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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!

Poecilia
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4 Answers4

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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.

Patrick Stevens
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  • (+1) Why is the substitution necessary? Could he not jump straight to differentiating? – gen-ℤ ready to perish Mar 16 '18 at 23:37
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    @ChaseRyanTaylor If you do, then you will have partials and can find the zeroes of the partials. However, you still need to map them to points on the circle (and not in the interior of the disc, or somewhere else). Easier to first do the parameterization and differentiate with respect to the one variable. – Doug M Mar 16 '18 at 23:42
  • @Patrick, We can easily avoid differentiation using Double angle formula and https://math.stackexchange.com/questions/1633972/put-a-cosx-b-sinx-into-form-a-sinx-theta – lab bhattacharjee Mar 17 '18 at 01:50
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$\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix} 1&\frac 12\\\frac 12&0\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}$

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.

Doug M
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2

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.

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

Bumblebee
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