Why Product Contours Have Distinct Shapes
A product can change sign depending on the signs of its factors. That simple fact has a visible effect on the contours of a function such as \(f(x,y)=xy\). A positive level occurs where the coordinates have matching signs; a negative level occurs where they have opposite signs. At level zero, either coordinate may be zero, so the contour changes into two entire lines.
Understanding these cases gives a reliable way to sketch the contours: first solve the level equation, then use the sign of the level to locate its branches, and finally test points in the original function. The goal here is to analyze product equations, especially \(xy=c\), rather than to infer a contour from a surface sketch.
What Is a Contour?
For a function \(f\) of two variables and a real number \(c\), the contour at level \(c\) is the set of points \((x,y)\) in the domain of \(f\) for which the function value is \(c\). It is found by solving the equation \(f(x,y)=c\) in the \(xy\)-plane. The number \(c\) is the height assigned to every point on that contour.
For \(f(x,y)=xy\), the level equation is \(xy=c\). When \(c\ne 0\), no point on the level curve can have \(x=0\), because then \(xy=0\), and likewise no point can have \(y=0\). Thus, for a nonzero level, division by \(x\) is valid at every point of the curve, giving \(y=c/x\). This description gives two branches, one on each of the two possible \(x\)-intervals: \(x>0\) and \(x<0\).
The expression \(c/x\) also shows that a nonzero contour does not meet either coordinate axis. As \(x\) approaches zero, the magnitude of \(c/x\) grows without bound; as the magnitude of \(x\) grows, \(c/x\) approaches zero. The curve can get arbitrarily close to either axis, but it never lies on one. This is the familiar shape of a hyperbola, with the coordinate axes as its asymptotes.
How the Sign of the Level Determines the Branches
The signs of \(x\) and \(y\) determine the sign of their product. If \(xy>0\), the two coordinates must both be positive or both be negative. Therefore a positive-level contour lies in quadrants I and III. If \(xy<0\), one coordinate is positive and the other is negative, so a negative-level contour lies in quadrants II and IV.
| Level | Sign relationship | Quadrants containing the contour |
|---|---|---|
| \(c>0\) | \(x\) and \(y\) have matching signs | I and III |
| \(c<0\) | \(x\) and \(y\) have opposite signs | II and IV |
| \(c=0\) | At least one coordinate is zero | The two coordinate axes |
For every nonzero \(c\), both branches correspond to the same value of the product. For example, when \(c>0\), a point in quadrant I has two positive factors, while a point in quadrant III has two negative factors. The product is positive in both cases.
Why the Zero Contour Is Different
When \(c=0\), the equation \(xy=c\) becomes \(xy=0\). The zero-product property says that a product of two real numbers is zero exactly when at least one of its factors is zero. Thus \(x=0\) or \(y=0\). The zero contour is not a single hyperbola branch; it is the union of the entire \(y\)-axis and the entire \(x\)-axis.
This is also why the formula \(y=c/x\) should only be used for nonzero levels. At \(c=0\), dividing by \(x\) would discard all points with \(x=0\), even though those points are part of the contour. The zero level must be solved directly in factored form.
Worked Examples
Worked Example: Sketching a Positive Level of \(xy\)
Find and describe the contour of \(f(x,y)=xy\) at level \(c=6\). First set the function equal to the level:
Since \(6\ne 0\), solve for \(y\) by dividing by \(x\); points on the curve cannot have \(x=0\):
The level is positive, so \(x\) and \(y\) must have the same sign. The branches lie in quadrants I and III. Choose values of \(x\) to obtain points on the branches. For \(x=1,2,3\), the corresponding \(y\)-values are \(6,3,2\). For \(x=-1,-2,-3\), they are \(-6,-3,-2\). Thus six points to plot are \((1,6),(2,3),(3,2),(-1,-6),(-2,-3),(-3,-2)\).
Check each group in the original equation: \(1\cdot6=6\), \(2\cdot3=6\), \(3\cdot2=6\), \((-1)(-6)=6\), \((-2)(-3)=6\), and \((-3)(-2)=6\). All six points lie on the contour. The curve continues through each quadrant branch, approaches but does not touch either axis, and has the coordinate axes as asymptotes.
Worked Example: Sketching a Negative Level of \(xy\)
Find and describe the contour at level \(c=-4\):
Here \(x\ne0\), and solving for \(y\) gives:
The level is negative, so \(x\) and \(y\) must have opposite signs. The branches lie in quadrants II and IV. In quadrant II, take \(x=-1\) and \(x=-2\). The corresponding values are \(y=4\) and \(y=2\), producing \((-1,4)\) and \((-2,2)\). In quadrant IV, take \(x=1\) and \(x=2\), giving \(y=-4\) and \(y=-2\), so the points are \((1,-4)\) and \((2,-2)\).
Check the products: \((-1)(4)=-4\), \((-2)(2)=-4\), \(1(-4)=-4\), and \(2(-2)=-4\). The points verify the contour equation. As before, the contour has two branches and does not meet either axis; unlike the positive-level example, its branches occupy quadrants II and IV.
