Constrained extrema with Lagrange multipliers

The functions $f$ and $g(x,y)=x^2+y^2-25$ are polynomials, and $\nabla g=(2x,2y)$ vanishes only at the origin, which does not lie on the constraint; so at every constrained extremum the condition of the Lagrange multiplier theorem holds. In the first equation $\lambda$ cannot be $0$, because that would give $3=0$; therefore $x$ and $y$ are obtained as functions of $\lambda$ and substituted into the constraint, which gives $\lambda =\pm \frac{1}{2}$.

The candidates are $(3,4)$ and $(-3,-4)$. The circle is closed and bounded and $f$ is continuous, so by the Weierstrass theorem $f$ has a maximum and a minimum on the constraint, which must be among the candidates. The maximum is $25$ at $(3,4)$ and the minimum is $-25$ at $(-3,-4)$. At the maximum point the gradient $\nabla f=(3,4)$ points in the direction of the radius, so the level line $3x+4y=25$ is tangent to the circle, as the figure shows.

As in the previous exercise, $\nabla g=(2x,2y)$ does not vanish on the constraint, so the extrema are among the solutions of the Lagrange system. The first equation can be written $2x(y-\lambda )=0$ and splits into two cases.

If $x=0$, the constraint gives $y=\pm \sqrt{3}$, and the second equation is satisfied with $\lambda =0$; at these points $f=0$. If $y=\lambda$, the second equation becomes $x^2=2y^2$, and substituting into the constraint gives $3y^2=3$, that is, $y=\pm 1$ and $x^2=2$. One obtains four points, at which $f=x^{2}y=2y$ equals $2$ or $-2$. The circle is closed and bounded, so the constrained maximum is $2$, at the points $\left(\pm \sqrt{2},1\right)$, and the minimum is $-2$, at the points $\left(\pm \sqrt{2},-1\right)$. The points $\left(0,\pm \sqrt{3}\right)$ satisfy the Lagrange condition but are not absolute extrema on the constraint.

The gradient of the constraint is $\nabla g=(2x,8y)$, zero only at the origin, which does not lie on the ellipse; the Lagrange multiplier theorem therefore applies at every point of the constraint. Substituting the second equation into the first gives $y=2\lambda \cdot 8\lambda y=16{\lambda }^{2}y$. The case $y=0$ would lead to $x=0$, which is excluded, therefore $y\ne 0$ and one can divide by $y$, obtaining $\lambda =\pm \frac{1}{4}$.

With $\lambda =\frac{1}{4}$ the first equation gives $y=\frac{x}{2}$, and the constraint becomes $x^2+x^2=8$, that is, $x=\pm 2$. With $\lambda =-\frac{1}{4}$ we have $y=-\frac{x}{2}$, and one obtains the other two points. The ellipse is closed and bounded, so comparing the values gives the maximum $2$ and the minimum $-2$.