$$ x^2 + xy + y^2 = t^2 $$

Find the maximum value of $ax + by$

One way of doing this is substituting

$ x = r \cos w $ and $ y= r \sin w $

Then using calculus we can find the maximum value but this is a very lengthy process

So I wanted to know if there is a shorter way of doing this


closed as off-topic by user370967, MathOverview, steven gregory, The Phenotype, Parcly Taxel Feb 18 '18 at 11:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, MathOverview, steven gregory, The Phenotype, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Is there a maximum value? Can't I just pick $a$ and $b$ as large as I want? $\endgroup$ – Arthur Feb 7 '18 at 16:51
  • $\begingroup$ a,b are arbitrary constants $\endgroup$ – user481779 Feb 7 '18 at 17:33
  • $\begingroup$ It looks like only $x$ and $y$ are constrained in how large they can be by the given equation. I'm also confused—couldn't you pick $a$ and $b$ to be infinitely large, since they are not bounded by any constraints? $\endgroup$ – AleksandrH Feb 7 '18 at 18:05
  • $\begingroup$ Use Lagrange multiplier, if $t$ is constant $\endgroup$ – Narasimham Feb 7 '18 at 19:16

you can use the Lagrange Multiplier Method $$f(x,y,\lambda)=ax+by+\lambda(x^2+xy+y^2-t^2)$$


It might be pretty lengthy too, but another way is using a Langrange multiplier.

You would get need to solve $(a,b)=\lambda (2x+y,x+2y)$ with the constraint $x^2 + xy + y^2 = t^2$, so after solving the linear equations you would get $x=\frac{2a-b}{3\lambda}$ and $y=\frac{-a+2b}{3\lambda}$, then plugging this in the contraint would give the needed $\lambda$.

So if you do it by hand, it is also lengthy (unless of course if you know the inverse of a $2\times 2$-matrix by heart, which isn't difficult to find), but using some program to solve linear equations you are done in no time.


$$(ax+by)^2\leq\frac{4}{3}(a^2-ab+b^2)(x^2+xy+y^2)$$ it's $$((a-2b)x+(2a-b)y)^2\geq0.$$ The equality occurs for $(a-2b)x+(2a-b)^2y=0$ and $x^2+xy+y^2=t^2.$

Thus, $$\max\limits_{x^2+xy+y^2=t^2}(ax+by)=\frac{2}{\sqrt3}|t|\sqrt{a^2-ab+b^2}.$$

  • 1
    $\begingroup$ Its quite interesting . Btw, how did you know exactly which perfect square to make? $\endgroup$ – user481779 Feb 7 '18 at 20:29
  • $\begingroup$ Can you please reply @michael $\endgroup$ – user481779 Feb 14 '18 at 14:53
  • $\begingroup$ I don't remember. I am sorry. I remember that I used C-S, but I can't restore it. $\endgroup$ – Michael Rozenberg Feb 15 '18 at 9:25

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