calculuslagrange multipliermultivariable-calculusoptimization
My book tells me that of the solutions to the Lagrange system, the smallest is the minimum of the function given the constraint and the largest is the maximum given that one actually exists.
But what if we only have one point as a solution? How to know whether Lagrange multipliers gives maximum or minimum?
When you solve for $\lambda$ using the systems:
$\begin{cases} 2x = \lambda 4x^3 & (1) \\ 2y = \lambda 4y^3 & (2) \\ x^4 + y^4 = 18 & (3) \end{cases}$
You cancel out $x$ in (1) to get $x^2 = 1/(2\lambda)$, you missed a case that $x=0$.
If $x=0$, then $y^4 = 18$, yields $y = \pm {18}^{\frac{1}{4}}$.
The minimum value of $f(x,y)$ would be $\sqrt{18}$, occurs at $(0,\pm {18}^{\frac{1}{4}})$ or $(\pm {18}^{\frac{1}{4}},0)$.
Here is an illustration of the problem at hand.
You want to find the the volume of the tetrahedron formed by the $x=y=z=0$ planes, and the plane tangent to the ellipsoid at some point on its surface, $(x,y,z)$. This volume, of course, will vary with different points on the surface of the ellipse, so the constraint function must be the following equality
$$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2=1$$
Now the vector normal to the surface of the ellipsoid at some point $(x,y,z)$ is given by the gradient of its equation.
$$\nabla g (x,y,z) = \left<\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2}\right>$$
The equation for the plane is tangent to the ellipsoid at this point is easily worked out to be
$$\frac{x}{a^2}x'+\frac{y}{b^2}b'+\frac{z}{c^2}z'=1$$
Now, finally, for the volume that this plane bounds within the first quadrant, we make use of the fact that this volume is simply $1/6$ the volume of the rectangular prism whose side lengths are given by the points of where the plane intersects the three axes.
$$V=\frac{(abc)^2}{6xyz}$$
And that is your function $f$ which you are trying to minimize.
$$f=V$$
$$g\rightarrow \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2=1$$
Best Answer
As Om(nom)$^3$ said in the comments, if you're working on a closed and bounded region then it's not possible to get only one critical point.
If you're not on a closed and bounded region then it's no longer guaranteed that you'll have more than one critical point. If you only have one critical point then you can use the Bordered Hessian technique. (Thanks to ziggurism for clearing that up.)