Extreme values on a triangle: throwing away the interior critical point
A candidate must find the extreme values of f(x, y) = x² + y² − 2x − 2y on the closed triangular region T with vertices O(0,0), A(3,0) and B(0,3). The submitted report reads: 'f has positive definite Hessian, so it is convex; a convex function on a polygon attains its maximum at a vertex, and a convex function cannot have an interior minimum unless the gradient vanishes there. Solving ∇f = (2x − 2, 2y − 2) = 0 gives only (1,1), with value −2, but an interior minimum is impossible for a function that grows outwards from the boundary. Hence the minimum is at a vertex: f(0,0) = 0 and f(3,0) = f(0,3) = 3, so the minimum is 0 and the maximum is 3.' The tutor points out that f(1,1) = −2 is smaller than every vertex value, and that (1,1) lies strictly inside T because 1 + 1 < 3.
1. Find the true global minimum and maximum of f on T, with a complete check that no candidate point is missed.
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Complete the square: f = (x − 1)² + (y − 1)² − 2, the squared distance from (1,1) minus 2.
Interior: ∇f = (2x − 2, 2y − 2) = 0 only at (1,1), which is interior since 1 + 1 < 3; f(1,1) = −2.
Edges: on y = 0, 0 ≤ x ≤ 3, f = x² − 2x, with minimum −1 at x = 1 and maximum 3 at x = 3; on x = 0 the same by symmetry. On the hypotenuse x + y = 3, f = x² + (3 − x)² − 6 = 2x² − 6x + 3, whose derivative 4x − 6 vanishes at x = 3/2 giving f = −3/2; the endpoints give 3.
Vertices: 0, 3, 3. Comparing all interior, edge and vertex values (−2, −3/2, −1, 0, 3): minimum −2 at (1,1), maximum 3 at (3,0) and (0,3). The distance reading confirms it: the farthest points of T from (1,1) are those two vertices, each at distance √5, giving 5 − 2 = 3.
2. Which sentences of the report are valid and which are false? Address both the 'convex ⇒ maximum at a vertex' statement and the 'convex ⇒ no interior minimum' statement.
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Valid: f is convex (Hessian 2I), and a convex function on a compact convex polygon does attain its maximum at an extreme point, so the reported maximum 3 is correct.
False: 'a convex function cannot have an interior minimum unless the gradient vanishes there'. Convexity works the other way round - for a differentiable convex function, a vanishing gradient at an interior point does guarantee a global minimum, and such interior minima are entirely normal. The report located (1,1), then discarded it by reversing the logic, and finally inferred a minimum from vertex values alone.
The deeper structural point: optimising over a closed region requires three separate searches - interior critical points, each smooth edge piece (including its endpoints), and the vertices. Here the true minimum is interior and the true maximum is at vertices, so any method that inspects only one of the three sets gives at least one wrong answer.
3. Suppose the region is replaced by the square 0 ≤ x ≤ 3, 0 ≤ y ≤ 3, with f unchanged. Without recomputing from scratch, state the new maximum and where it is attained, and explain what carries over from the triangle case.
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Since f = |(x, y) − (1,1)|² − 2, the minimum is still −2, attained at the interior point (1,1), which lies in the square.
The maximum is attained at the point of the square farthest from (1,1). Check the four vertices: (3,3) gives 2² + 2² = 8; (3,0) and (0,3) give 5 each; (0,0) gives 2. So the maximum is 8 − 2 = 6, attained only at (3,3).
What carries over: the interior critical point must always be evaluated and here it decides the minimum; the maximum again sits at an extreme point, but which extreme point is fixed by geometry, not by any symmetry of the region. The triangle case was misleading precisely because its two farthest vertices happened to tie, which can hide the fact that the maximum need not be attained at every vertex.