Find minimum and maximum values of a function in a closed and bounded subset

Question

I need to find minimum and maximum for this one:

$f(x,y)=xy$, $(|x|^a+|y|^a\leq1)$, $a>0$

I tried to use Lagrange multipliers but got stuck.

Answer 1

You're set is

$$ \mathbb{D}_\alpha \triangleq \mathbb{B}(0,1) $$

The disc associated to the following norm:

$$ || \ ||_{\alpha} :(x,y)\to |x|^\alpha+|y|^\alpha $$

So first determine the reunion of the open set which constitute your disc then add the border.

So you have in a open :

$$ \partial_1f(x,y)=y$$ $$ \partial_2f(x,y)=x$$

The only critical point in the open is $(0,0)$

On the border we get for instance (I consider first case of $ 0 \leq x $ and $ 0 \leq y$ ) so $ y=(1-x^\alpha)^\frac{1}{\alpha} $

So in these conditions:

$$ f(x,y)=g(x)\triangleq x(1-x^\alpha)^\frac{1}{\alpha} $$

Differentiating $g$:

$$ g'(x)= (1-x^\alpha)^\frac{1}{\alpha}(1-\dfrac{\alpha x^\alpha}{1-x^{\alpha}}) $$

You shall find roots ! (not complicated)

Answer 2

By AM-GM $$xy\leq|xy|\leq\left(\frac{|x|^a+|y|^a}{2}\right)^{\frac{2}{a}}\leq\left(\frac{1}{2}\right)^{\frac{2}{a}}.$$ The equality occurs for $x=y=\left(\frac{1}{2}\right)^{\frac{1}{a}},$ which says that we got a maximal value.

By the same way we can got a minimal value: $-\left(\frac{1}{2}\right)^{\frac{2}{a}}.$

Answer 3

In this solution, the objective is to optimize the value of $x_1x_2\cdots x_n$ subject to $$\sum_{i=1}^n\,\left|x_i\right|^{a_i}\leq m\,,$$ where $m,a_1,a_2,\ldots,a_n\in\mathbb{R}_{>0}$. The OP's question is a particular case where $n=2$, $x_1=x$, $x_2=y$, $a_1=a_2=a$, and $m=1$.

By the Weighted AM-GM Inequality, $$\begin{align} m&\geq \sum_{i=1}^n\,\left|x_i\right|^{a_i}=\sum_{i=1}^n\,\frac{\frac{1}{a_i}}{\sum_{j=1}^n\,\frac{1}{a_j}}\Biggl(a_i\,\Big(\sum_{j=1}^n\,\frac{1}{a_i}\Big)\,\left|x_i\right|^{a_i}\Biggr) \\&\geq \prod_{i=1}^n\,\Biggl(a_i\,\Big(\sum_{j=1}^n\,\frac{1}{a_i}\Big)\,\left|x_i\right|^{a_i}\Biggr)^{\frac{\frac{1}{a_i}}{\sum_{j=1}^n\,\frac{1}{a_j}}}\\&=\left(\Big(\sum_{j=1}^n\,\frac{1}{a_j}\Big)^{{\sum_{j=1}^n\,\frac{1}{a_j}}}\,\Big(\prod_{i=1}^n\,a_i^{\frac{1}{a_i}}\Big)\,\Big|\prod_{i=1}^n\,x_i\Big|\right)^{\frac{1}{\sum_{j=1}^n\,\frac{1}{a_j}}}\,. \end{align}$$ Thus, $$\Big|\prod_{i=1}^n\,x_i\Big|\leq \frac{m^{\sum_{j=1}^n\,\frac{1}{a_j}}}{\Big(\sum_{j=1}^n\,\frac{1}{a_j}\Big)^{\sum_{j=1}^n\,\frac{1}{a_j}}\,\Big(\prod_{i=1}^n\,a_i^{\frac{1}{a_i}}\Big)}\,,$$ or $$-\frac{m^{\sum_{j=1}^n\,\frac{1}{a_j}}}{\Big(\sum_{j=1}^n\,\frac{1}{a_j}\Big)^{\sum_{j=1}^n\,\frac{1}{a_j}}\,\Big(\prod_{i=1}^n\,a_i^{\frac{1}{a_i}}\Big)}\leq \prod_{i=1}^n\,x_i\leq +\frac{m^{\sum_{j=1}^n\,\frac{1}{a_j}}}{\Big(\sum_{j=1}^n\,\frac{1}{a_j}\Big)^{\sum_{j=1}^n\,\frac{1}{a_j}}\,\Big(\prod_{i=1}^n\,a_i^{\frac{1}{a_i}}\Big)}\,.$$ The equality holds if and only if $$x_i=\pm\left(\frac{m}{a_i\,\sum_{j=1}^n\,\frac{1}{a_j}}\right)^{1/a_i}\,.$$

In the OP's case, we have $$|xy|\leq 2^{-2/a}\text{ or }-2^{-2/a}\leq xy\leq +2^{-2/a} \,.$$ The minimum occurs iff $(x,y)=\pm\left(2^{-1/a},-2^{-1/a}\right)$, whereas the maximum occurs iff $(x,y)=\pm\left(2^{1/a},2^{1/a}\right)$.