Science:Math Exam Resources/Courses/MATH100 C/December 2024/Question 10 (b)/Solution 1

We use Lagrange multipliers with the constraint $g (x, y) = x^{2} + y^{2} = 1$ .

Compute partial derivatives: $f_{x} = y, f_{y} = x, g_{x} = 2 x, g_{y} = 2 y .$

Set up the system: $f_{x} = λ g_{x} \Rightarrow y = 2 λ x,$ $f_{y} = λ g_{y} \Rightarrow x = 2 λ y .$

From these, $\frac{y}{2 x} = λ = \frac{x}{2 y} .$

Along the domain, we can’t have $x = y = 0$ . Thus, $\frac{y}{2 x} = \frac{x}{2 y} \Rightarrow 2 y^{2} = 2 x^{2} \Rightarrow x^{2} = y^{2},$ so $x = \pm y .$

Using the constraint $x^{2} + y^{2} = 1$ :

If $x = y$ , then $2 x^{2} = 1 \Rightarrow x = \pm \frac{1}{\sqrt{2}}, y = \pm \frac{1}{\sqrt{2}} .$

If $x = - y$ , then $2 x^{2} = 1 \Rightarrow x = \pm \frac{1}{\sqrt{2}}, y = \mp \frac{1}{\sqrt{2}} .$

Therefore, the points of interest on the boundary are $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (- \frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}}), (\frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}}), (- \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) .$