Tutorials › Calculus III › Critical-Point Systems in Three Variables

Home  /  Tutorials  /  Critical Points and Local Extrema

Critical Points and Local Extrema · Tutorial 413 of 999

Critical-Point Systems in Three Variables

Learn to solve coupled systems of three first-partial equations systematically, while keeping track of every real stationary point.

Intermediate 10 min read

What You'll Learn

  • Define stationary points of differentiable functions of three variables using all three first partial derivatives.
  • Translate a polynomial function into its coupled gradient system.
  • Use substitution and elimination to reduce a three-equation system.
  • Factor equations and split into cases without losing zero-coordinate solutions.
  • Check candidate points in every original partial-derivative equation.
  • Distinguish finding stationary points from classifying their local behaviour.

From a Function to Three Equations

A function of three variables can change as any one of its coordinates changes. To locate a point where the function has no first-order change in any coordinate direction, we must set all three first partial derivatives equal to zero. The resulting equations are often coupled: each equation may involve more than one variable, so solving them requires careful substitution, elimination, and case-splitting.

This tutorial is about finding those solutions, not deciding whether they are local maxima, local minima, or saddle points. A point that satisfies all three equations is a stationary point and a candidate for further analysis; the equations alone do not classify its local behaviour.

Definition: Let \(f\) be differentiable on an open subset of \(\mathbb{R}^3\). A point \((a,b,c)\) in that subset is a stationary point of \(f\) if all three first partial derivatives vanish there: $$ f_x(a,b,c)=0,\qquad f_y(a,b,c)=0,\qquad f_z(a,b,c)=0. $$ For a polynomial function, these partial derivatives exist everywhere, so finding its stationary points means solving this system over the real numbers.

The three equations are the components of the gradient equation \(\nabla f(a,b,c)=\mathbf{0}\). A solution must satisfy every equation at the same triple of coordinates. Solving one or two equations alone does not establish that a point is stationary.

A Systematic Approach

Start by computing each partial derivative separately, treating the other two coordinates as constants. Then write down all three equations before simplifying. Look for equations that are linear in one variable, expressions that can be substituted into the others, and differences or combinations of equations that factor. If a factor can be zero, keep that case: dividing by it may discard solutions.

1
Compute all three partial derivatives.
Keep each derivative paired with its own variable and check signs, especially in coupled terms.
2
Set all three derivatives to zero.
The stationary points are the simultaneous real solutions of the resulting system.
3
Reduce the system carefully.
Substitute or factor, splitting into cases whenever a product is zero.
4
Check every candidate in the original equations.
This catches algebraic errors and confirms that no equation was lost during elimination.

Worked Example: A Coupled Linear System

Worked Example: A Coupled Linear System

Find the stationary point of \(f(x,y,z)=x^2+y^2+z^2+xy+yz+xz-x-2y-3z\).

Differentiate with respect to each variable:

$$ \begin{aligned} f_x&=2x+y+z-1,\\ f_y&=x+2y+z-2,\\ f_z&=x+y+2z-3. \end{aligned} $$

Setting all three expressions to zero gives a simultaneous linear system. Let \(S=x+y+z\). The equations can then be written as

$$ S+x=1,\qquad S+y=2,\qquad S+z=3. $$

Adding these equations gives \(3S+(x+y+z)=1+2+3\). Since \(x+y+z=S\), this becomes \(4S=6\), so \(S=\frac{3}{2}\). Substituting back gives

$$ x=1-\frac32=-\frac12,\qquad y=2-\frac32=\frac12,\qquad z=3-\frac32=\frac32. $$

Check the candidate in the original partial derivatives:

$$ \begin{aligned} f_x\left(-\frac12,\frac12,\frac32\right) &=-1+\frac12+\frac32-1=0,\\ f_y\left(-\frac12,\frac12,\frac32\right) &=-\frac12+1+\frac32-2=0,\\ f_z\left(-\frac12,\frac12,\frac32\right) &=-\frac12+\frac12+3-3=0. \end{aligned} $$

Thus the stationary point is \(\left(-\frac12,\frac12,\frac32\right)\).

Why Substitution Preserves the Solutions

Substitution is useful because an equation can express one variable in terms of the others. If, for example, one equation gives \(z=xy\), then any solution of the original system must have that value of \(z\). Replacing \(z\) by \(xy\) in the other equations therefore gives necessary conditions on \(x\) and \(y\). Conversely, if \(x,y\) satisfy the substituted equations and we set \(z=xy\), all three original equations hold, provided the substitution was made in each required equation.

Key point: A substitution reduces the number of variables, but it does not remove the need to satisfy the original system. After solving the reduced equations, recover the substituted variable and verify the resulting triples in all three original equations.

Worked Example: Factoring After Substitution

Worked Example: Factoring After Substitution

Find every stationary point of \(f(x,y,z)=\frac12(x^2+y^2+z^2)-xyz+x\).

