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Maximum of: $$ xy+y^2 $$ Domain: $$ x \ge 0, x^2+y^2 \le1 $$

I know that the result is: $$ \frac{1}{2}+\frac{1}{\sqrt{2}} $$ for $$ (x,y)=\left(\frac{1}{\sqrt{2(2+\sqrt{2})}},\frac{\sqrt{2+\sqrt{2}}}{2}\right) $$ But I don't know how to get this result.

I know that: $$ xy+y^2 \le \frac{1}{2}+y^2 $$ so: $$ xy \le \frac{1}{2} $$ And also: $$ xy+y^2 \le xy+1-x^2 \equiv 1+x(y-x) $$ But I don't know what to do next...

fonini
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Diana
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3 Answers3

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Another way: WLOG $x=\sin t,y=\cos t$ where $0\le t\le\pi$

$$xy+y^2=\dfrac{\sin2t+\cos2t+1}2=\dfrac{\sqrt2\sin(2t+\pi/4)+1}2$$

Clearly, the maximum value occurs if $2t+\dfrac\pi4=2m\pi+\dfrac\pi2\iff t=m\pi+\dfrac\pi8$

lab bhattacharjee
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Hint$$x^2+y^2=x^2+(3-2\sqrt{2})y^2+(2\sqrt{2}-2)y^2 $$

Now notice $$x^2+(3-2\sqrt{2})y^2+(2\sqrt{2}-2)y^2 \ge 2(3-2\sqrt{2})^{\frac{1}{2}}xy+(2\sqrt{2}-2)y^2 (\because \text{AM-GM})$$Now note $(\sqrt{2}-1)^2=3-2\sqrt{2}$.

S.C.B.
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  • Thank you, but what does AM-GM means? – Diana Apr 30 '16 at 16:32
  • @Diana Arithematic mean-geometric mean inequality. Look it up on wikipedia. Basically, $$\frac{x_1+x_2+\dots+x_n}{n} \ge \sqrt[n]{x_1x_2\dots x_n}$$It's a often used abreviation. – S.C.B. Apr 30 '16 at 16:32
  • @Diana Where, $x_1, x_2, x_3, \dots x_n$ are positive real numbers and $n$ is a natural number. Do you understand now? I basically used the case $n=2$, which implies $$x^2+y^2 \ge 2xy$$ – S.C.B. Apr 30 '16 at 16:39
  • Oh, yeah, when you told me what AM-GM means, I looked up and I understood, thank you. – Diana Apr 30 '16 at 17:24
  • How does this help? It shows that the solution has an upper bound of ${1 \over 2(\sqrt{2}-1)}$, but I don't see how you roll this into a solution or a maximum value? – copper.hat Apr 30 '16 at 18:45
  • @copperhat THe last line shows that $(xy+y^2)(2\sqrt{2}-2) \le 1$, and does this not tell you an maximum value of $xy+y^2$? What solution are you talking about? I fail to understand what you are implying. With the above inequalities we can find the maximum of $xy+y^2$. And $\frac{1}{2(\sqrt{2}-1)}=\frac{1}{2}+\frac{1}{\sqrt{2}}$, but I don't think that is what you mean. Illustrate your point more clearly for me. – S.C.B. Apr 30 '16 at 21:54
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Here is another approach:

Let $A=\begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix}$ and write the problem as $\max \{ {1 \over 2} x^T Ax | \|x\| = 1, x_1 \ge 0 \}$ (where $x=(x_1,x_2)^T$ for notational convenience). Note that $A$ is real and symmetric.

Note that the constraint $x_1 \ge 0$ is unnecessary since if $x$ solves $\max \{ {1 \over 2} x^T Ax | \|x\| \le 1 \}$, then $(\operatorname{sgn} x_1)x$ is also a solution and solves the original problem.

Hence we can look for solutions of $\max \{ {1 \over 2} x^T Ax | \|x\| \le 1 \}$.

Since $A$ is symmetric, it has an orthonormal basis of eigenvectors $v_1,v_2$ and with $V=[v_1 \ v_2]$, and $y=V^{-1} x$ we have $x^T Ax = \lambda_1 y_1^2+\lambda_2 y_2^2$ and $\|x\| = \|y\|$. Without loss of generality, take $\lambda_1 \ge \lambda_2$.

It is straightforward to check that $\lambda_1= 1+ \sqrt{2}>0$, and that $\lambda_1 \|y\|^2 =\lambda_1 \ge \lambda_1 y_1^2+\lambda_2 y_2^2$, and so $\max \{ {1 \over 2} (\lambda_1 y_1^2+\lambda_2 y_2^2) | \|y\| \le 1 \} = {1 \over 2} \lambda_1$ and a (in fact, the) maximising $y$ is $(1,0)^T$.

Hence the problem reduces to finding a unit eigenvector corresponding to $\lambda_1$, and it is easy to check that $x= ({ 1\over \sqrt{2} \sqrt{2+ \sqrt{2}}}, { 1+ \sqrt{2}\over \sqrt{2} \sqrt{2+ \sqrt{2}}} )^T$ is a solution.

copper.hat
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