Maximum value of equations #3

We can actually solve this geometrically.

The constraint $a^2+b^2=2$ is a circle of radius $\sqrt2$ centred at the origin of the $a,b$ plane.

Plotting $a+b=C$ gives a straight line. This intersects the circle either twice, once or not at all, depending on the value of $C$ .

The line corresponding to the maximum value of $C$ that gives an intersection must be tangent to the circle at the point $(1,1)$ , and this maximum value is $\boxed2$ .

Given that $a^2 + b^2 = 2$ , $\implies \dfrac {a^2}2 + \dfrac {b^2}2 = 1$ , a equation of unit circle centered at the origin $(0,0)$ . We can substitute $\dfrac {a^2}2 = \cos^2 \theta$ and $\dfrac {b^2}2 = \sin^2 \theta$ or $a = \sqrt 2 \cos \theta$ and $b=\sqrt 2 \sin \theta$ . Then we have:

$\begin{aligned} a + b & = \sqrt 2 \cos \theta + \sqrt 2 \sin \theta \\ & = 2 \left(\frac 1{\sqrt 2}\cos \theta + \frac 1{\sqrt 2}\sin \theta \right) \\ & = 2 \sin \left(\theta + \frac \pi 4\right) \\ \implies \max (a+b) & = \max \left(2 \sin \left(\theta + \frac \pi 4\right) \right) \\ & = 2 \color{#3D99F6} \max \left(\sin \left(\theta + \frac \pi 4\right) \right) & \small \color{#3D99F6} \text{Note that } \max \left(\sin \left(\theta + \frac \pi 4\right) \right) = 1 \text{ when }\theta = \frac \pi 4 \implies a = b = 1 \\ & = \boxed 2 \end{aligned}$

Note that $\left(a+b\right)^2=a^2+b^2+2ab\le 2\left(a^2+b^2\right) =4$ . This means that $a+b \le 2$ .