Optimization in two variables and Lagrange multipliers, question 8

Topic: Optimization in two variables and Lagrange multipliers
Difficulty: Challenging
Suggested time: 7 min
Marks: 4
Theory: UBC Math 100 §16.2Key results

Find the absolute minimum value of $f(x, y) = x^{2} + y^{2} - 6x - 4y$ on the closed triangular region with vertices $(0, 0)$ , $(4, 0)$ and $(0, 4)$ .

Your answer

A fraction or a decimal.

How sure are you?

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Theory UBC Math 100 §16.2Key results

Worked solution

Strategy

f is continuous on a closed, bounded region, so it has an absolute minimum, either at a critical point inside or on the boundary. Check the inside, reduce each edge to a one-variable closed-interval problem, and compare everything.

Working

$f_x = 2x - 6 = 0, \quad f_y = 2y - 4 = 0 \ \Rightarrow\ (3, 2); \qquad 3 + 2 = 5 > 4$
The only critical point lies outside the triangle, so the minimum is on the boundary.
$y = 0:\ f = x^{2} - 6x, \ \min -9 \text{ at } x = 3; \qquad x = 0:\ f = y^{2} - 4y, \ \min -4 \text{ at } y = 2$
The two edges along the axes, each on [0, 4].
$y = 4 - x:\ f = x^{2} + (4 - x)^{2} - 6x - 4(4 - x) = 2x^{2} - 10x, \ \min -\tfrac{25}{2} \text{ at } x = \tfrac{5}{2}$
The slanted edge: 4x − 10 = 0 at x = 5/2, the point (5/2, 3/2).
$f(0, 0) = 0, \qquad f(4, 0) = -8, \qquad f(0, 4) = 0$
The corners, which are the endpoints of the edges.
$\min\left\{-9,\ -4,\ -\tfrac{25}{2},\ 0,\ -8\right\} = -\tfrac{25}{2}$
The minimum is partway along the slanted edge, at (5/2, 3/2).

Answer

f_{\min} = -\frac{25}{2} \ \text{at}\ \left(\frac{5}{2}, \frac{3}{2}\right)

Check

As f = (x − 3)² + (y − 2)² − 13, the least value is at the point of the triangle nearest (3, 2), which is (2.5, 1.5), at squared distance 0.5: −12.5.

Where this goes wrong

Reporting f(3, 2) = −13 from the critical point without checking that it is in the region. It is not, since 3 + 2 > 4. The least value on the triangle is on the edge nearest that point, x + y = 4.

Leans on

The closed-interval method, from Optimization

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Optimization in two variables and Lagrange multipliers practice question 8