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

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

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

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

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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.

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$$(ax+by)^2\leq\frac{4}{3}(a^2-ab+b^2)(x^2+xy+y^2)$$ 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, $$\max\limits_{x^2+xy+y^2=t^2}(ax+by)=\frac{2}{\sqrt3}|t|\sqrt{a^2-ab+b^2}.$$

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    $\begingroup$ 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

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