How to maximize $x^2+xy$?

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

$\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix} 1&\frac 12\\\frac 12&0\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}$

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.

Answer 4

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 $$\dfrac{1-\sqrt2}{2}\le A\le\dfrac{1 +\sqrt2}{2}.$$