0
$\begingroup$

Let $p \in \Bbb R$ and $a_1,a_2 > 0$. For $f(x,y)=a_1x+a_2y$ , show that under the constraint $x^p + y^p = 1$ it gets a maximum value of $(a_1^q+a_2^q)^{\frac{1}{q}}$ for $q=\frac{1}{1-\frac{1}{p}}$.

My attempt: using lagrange multipliers method, I have found that the only critical point is $x= \Big( \frac{a_1^q}{a_1^q+a_2^q} \Big)^{\frac{1}{p}} , $ $y= \Big( \frac{a_2^q}{a_1^q+a_2^q} \Big)^{\frac{1}{p}}$ which yields the wanted value for $f$. My question is, how can I be sure it is a maximum? I tried another point $(1,0)$ and saw that $f$ gets a lower value, but the constraint is not a compact set, so maybe we don't get a maximum at all?

$\endgroup$
0
$\begingroup$

i would write $$y=(1-x^p)^{1/p}$$ then you will have $$f(x,(1-x^p)^{1/p})=a_1x+a_2(1-x^p)^{1/p}$$ and this is easier to handle also for the second derivative. you will get $$f'(x)=a_1-\frac{a_2(1-x^p)^{1/p}x^p}{x(1-x^p)}$$ and we get $$f'(x)=0$$ if $$a_1^{p/(1-p)}+a_2^{p/(1-p)}x^p=a_2^{p/(1-p)}$$

$\endgroup$
  • $\begingroup$ I'm not sure I understand how that helps me. How can I know I even get an absolute maximum at all? $\endgroup$ – user401516 Nov 25 '17 at 10:49

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.