if $x,y\in R$ and $x^3+y^3=2\;,$ then maximum and minimum value of $x+y$

using $\displaystyle \frac{x^3+y^3}{2}\geq \left(\frac{x+y}{2}\right)^3$

So $(x+y)^3\leq 2^3$ so $x+y\leq 2$

could some help me to find minimum value, thanks

  • $\begingroup$ You could try to judge by symmetry. $\endgroup$ – Dr. Wolfgang Hintze Jan 12 '17 at 15:26

From $$ 2 = x^3 + y^3 = (x+y)(x^2-xy+y^2) = (x+y) \frac{(x-y)^2+x^2+y^2}2 $$ it follows that $x+y> 0$. On the other hand, for arbitrary $t > 0$ $$ x =-t \quad , \quad y = \sqrt[3]{2+t^3} $$ satisfies $x^3+y^3 =2$ and $$ x + y = \sqrt[3]{2+t^3} - t = \frac{2}{(\sqrt[3]{2+t^3})^2 + t \sqrt[3]{2+t^3} + t^2} \to 0 $$ for $t \to \infty$. Therefore the infimum is zero and a minimum does not exist.

enter image description here

  • $\begingroup$ (+1) Sorry, I didn't see your answer when I was writing mine. Even so, they look quite different although they are essentially the same. $\endgroup$ – robjohn Jan 12 '17 at 16:01
  • $\begingroup$ @robjohn: No problem. Actually the first half of Stefan's answer is also the same (only in condensed form) and appeared only shortly before I was finished writing mine. $\endgroup$ – Martin R Jan 12 '17 at 16:03
  • $\begingroup$ Indeed, I just noticed that. Looking at the end of his answer, I thought he was finding a minimum, and I was about to write a comment. Then I read the beginning more closely. The only answer I saw before I posted was Dr. Sonnhard Graubner's. $\endgroup$ – robjohn Jan 12 '17 at 16:11

Note that Since $$ 3x^2+3y^2y'=0 $$ we have $$ y'=-\frac{x^2}{y^2} $$ Therefore, $$ \begin{align} 0 &=(x+y)'\\[6pt] &=1+y'\\ &=1-\frac{x^2}{y^2}\\ \end{align} $$ At $x=y=1$, we get a maximum of $2$.

$x=-y$ doesn't happen, but $\frac xy\to-1$ as $x\to\pm\infty$. Since $xy\le\frac{x^2+y^2}2$, we have $x^2-xy+y^2\ge\frac{x^2+y^2}2$. Therefore, $$ \begin{align} x+y &=\frac{x^3+y^3}{x^2-xy+y^2}\\ &\le\frac4{x^2+y^2}\\[6pt] &\to0 \end{align} $$

  • $\begingroup$ Could I ask why $x=-y$ doesn't happen, while $\frac{x}{y} \rightarrow -1$ does as $x \rightarrow \pm \infty$? $\endgroup$ – learning Jan 13 '17 at 2:19
  • $\begingroup$ OK, $x=-y$ doesn't happen because of the constraint. I don't get the second part. $\endgroup$ – learning Jan 13 '17 at 2:35
  • $\begingroup$ Which second part? $\frac xy\to-1$? $x^2-xy+y^2\ge\frac{x^2+y^2}2$? $x+y\le\frac4{x^2+y^2}$? $\endgroup$ – robjohn Jan 13 '17 at 5:09
  • $\begingroup$ I meant the second part of my question, so $\frac{x}{y} \rightarrow -1$. But I figured it out now. Thanks $\endgroup$ – learning Jan 13 '17 at 6:28
  • $\begingroup$ Oh, that makes sense. Sorry, it was late and I was tired. Glad you've got it all now. $\endgroup$ – robjohn Jan 13 '17 at 9:14

If $x,y$ are real numbers then there is no absolute minimum. In fact the value of $x+y$ can be made as close to zero as we like. We have $y = \sqrt[3]{2 - x^3}$

Obviously for very large $x$ we have that $y \approx -x$. Additionally the value must be positive, as $y > -x$, from the condition.

On the other side if we add the condition that $x,y$ are positive, then we have:

$$(x+y)^3 \ge x^3 + y^3 = 2 \implies x+y \ge \sqrt[3]{2}$$

And the minimum is obtained for $(x,y) = \{(\sqrt[3]{2},0), (0,\sqrt[3]{2})\}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.