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I'm trying to find the maximum value of the function $f(x,y)=(ax+by)^p+x^p$ subject to the constraint $x^p+y^p=1$. Here, $a,b$ and $p$ are constants with $a,b>0$ and $p>1$, and $x,y>0$. I have found the maximum in the special case $p=2$ and tried to use Lagrange multipliers in the general case but couldn't success. Any help?

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    $\begingroup$ We can't comment on your work if we don't see it. $\endgroup$ – Ross Millikan Jan 4 '15 at 15:53
  • $\begingroup$ In the case $p=2$ I have used the substitutions $x=cos(t)$ and $y=sin(t)$, $0<t<Pi/2$. The problem reduces to maximization the function $g(t)=.5(a^2+1)(1+cos(2 t))+.5 b^2(1-cos(2 t))+ab sin(2 t)$. It is easily verified that the maximum of this function occurs when $t=arctan(2ab/(1+a^2+b^2))$ and that this maximum equals $.5(1+a^2+b^2+((1+a^2+b^2)^2-4 b^2)^.5).$ $\endgroup$ – Amer Jan 4 '15 at 21:26
  • $\begingroup$ What did you try in the general case? $\endgroup$ – user856 Jan 5 '15 at 9:25
  • $\begingroup$ I have tried the Lagrange multipliers method but couldn't solve the resulted equations. $\endgroup$ – Amer Jan 5 '15 at 12:23
  • $\begingroup$ I'll answer your question from your other thread here, to avoid duplicating. If your function is concave, then its local maxima are all global maximal. Therefore, you can just set $\nabla f = 0$ and solve for the stable points. However, even in the case that $p=2$, I don't think that is true for you for general $a$ and $b$. Also, since you are treating $a$ and $b$ as abstract parameters, you can't use a numerical solver like CVX :( I'm not sure what an analytic solution would be for arbitrary $p$, though. $\endgroup$ – Y. S. Jan 6 '15 at 21:36

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