Optimization in two variables and Lagrange multipliers, question 4: UBC MATH 100 practice

Question 4 of 14Introductory2 marks · 3 minTheory UBC Math 100 §16.1.1Key results

The graph of $f(x, y) = \sqrt{(x - 1)^{2} + (y + 2)^{2}}$ is a cone with its tip on the $xy$ -plane. Which statement about the critical points of $f$ is correct?

f has no critical points, because f_x and f_y are never both 0.
The only critical point is (1, −2), where f_x and f_y are both 0.
The only critical point is (1, −2), where f_x and f_y do not exist, and f has its smallest value, 0, there.
The only critical point is (1, −2), and f has a local maximum there, because the surface comes to a point.

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Theory UBC Math 100 §16.1.1Key results

Worked solution

Strategy

Find the partial derivatives and look for both kinds of critical point: where both are 0, and where one fails to exist. Then compare f there with its values nearby.

Working

$f_x = \frac{x - 1}{\sqrt{(x - 1)^{2} + (y + 2)^{2}}}, \qquad f_y = \frac{y + 2}{\sqrt{(x - 1)^{2} + (y + 2)^{2}}}$
The chain rule on the square root, at every point except (1, −2), where the denominator is 0.
$f_x = f_y = 0 \iff x = 1 \text{ and } y = -2$
Both tops vanish only at (1, −2), and the formulas do not apply there. So no point has both partials equal to 0: the cone has no flat spot.
$f(x, -2) = \sqrt{(x - 1)^{2}} = |x - 1|$
Along y = −2 the surface is a V with a corner at x = 1, so f_x(1, −2) does not exist; likewise f_y. That makes (1, −2) a critical point.
$f(x, y) \geq 0 = f(1, -2)$
A square root is never negative, so the tip of the cone is the lowest point: the absolute minimum.

Answer

\text{one critical point, } (1, -2), \text{ where } f_x, f_y \text{ do not exist; the minimum } 0

Check

f is the distance from (x, y) to (1, −2), which is 0 at that point and positive everywhere else.

Where this goes wrong

Looking only for points where both partial derivatives are 0. A critical point also includes any point where a partial derivative does not exist, and here that point, the tip of the cone, is the minimum.