How to maximize $x^2+xy$? I am looking for $x\in\mathbb{R}$ and $y\in\mathbb{R}$ such that:


*

*$x^2+y^2=1$

*$x^2+xy$ is maximized.


How can I find them?
Thank you!
 A: Substitute $x = \cos(\theta), y = \sin(\theta)$. Then you wish to maximise $\cos^2(\theta) + \cos(\theta) \sin(\theta)$, which you can do by differentiating.
A: $\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix} 1&\frac 12\\\frac 12&0\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}$
The egeinvalues of that matrix are $\frac {1+\sqrt{2}}2$ and $\frac {1-\sqrt{2}}2$
$\lambda_1 \|\mathbf x\|<\mathbf x^T A\mathbf x \le \lambda_2 \|\mathbf x\|$
The constraint implies $\|(x,y)\| = 1$
Which means that the eigenvalues will be the maximal and minimal values of your objective function.
A: You want to maximise $x^2+xy$
on the unit circle.
Let $x=\cos(t)$ and $y=\sin (t)$ 
$$ 0\le t \le 2\pi $$
$$ x^2+xy=\cos ^2(t) + \cos (t) \sin (t) =f(t)$$
We use double angle formula to get $$ f(t) = 1/2[1+\cos (2t) + \sin (2t)]$$
$$ f'(t)=0 \implies  tan(2t)=1 \implies$$
$$ 2t = \pi /4 $$ or $$ 2t = 5\pi /4 $$ 
Thus $$f(\pi /8) = 1/2[1+\cos (\pi /4) + \sin (\pi /4)]
     =\frac {1+\sqrt 2}{2}$$
is the maximum value of  $ x^2+xy$ on the unit circle. 
And  $$f(5\pi /8) = 1/2[1+\cos (5\pi /4) + \sin (5\pi /4)]=\frac {1-\sqrt 2}{2}$$
is the minimum value of  $ x^2+xy$ on the unit circle. 
A: Let $A=x^2+xy$ and $1=x^2+y^2.$ Since here we have at most quadratic terms, we can use the following completely ad hoc method: Lets determine $a\in\Bbb{R}$ such that $$aA+1=(a+1)x^2+axy+y^2$$ is a perfect square. For that we need the discriminant $\Delta=a^2-4(a+1)=(a-2)^2-8=0.$ Thus $a=2(1\pm\sqrt2).$ Then $2(1\pm\sqrt2)A+1\ge0$ and this implies $$\dfrac{1-\sqrt2}{2}\le A\le\dfrac{1
+\sqrt2}{2}.$$ 