Worked Example: Finding the Zero Contour
Find the contour at level \(c=0\) for \(f(x,y)=xy\), and determine whether the point \((0,5)\) belongs to it.
Set the product equal to zero and use the zero-product property:
The full contour is therefore the union of both coordinate axes: all points \((0,y)\) and all points \((x,0)\). At \((0,5)\), the function value is \(f(0,5)=0\cdot5=0\), so the point belongs to the contour. For another check, \(f(3,0)=3\cdot0=0\), so \((3,0)\) also belongs. In contrast, \(f(3,5)=15\), so \((3,5)\) does not lie on the zero contour.
A Short Derivation of the Contour Shape
The branch locations and the curve equation follow directly from the product equation. For \(c\ne0\), suppose \(xy=c\). Neither factor can be zero, and division by \(x\) gives \(y=c/x\). Conversely, if \(x\ne0\) and \(y=c/x\), then multiplying by \(x\) gives \(xy=x(c/x)=c\). Thus the equation \(y=c/x\), with \(x\ne0\), describes exactly the same points as \(xy=c\), not merely some points on the contour.
Now consider the sign. If \(c>0\), then \(xy>0\). A product of two nonzero real numbers is positive precisely when both are positive or both are negative, giving quadrants I and III. If \(c<0\), the factors have opposite signs, giving quadrants II and IV. At \(c=0\), the zero-product property instead gives the axes. These cases account for every real value of \(c\).
The contours also reflect the numerical size of the level. At a fixed nonzero \(x\), the point on the contour has \(y=c/x\), so changing \(c\) changes \(y\) in direct proportion. For instance, at \(x=2\), levels \(c=2\), \(c=4\), and \(c=6\) give \(y=1\), \(y=2\), and \(y=3\), respectively. The contours in a contour map are therefore distinct curves, and their positions depend on the chosen levels.
Products with Scale or Shifted Factors
The same reasoning applies when the product has a constant scale factor or when its factors are shifted coordinates. The first step is to keep the product factored and isolate it before interpreting its sign. A scale factor changes which product value is required; shifted factors change which lines make the product zero.
Worked Example: A Product with Shifted Factors
Consider \(g(x,y)=2(x-1)(y+2)\). Find the level curve at level \(8\), and give points on both branches.
Set the function equal to the level and divide both sides by \(2\):
The product is nonzero, so neither factor can be zero. Solving for \(y\) gives:
The product of the two shifted factors is positive. They must therefore both be positive or both be negative. If \(x-1=2\), then \(y+2=2\), giving \((x,y)=(3,0)\). If \(x-1=-2\), then \(y+2=-2\), giving \((x,y)=(-1,-4)\). Substitution into the original function checks both points: \(g(3,0)=2(2)(2)=8\), and \(g(-1,-4)=2(-2)(-2)=8\).
The factors are zero on the lines \(x=1\) and \(y=-2\), and this nonzero contour does not meet either line. Indeed, if \(x=1\) or \(y=-2\), then \(g(x,y)=0\), not \(8\). Those lines play the role of the axes in the unshifted product \(xy\).
A Common Mistake: Dividing Away Part of a Contour
A frequent error is to divide by a variable before checking whether it might be zero. For example, starting from \(xy=0\), one might divide both sides by \(x\) and conclude \(y=0\). That calculation is valid only under the added assumption \(x\ne0\). It loses every point with \(x=0\), including \((0,1)\), where the original product is \(0\cdot1=0\).
For nonzero levels, division by \(x\) is safe because \(xy=c\ne0\) already implies \(x\ne0\). For the zero level, use factoring and the zero-product property instead. This distinction prevents a sketch from omitting an entire axis.
How to Read or Sketch a Product Contour
A dependable sketch begins with the equation, not with a memorized picture. Identify whether the level is positive, negative, or zero. For a nonzero level, solve for one variable only after establishing that the divisor cannot be zero. Then use the signs of the factors to place the branches, calculate a few exact points, and check each point in the original function.
For \(xy\), write \(xy=c\).
If \(c=0\), factor the equation. If \(c\ne0\), neither factor can vanish.
Matching signs give a positive product; opposite signs give a negative product.
Substitute each proposed point into the original function and confirm that its value is \(c\).
For the basic product \(xy\), the nonzero contours are hyperbolas, with the sign of \(c\) selecting the pair of quadrants. The zero contour is the two coordinate axes. For shifted factors, the corresponding zero lines are where either factor vanishes. In every case, the contour equation tells exactly which points have the chosen function value; the sketch is a picture of that set.
Check Your Understanding
Answer these questions by writing the level equation and using the signs of its factors.
- In which quadrants does the contour \(xy=9\) lie? Give two points on it and verify them.
- In which quadrants does \(xy=-6\) lie? Explain why it does not cross either coordinate axis.
- Describe the complete contour \(xy=0\). Why is it not enough to report just \(y=0\)?
- For \(h(x,y)=(x+1)(y-3)\), identify the two lines on which \(h(x,y)=0\).
- For \(xy=12\), solve for \(y\) and state the restriction on \(x\). Why is division by \(x\) valid here?