Why Check a Triangle Edge by Edge?
A triangle is a bounded region with three straight sides and three vertices. A continuous function on a closed triangle attains both an absolute maximum and an absolute minimum, but those values need not occur at a vertex. They may occur at an interior critical point or somewhere along one of the sides.
The practical method is to examine the interior and then each edge separately. On an edge, the two-variable function becomes a one-variable function. Its critical points and endpoint values give the boundary candidates for comparison.
The Edge-by-Edge Candidate Method
Suppose a triangle has vertices \(P_1=(x_1,y_1)\), \(P_2=(x_2,y_2)\), and \(P_3=(x_3,y_3)\). To traverse an edge from \(P_i\) to \(P_j\), use the parameterization
Here the parameter interval includes both endpoints: \(t=0\) gives \(P_i\), and \(t=1\) gives \(P_j\). If the function is \(f(x,y)\), its restriction to this edge is the one-variable function \(g(t)=f(\mathbf{r}(t))\). Find its critical points for \(0<t<1\), then evaluate it at those points and at \(t=0,1\).
Repeat this procedure for all three edges. The vertices occur as endpoints of the parameter intervals, so they are included in the comparison. Also find any interior critical points by solving \(f_x=0\) and \(f_y=0\) within the triangle. The largest and smallest values among all these candidates are the absolute maximum and minimum.
Solve \(f_x(x,y)=0\) and \(f_y(x,y)=0\), retaining only solutions in the triangle’s interior.
Write its points as a function of \(t\), with \(0\le t\le1\), and substitute into \(f\).
Differentiate each restricted function, solve for critical parameters in \(0<t<1\), and evaluate at those parameters and both endpoints.
The greatest candidate value is the absolute maximum; the least is the absolute minimum.
Why This Search Is Complete
The method works because any point where an absolute extremum occurs is either in the triangle’s interior or on its boundary. At an interior extremum of a differentiable function, both first partial derivatives must vanish. On the boundary, a point lies on an edge or is a vertex. An extremum in the open part of an edge must be a critical point of the function restricted to that edge; otherwise, the one-variable derivative test would allow a nearby increase and decrease. A vertex is an endpoint of each incident edge and is checked directly.
The derivative of an edge restriction follows from the chain rule. If \(\mathbf{r}(t)=(x(t),y(t))\), then
Thus, solving \(g'(t)=0\) checks for stationary values in the direction of that edge. It does not replace the interior test, which checks changes in two independent directions.
Worked Example: A Triangle With Unequal Side Lengths
Worked Example: Compare Interior and Edge Candidates
Find the absolute extrema of \(f(x,y)=x^2+y^2-2x-2y\) on the triangle with vertices \((0,0)\), \((4,0)\), and \((0,3)\). Its interior is described by \(x>0\), \(y>0\), and \(x/4+y/3<1\).
Interior candidates. The partial derivatives are \(f_x=2x-2\) and \(f_y=2y-2\). Setting both equal to zero gives \(x=1\), \(y=1\). This point is interior because \(1/4+1/3=7/12<1\). Its value is \(f(1,1)=1+1-2-2=-2\).
Edge from \((0,0)\) to \((4,0)\). Parameterize it by \((x,y)=(4t,0)\), \(0\le t\le1\). Then \(g(t)=16t^2-8t\), so \(g'(t)=32t-8\). The critical parameter is \(t=1/4\), giving \(g(1/4)=1-2=-1\). The endpoint values are \(g(0)=0\) and \(g(1)=16-8=8\).
Edge from \((0,0)\) to \((0,3)\). Use \((x,y)=(0,3t)\). Then \(g(t)=9t^2-6t\), so \(g'(t)=18t-6\) vanishes at \(t=1/3\). There \(g(1/3)=1-2=-1\). The endpoint values are \(g(0)=0\) and \(g(1)=9-6=3\).
Edge from \((4,0)\) to \((0,3)\). Use \((x,y)=(4-4t,3t)\). Substitution gives
Thus \(g'(t)=50t-30=0\) at \(t=3/5\), where \(g(3/5)=25(9/25)-18+8=9-18+8=-1\). The endpoints give \(g(0)=8\) and \(g(1)=25-30+8=3\).
The candidate values are \(-2,-1,0,3,8\). Therefore the absolute minimum is \(-2\) at the interior point \((1,1)\), and the absolute maximum is \(8\) at \((4,0)\). Substitution checks these locations: \(f(1,1)=-2\) and \(f(4,0)=16-8=8\).
