Find the maximum value of the function $f(x,y) = (x+2y)^2$ on the unit circle $x^2+y^2=1$.
I defined a new function $g(x)= x^2 +4x\sqrt{1-x^2} +4(1-x^2) $ and let its derivative equal to zero. Solving this, I got a three relevant different values for $x$, $-0.39, 0.49,$ and $0.86$. The maximum value of the functions from these three is at the point $(0.49, 0.87)$ where the function is roughly equal to $5$.
How can I tell if this is truly the actual maximum value? I checked on Wolframalpha and the value seems to be correct, but the $x$ and $y$ values are different $(\frac{1}{\sqrt5}, \frac{2}{\sqrt5})$.
The question also then asks to find a linear homomorphism $\phi: \mathbb{R^2} \to \mathbb{R^2}$, $\phi(x,y) = (x', y')$ such that $f \phi(x,y) = \lambda_1 x'^2 +\lambda_2 y'^2$ for some $\lambda_1, \lambda_2 \in \mathbb{R}$, which I am not sure how to start. Any help would be great, thanks.
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$\begingroup$Observe first that $f(x,y) = f(-x, -y)$, so it suffices to consider only the half-circle $y \geq 0$.
Let's rewrite $f$ as$$f(x,y) = (x+2y)^2 = x^2 + y^2 + 4xy + 3y^2$$which, on the unit circle, becomes$$f(x,y) = 1 + 3y^2 + 4xy$$Since we're interested in the region $y \geq 0$, this is clearly maximised when $x \geq 0$, so we can equivalently find the maximum value of $f(x,y)$ subject to $x, y \geq 0$.
On $x, y \geq 0$, observe that $x+2y \geq 0$, so $$(x+2y)^2 \text{ maximised} \iff g(x,y) = x+2y \text{ maximised}$$This is a somewhat easier problem to solve - let $x = \cos \theta$ and $y = \sin \theta$, so that$$g(\theta) = \cos \theta + 2 \sin \theta$$Then$$g'(\theta) = -\sin \theta + 2 \cos \theta = 0 \implies \tan \theta = 2$$By drawing a suitable right-angle triangle (choose an angle that isn't a right angle, let the non-hypotenuse side be $1$ so that the remaining sides are $\tan \theta$ and $\sqrt{1 + \tan^2 \theta}$), one can conclude that, for $0 \leq \theta < \frac{\pi}{2}$ (as in our case),$$\cos \theta = \frac{1}{\sqrt{1 + \tan^2 \theta}}, \quad \sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}}$$and hence we find that, for $\tan \theta = 2$,$$x = \cos \theta = \frac{1}{\sqrt{5}}, \quad y = \sin \theta = \frac{2}{\sqrt{5}}$$so that$$f(x,y) = (x+2y)^2 = \left( \frac{1 + 2 \times 2}{\sqrt{5}} \right)^2 = 5$$is the maximum value of $f$.
For the homomorphism, note that $x+2y$ looks like a shear (which is invertible, and clearly linear as it's a matrix transformation),$$\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + 2y \\ y \end{pmatrix}$$If you let $x' = x + 2y$, $y' = y$, (i.e. let this define $\phi$ appropriately) then$$f(x, y) = (x')^2$$which is of the required form with $\lambda_1 = 1, \, \lambda_2 = 0$.
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