Problem 1 - Analysis

Compute the maximal and minimal values of the following functions.

1.

$f(x,y) = x^3 - 3axy + y^3 \quad (a>0)$

2.

$f(x,y) = (ax^2 + by^2)\exp(- x^2 - y^2) \quad (b>a>0)$

Taylor-Young formula with $f \in \mathcal{C}^2(U, \mathbb{R})$ with $U\in \mathcal{T}(E)$:

$$f(a+u)-f(a)=\sum_{i=1}^n \frac{\partial f}{\partial x_i}(a)u_i + ... ..._j}(a)u_i u_j \right) + o_{u\rightarrow 0}\left(\vert\vert u\vert\vert^2\right)$$

We call Monge notation the particular case where $\dim E = 2$:

$$f(a+h,b+k) - f(a,b) = ph + qk + \frac{1}{2}\left( rh^2 + 2shk + tk^2\right) + o(h^2+k^2)$$

where

$$p=\frac{\partial f}{\partial x}(a,b) \quad q=\frac{\partial f}{\partial y}(a,b)$$

$$r = \frac{\partial^2 f}{\partial x^2}(a,b) \quad s=\frac{\partial^2 f}{\partial x \partial y}(a,b) \quad t=\frac{\partial^2 f}{\partial y^2}(a,b)$$

If $f \in \mathcal{C}^1(U, \mathbb{R})$ where $U\in \mathcal{T}(E)$, we say that $a$ is a critical point when $df_a=0$.

We assume that $\dim E = 2$. Let $f \in \mathcal{C}^2(U, \mathbb{R})$ with $U\in \mathcal{T}(E)$ and let $(a,b)\in U$ be a critical point of $f$ such that $s^2-rt\neq 0$ at $a$:

• If $s^2 - rt < 0, r > 0$, $f$ has a minimal value at $(a,b)$.
• If $s^2 - rt < 0, r < 0$, $f$ has a maximal value at $(a,b)$.
• If $s^2 - rt > 0$, $f$ has neither a minimal value nor a maximal value at $(a,b)$.

1.

$\frac{\partial f}{\partial x}(x,y) = 3x^2 - 3ay$
$\frac{\partial f}{\partial y}(x,y) = 3y^2 - 3ax$
$\frac{\partial^2 f}{\partial x \partial y}(x,y) = -3a$
$\frac{\partial^2 f}{\partial x^2}(x,y) = 6x$
$\frac{\partial^2 f}{\partial y^2}(x,y) = 6y$

$3x^2 - 3ay = 0$

$3y^2 - 3ax = 0$

$\Leftrightarrow$ $(x,y)=(0,0)$
At $(0,0), s^2 - rt = 9a^2 > 0$
Therefore, $f$ has no extremum.

2.

$\frac{\partial f}{\partial x}(x,y) = 2x\exp(-x^2-y^2)(a - ax^2 - by^2)$
$\frac{\partial f}{\partial y}(x,y) = 2y\exp(-x^2-y^2)(b - by^2 - ax^2)$
$\frac{\partial^2 f}{\partial x \partial y}(x,y) = 4xy\exp(-x^2-y^2)(-b-a+ax^2+by^2)$
$\frac{\partial^2 f}{\partial x^2}(x,y) = 2\exp(-x^2-y^2)(a-5ax^2+2ax^4+2bx^2y^2-by^2)$
$\frac{\partial^2 f}{\partial y^2}(x,y) = 2\exp(-x^2-y^2)(b-5by^2+2by^4+2ax^2y^2-ax^2)$

$2x(a-ax^2-by^2)=0$

$2y(b-ax^2-by^2)=0$

$\Leftrightarrow$

$x=0$ or $a=ax^2+by^2$

$y=0$ or $b=ax^2+by^2$

$\Leftrightarrow$ $(x,y) \in \{ (0,0), (0,1), (0,-1), (1,0), (-1,0)\}$

• At $(0,0), s^2 - rt = -4a^2b^2 < 0$, we a minimum (because $r = 2a > 0$) of value 0.
• At $(0,1), s^2 - rt = \frac{8b}{e^2}(a-b) < 0$, we have a maximum (because $r < 0$) of value $\frac{b}{e}$.
• At $(0,-1), s^2 - rt = \frac{8b}{e^2}(a-b) < 0$, we have a maximum (because $r < 0$) of value $\frac{b}{e}$.
• At $(1,0), s^2 - rt = \frac{8a}{e^2}(b-a) > 0$, we don't have any extremum.
• At $(-1,0), s^2 - rt = \frac{8a}{e^2}(b-a) > 0$, we don't have any extremum.