Worked Example: An Edge Critical Point Gives the Maximum
Worked Example: A Product on a Right Triangle
Find the absolute extrema of \(f(x,y)=xy\) on the triangle with vertices \((0,0)\), \((4,0)\), and \((0,3)\).
For an interior critical point, \(f_x=y\) and \(f_y=x\). Both derivatives vanish only at \((0,0)\), which is a vertex, not an interior point. Thus there is no interior critical point.
On the edge \((4t,0)\), the restriction is \(g(t)=0\), so its endpoint and interior values are all zero. On the edge \((0,3t)\), the restriction is also \(g(t)=0\), again giving only zero values.
On the edge \((4-4t,3t)\), the restriction is
Its derivative is \(g'(t)=12-24t\), which vanishes at \(t=1/2\). The corresponding point is \((4-4(1/2),3(1/2))=(2,3/2)\), and its value is \(g(1/2)=6-3=3\). The edge endpoints have values \(g(0)=0\) and \(g(1)=12-12=0\).
All candidate values are either zero or three, so the absolute minimum is zero, attained on both coordinate-axis edges, and the absolute maximum is three at \((2,3/2)\). Direct substitution confirms \(f(2,3/2)=2(3/2)=3\).
Worked Example: An Interior Point Beats Every Boundary Minimum
Worked Example: Comparing Three Edges With an Interior Point
Find the absolute extrema of \(f(x,y)=x^2+y^2+xy-2x-2y\) on the triangle with vertices \((0,0)\), \((2,0)\), and \((0,2)\).
The interior equations are \(2x+y-2=0\) and \(x+2y-2=0\). Subtracting the second equation from the first gives \(x-y=0\), so \(x=y\). Substitution gives \(3x=2\), hence the critical point is \((2/3,2/3)\), which is interior because \(2/3+2/3<2\). Its value is
On the edge \((t,0)\), \(0\le t\le2\), the restriction is \(g(t)=t^2-2t\). Its derivative \(2t-2\) vanishes at \(t=1\), with value \(-1\); the endpoints give \(g(0)=0\) and \(g(2)=0\). The edge \((0,t)\) has the same restriction, so it has the same critical value and endpoint values.
For the third edge, use \((x,y)=(t,2-t)\). Substitution yields
Again \(g'(t)=2t-2\), so \(t=1\) gives value \(-1\), while \(g(0)=g(2)=0\). Comparing the interior value \(-4/3\) with the boundary values \(-1\) and \(0\), the absolute minimum is \(-4/3\) at \((2/3,2/3)\), and the absolute maximum is \(0\) at the three vertices. As checks, substituting \((2/3,2/3)\) gives \(-4/3\), and substituting any vertex gives zero.
A Common Mistake: Checking Only the Vertices
Vertices are necessary candidates, but they are not the only boundary candidates. In the product example, all three vertex values are zero. Concluding from those values alone that the maximum is zero is incorrect: the restriction to the slanted edge has derivative \(12-24t\), which vanishes at \(t=1/2\), and the function value there is three. The correction is to test each edge’s interior critical points as well as its endpoints.
The opposite omission can also matter. In the final example, the least boundary value is \(-1\), but the interior critical point has value \(-4/3\). A boundary-only search would miss the absolute minimum. The complete search requires both the interior test and all three edge tests.
Using the Method Reliably
For each edge, keep its own parameterization and interval visible during the calculation. Solve \(g'(t)=0\) only for parameters strictly inside the interval; the endpoints are checked separately. Then record the point and function value for every candidate, including interior critical points. This makes it easier to verify that no vertex, edge, or interior candidate has been overlooked.
Triangle optimization is useful whenever a model or constraint limits two variables to a triangular region. The edge-by-edge method turns the boundary part of the problem into three one-variable calculations, while keeping the interior search separate. The final comparison—not a local classification on one edge—is what determines the absolute extrema on the whole region.
Check Your Understanding
For each question, identify which candidates must be checked or carry out the indicated calculation.
- For an edge parameterized on \(0\le t\le1\), which parameter values represent its vertices?
- Why does an interior extremum of a differentiable function require both first partial derivatives to vanish?
- For \(g(t)=15t-15t^2\) on \(0\le t\le1\), find its critical parameter and evaluate the function there and at both endpoints.
- Why can checking only the three vertices fail to find the absolute maximum on a triangle?
- List the full set of candidate types that must be compared to find absolute extrema on a closed triangle.