The partial derivatives and their zero equations are

$$ \begin{aligned} f_x=x-yz+1&=0,\\ f_y=y-xz&=0,\\ f_z=z-xy&=0. \end{aligned} $$

The third equation gives \(z=xy\). Substitute this into the first two:

$$ \begin{aligned} x-y(xy)+1&=0 &&\Longrightarrow& x(1-y^2)+1&=0,\\ y-x(xy)&=0 &&\Longrightarrow& y(1-x^2)&=0. \end{aligned} $$

The second reduced equation is a product equal to zero, so consider both cases.

Case 1: \(y=0\). The first reduced equation becomes \(x+1=0\), so \(x=-1\). The substitution \(z=xy\) gives \(z=0\). Thus one candidate is \((-1,0,0)\).

Case 2: \(y\ne 0\). Then \(y(1-x^2)=0\) implies \(x^2=1\), so \(x=1\) or \(x=-1\). If \(x=1\), the first reduced equation becomes \(1-y^2+1=0\), hence \(y^2=2\). This gives \(y=\sqrt2\) or \(y=-\sqrt2\), and \(z=xy=y\). If \(x=-1\), the first reduced equation becomes \(-1+y^2+1=0\), so \(y^2=0\), contradicting the case assumption \(y\ne0\). There are no solutions in this subcase.

Check the three candidates in the original equations. For \((-1,0,0)\), the left sides \(x-yz+1\), \(y-xz\), and \(z-xy\) are respectively \(0,0,0\). For \((1,\sqrt2,\sqrt2)\), they are \(1-2+1=0\), \(\sqrt2-\sqrt2=0\), and \(\sqrt2-\sqrt2=0\). For \((1,-\sqrt2,-\sqrt2)\), they are \(1-2+1=0\), \(-\sqrt2-(-\sqrt2)=0\), and \(-\sqrt2-(-\sqrt2)=0\).

The stationary points are therefore \((-1,0,0)\), \((1,\sqrt2,\sqrt2)\), and \((1,-\sqrt2,-\sqrt2)\).

Factoring and Keeping Every Case

In polynomial systems, products equal to zero are especially valuable. If \(AB=0\) for real numbers \(A\) and \(B\), then \(A=0\) or \(B=0\). Both possibilities must be examined. A common source of missing solutions is to divide by a variable or factor without first considering the case where it is zero.

Worked Example: A Product Argument

Worked Example: A Product Argument

Find every stationary point of \(f(x,y,z)=x^2+y^2+z^2-xyz\).

The partial derivatives give the system

$$ 2x-yz=0,\qquad 2y-xz=0,\qquad 2z-xy=0. $$

First consider the possibility that at least one coordinate is zero. If \(x=0\), the second equation gives \(2y=0\), so \(y=0\); the third then gives \(2z=0\), so \(z=0\). The same conclusion follows if \(y=0\) or \(z=0\), by the corresponding equations. Thus the origin is a solution.

Now suppose all three coordinates are nonzero. The equations become \(2x=yz\), \(2y=xz\), and \(2z=xy\). Multiplying them gives

$$ 8xyz=(xyz)^2. $$

Because \(xyz\ne0\), divide by \(xyz\) to obtain \(xyz=8\). Multiply \(2x=yz\) by \(x\); then \(2x^2=xyz=8\), so \(x^2=4\). In the same way, \(y^2=4\) and \(z^2=4\). Each coordinate is therefore either \(2\) or \(-2\), and their product must be \(8\). The four possibilities are \((2,2,2)\), \((2,-2,-2)\), \((-2,2,-2)\), and \((-2,-2,2)\).

For the origin, each partial derivative is zero. At each of the four other candidates, \(yz=2x\), \(xz=2y\), and \(xy=2z\), so all three partial derivatives vanish. Therefore the complete list is the origin together with those four sign combinations.

A Common Mistake: Dividing Too Soon

In the preceding example, one might multiply the equations and then divide by \(xyz\) immediately. That division is valid only in the case \(xyz\ne0\). If it is done without first treating zero coordinates, the origin is lost. The resulting calculation may correctly find the four nonzero stationary points but still give an incomplete answer.

Correction: Before dividing by a variable or a product of variables, explicitly check whether it could be zero. Solve the zero case separately, then proceed under the nonzero assumption. Finally, check all candidates in the original equations.

What the Solution Set Does—and Does Not—Tell You

Solving the three first-partial equations identifies stationary points. Depending on the function, the real solution set may be empty, consist of a finite number of points, or contain infinitely many points. The algebra must establish which solutions exist; it is not enough to find one plausible point or to rely on a symmetry guess.

Finding stationary points is an important step in studying local extrema, but it is not the classification itself. At a stationary point, the first partial derivatives vanish; that fact by itself does not say whether nearby function values are larger, smaller, or both. Further analysis is needed to determine local behaviour. For this tutorial, the essential task is to solve the coupled equations completely and report every real solution.

Check Your Understanding

Use the methods in this tutorial to answer the following questions.

  1. What three equations must a point satisfy to be stationary for a differentiable function of three variables?
  2. Why must a candidate obtained by substitution be checked in all three original equations?
  3. When an equation factors as \(y(1-x^2)=0\), what cases must be considered?
  4. In the product argument example, why was the zero-coordinate case treated before dividing by \(xyz\)?
  5. Does satisfying all three first-partial equations, by itself, classify a point as a local maximum or minimum?

Previous

Next

Back to Tutorials