How can I calculate using Lagrange multipliers the minimum and maximum of the function $$F(x,y) = \frac{ 2(x^2 +y^2)}{x^2 + y^2 + 1} - {16x^2 + 1}$$
in the domain $D= \Big\{x,y \in \mathbb{R} \ ; \ { x^2 +y^2} \leq 1 \Big\}$.
I am having trouble calculating the partial derivatives of this function and would appreciate any help. Thanks.
On
[This solution requires no calculus, so it does not directly answer the OP's question, but I hope it will be enlightening.]
The function is even in both $x$ and $y$, so if you find one maximum and one minimum, you can reflect them on the x-axis and the y-axis to find the rest.
Rewriting the function
\begin{align} F(x, y) &= \frac{2(x^2 +y^2)}{x^2 + y^2 + 1} - 16x^2 + 1 \\ &= \frac{2(x^2 +y^2 + 1) -2 }{x^2 + y^2 + 1} - 16x^2 + 1\\ &= 2 - \frac{2}{x^2 + y^2 + 1} - 16x^2 + 1\\ &= 3 - \frac{2}{x^2 + y^2 + 1} - 16x^2 \end{align}
Looking at the second term only, we can see that in the domain of interest:
$$ 1 \leq \frac{2}{x^2 + y^2 + 1} \leq 2 $$ the minimum value being obtained on the circular boundary of the domain and the maximum value at the origin.
Similarly for the third term:
$$ 0 \leq 16x^2 \leq 16 $$ the minimum value being obtained at $x=0$ and the maximum value at $x = \pm 1$.
To maximize $F(x,y)$ we need to minimize the second and the third term (because they are positive quantities that are subtracted from 3, so they server to reduce the value of $F(x,y)$). Clearly then, we need the point(s) to be on the circular boundary to minimize the first term and we need $x=0$ to minimize the second. So the maximum is obtained at $(0, \pm 1)$.
Conversely, to minimize $F(x,y)$, we need to maximize the second and third terms. So we need to be at the origin for the second term and at $x = \pm 1$ for the third term. Unfortunately, we cannot do both, but given the relative magnitudes of the two terms, it's more important to maximize the third term even at the expense of not maximizing the second. Although this argument is a bit hand-wavy, it is true that the absolute minima are at $(\pm 1, 0)$.
If you plot this function, you'll see that it looks like a very shallow saddle: the origin is a saddle point: minimum in the y-direction and maximum in the x-direction with the saddle sides dropping off fast, but the front-to-back curve just slightly dipping down:
I can suggest the following solution: 1) calculate all local min/max at $\mathbb{R}^2$ using standard calculus tools - discard all points outside domain $D$, and then 2) calculate possible min/max using Lagrange multiplier with the constrain $x^2+y^2=1$
Finally, compare the results - the min/max may lie on the boundary line of the domain D.
The partial derivatives are:
$$\frac{\partial f}{\partial x}(x,y)=\frac{2(x^2+y^2)\cdot 2x-4x(x^2+y^2+1)}{(x^2+y^2+1)^2}-32x=\frac{4 x (1-8 (x^2+y^2+1)^2)}{(x^2+y^2+1)^2}$$ $$\frac{\partial f}{\partial y}(x,y)=\frac{4y(x^2+y^2+1)-2(x^2+y^2)\cdot 2y}{(x^2+y^2+1)^2}=\frac{4 y}{\left(x^2+y^2+1\right)^2}$$