same time. Once we have a critical point we want to determine if it is a maximum, minimum, or something else. The easiest way is to look at the graph near the critical point.

It is clear from the graph that this critical point is a local minimum.

It is easy to see that \(f(x,y)=x^2+y^2\) has a critical point at \((0,0)\) and that that point is a minimum for the function. Similarly, \(f(x,y)=-x^2-y^2\) has a critical point at \((0,0)\) and that that point is a maximum for the function. For some functions, like \(f(x,y)=x^2-y^2\text{,}\) which has a critical point at \((0,0)\text{,}\) we can have a maximum in one direction and a minimum in another direction. Such a point is called a saddle point. We note that we can have a saddle point even if the \(x\) and \(y\) slice curves both indicate a minimum.

Example6.3.5.A Saddle Point at a Minimum on Both Axes.

Show that \(f(x,y)=x^2-3xy+y^2\) has a critical point at \((0,0)\text{,}\) which is a minimum of both slice curves, but is not a local minimum.

Looking at the graph, we see that this graph does not have a minimum.

Subsection6.3.2Second Partial Derivatives

With only first derivatives, we can just find the critical points. To check if a critical point is maximum, a minimum, or a saddle point, using only the first derivative, the best method is to look at a graph to determine the kind of critical point. For some applications we want to categorize the critical points symbolically.

With functions of one variable we used the second derivative to test if a critical point was a maximum or minimum. In the two variable case we need to define the second derivatives and use them to define the discriminant of a function to test if a critical point is a minimum, maximum, or saddle point. We first need to define second partial derivatives.

Note that \(f_{xx}\) is simply the old second derivative of the curve \(f(x,y_0)\) and \(f_{yy}\) is simply the old second derivative of the curve \(f(x_0,y)\text{.}\) For functions with continuous second partial derivatives, the mixed partials, \(f_{yx}\) and \(f_{xy}\) are the same.

Let \((a,b)\) be a critical point of \(f(x,y)\text{.}\)

If \(D(a,b)>0\) and \(f_{xx} (a,b)>0\) then \((a,b)\) is a local minimum of \(f(x,y)\text{.}\)

If \(D(a,b)>0 \) and \(f_{xx} (a,b)\lt 0\) then \((a,b)\) is a local maximum of \(f(x,y)\text{.}\)

If \(D(a,b)\lt 0\) then \((a,b)\) is a saddle point of \(f(x,y)\text{.}\)

If \(D(a,b)=0\) we do not have enough information to classify the point.

Example6.3.9.Using the Discriminant to Classify Critical Points.

Based on the information given, classify each of the following points as a local maximum, local minimum, saddle point, not a critical point, or not enough information to classify.

We have critical points when both first partials are 0, so at \((1,2)\text{,}\)\((-1,2)\text{,}\)\((1,0)\text{,}\) and \((-1,0)\text{.}\)

At \((1,2)\text{,}\) both \(D\) and \(f_{xx}\) are positive, so we have a local minimum.

At \((-1,2)\) and \((1,0)\text{,}\)\(D\) is negative, so we have a saddle point.

At \((-1,0)\text{,}\)\(D\) is positive and \(f_{xx}\) is negative, so we have a local maximum.

Exercises6.3.4Exercises: Critical Points and Extrema Problems

1.Reading check, Wire Frames, Critical Points and Extrema.

This question checks your reading comprehension of the material is section 6.3, Critical Points and Extrema, of Business Calculus with Excel. Based on your reading, select all statements that are correct. There may be more than one correct answer. The statements may appear in what seems to be a random order.

If a function is a minimum in both the \(x\) and \(y\) directions, then it is a minimum.

We cannot have a maximum if the discriminant is zero.

If the discriminant of \(f\) is positive at a critical point, and \(f_{xx}\) is positive, then we have a local minimum.

The formula for the discriminant of \(f(x,y)\) is is \(f_{xx}f_{yy}-f^2_{xy}\text{.}\)

If the discriminant is negative at a critical point, then we have a saddle point.

If the discriminant of \(f\) is positive , and \(f_{xx}\) is negative, then we have a local maximum.

A saddle point has a minimum in one direction and a maximum in a different direction.

The point \((a, b)\) is a critical point for the multivariable function \(f(x,y)\text{,}\) if both partial derivatives are 0 at the same time.

We can use either the method of substitution (solve for \(x\) or \(y\) in one equation and substitute into the other and solve), or method by elimination (multiply both equations by carefully chosen numbers and add/subtract the equations from each other.)

To solve this problem we will rename Gizmos \(x\) and Widgets \(y\text{.}\) This will make using Wolfram Alpha slightly easier, and symbol manipulation a tad more straight forward.

for the region \(0\le \QuantityWidget\le 1500\text{,}\) and \(0\le \QuantityGizmo\le 1500\text{.}\) (Warning: There are several critical points.)

8.

Based on the information given, classify each of the following points as a local maximum, local minimum, saddle point, not a critical point, or not enough information to classify.

Based on the information given, classify each of the following points as a local maximum, local minimum, saddle point, not a critical point, or not enough information to classify.

p

\(f_x\)

\(f_y\)

\(f_{xx}\)

\(f_{xy}\)

\(f_{yy}\)

A

1

2

3

4

5

B

0

0

0

0

0

C

0

1

2

5

3

D

0

0

2

2

2

E

0

0

1

2

3

F

0

0

0

1

0

G

0

0

0

-1

0

10.

Using polynomials of the form \(f(x,y)=ax^3+bx^4+cy^3+dy^4\text{,}\) produce a function that has a critical point at \((0, 0)\text{,}\) of each type.

A local maximum.

A local minimum.

A saddle point where the function f(x,0) has a local maximum and f(0,y) has a local minimum.

A saddle point where the function f(x,0) and f(0,y) both have inflection points.

It helps to consider the question with only one variable.

\(f(x)=ax^3\) has an inflection point at \(x=0\) and neither a max nor a minimum.

\(f(x)=ax^4\) has minimum at \(x=0\) if \(a\gt 0\) and a maximum if \(a\lt 0\text{.}\)

Since all terms are of degree at least three, all second partial derivatives are zero at the origin, so the discriminant test fails.

A local maximum: \((x,y) = -x^4-y^4\text{.}\) Both \(x^4\) and \(y^4\) are nonnegative, so the function is negative everywhere except at the origin where it is 0.

A local minimum: \(f(x,y)= x^4+y^4\text{.}\) Both \(x^4\) and \(y^4\) are nonnegative, so the function is positive everywhere except at the origin where it is 0.

A saddle point where the function \(f(x,0)\) has a local maximum and \(f(0,y)\) has a local minimum: \((x,y)= -x^4+y^4\text{.}\)

A saddle point where the function \(f(x,0)\) and \(f(0,y)\) both have inflection points: \(f(x,y)= x^3+y^3\)