Vrftsjtryk

In a constrained optimization problem, let's consider the example $$begin{cases}f(x, y) = yx^2 Tiny(function to be maximized) \ g(x, y) = x^2 + y^2 = 1 Tiny(constraint)end{cases}$$ why does the answer not need to satisfy $f(x^*, y^*) = 1$? Geometrically, viewing $f(x, y) = yx^2$ and $g(x, y) = x^2 + y^2$ in $ℝ^3$ (which motivated this question), why aren't solutions required to be points where $f(x, y)$ and $g(x, y)$ intersect, or at least where $f(x, y)$ intersects $g(x, y) = 1$? The solutions turn out to be $f(x^*, y^*, f(x^*, y^*)) = (±frac{sqrt6}{3}, frac{sqrt3}{3}, frac{2sqrt3}{9})$, which both have a height or $z$-coordinate of $frac{2sqrt3}{9}$, while I would expect any point that satisfies $g(x, y) = 1$ to have a height or $z$-coordinate of $1$. Instead of lying within the within the flat slice of the graph of $g(x, y) = x^2 + y^2$ where $g(x, y) = 1$, the solutions lie within the slice representing $g(x, y) = frac{2sqrt3}{9}$, seemingly failing to satisfy the constraint.

This worry can be obfuscated by flattening $ℝ^3$ into a contour plot where the constraint and maximized function do intersect, but only by discarding a dimension of information from the original picture; being aware of the 3D graph the contour plot represents, I still find the matter conceptually troublesome.

One proposed idea has been to view $g(x, y)$ as living in $ℝ^2$, thus ignoring its height/$z$-coordinate/output altogether. However, this seems unsatisfactorily at odds with its deep symmetry with $f(x, y)$, which lives in $ℝ^3$. Perhaps the labels and terminology in constrained optimization problems give the impression that the function and the constraint are dissimilar animals, but I get the feeling from my trivially faint glimpse of Lagrangian duality that they're actually highly symmetric. One is $f(x, y) = yx^2 = ????$, and the other $g(x, y) = x^2 + y^2 = 1$, and in fact, once solved, I can forget the $x*$ and $y*$ parts of the solution and reframe the problem where $f(x, y) = yx^2 = frac{2sqrt3}{9}$ is the constraint, and $g(x, y) = x^2 + y^2 = ????$ is the function, and I'll rediscover the same $x^*$ and $y^*$, along with the original constraint constant $1$. I have a hard time convincing myself that expressions with such symmetricity aren't properly viewed as equal in dimension.

Some geometric ideas

In the attached plot we have in light red the surface $S_1(x,y,z) = x^2 y-z = 0$ and in light yellow the surface $S_2(x,y,z) = x^2+y^2-1 = 0$ In blue is depicted the intersection $S_1(x,y,z)cap S_2(x,y,z)$

We can obtain a surfaces $S_3$ containing the intersection curve, which is more handy

$$
S_3(x,y,z) = (S_1circ S_2)(x,y,x) = (1-y^2) y -z=0
$$

In gold color we have $S_3(x,y,z)$

Now the solutions for

$$
frac{d}{dy}((1-y^2) y) = 0\
$$

are contained into the set of stationary points in $S_1(x,y,z)cap S_2(x,y,z)$

NOTE

The stationary points for the problem are

$$
left[
begin{array}{ccc}
x & y & z \
-sqrt{frac{2}{3}} & -frac{1}{sqrt{3}} & -frac{2}{3 sqrt{3}} \
-sqrt{frac{2}{3}} & frac{1}{sqrt{3}} & frac{2}{3 sqrt{3}} \
sqrt{frac{2}{3}} & -frac{1}{sqrt{3}} & -frac{2}{3 sqrt{3}} \
sqrt{frac{2}{3}} & frac{1}{sqrt{3}} & frac{2}{3 sqrt{3}} \
end{array}
right]
$$

Those points are shown in red over the intersection

NOTE

The MATHEMATICA script associated to the first plot is

f = y x^2 - z
h = x^2 + y^2 - 1
gr1 = ContourPlot3D[{h == 0, f == 0}, {x, -1.5, 1.5}, {y, -1.5, 1.5},
{z, -1.5, 1.5},
MeshFunctions -> {Function[{x, y, z, g}, h - f]},
MeshStyle -> {{Thick, Blue}}, Mesh -> {{0}},
ContourStyle -> {Directive[Yellow, Opacity[0.5],
Specularity[White, 30]],
Directive[Red, Opacity[0.5], Specularity[White, 30]]}, PlotPoints -> 40]


Geometrically, viewing $f(x, y)=yx^2$ and $g(x, y)=x^2+y^2$ in $R^3$ (which motivated this question), why aren't solutions required to be points where $f(x, y)$ and $g(x, y)$ intersect, or at least where $f(x, y)$ intersects $g(x, y)=1$?

You are right, $g(x,y)=x^2+y^2$ is a two-variable function, whose graph is paraboloid in $mathbb R^3$. However, $g(x,y)=x^2+y^2=1$ is no longer two-variable function, but a contour curve of the parabaloid, which is a circle in $mathbb R^2$. So, the constraint $g(x,y)=x^2+y^2=1$ implies the points $(x,y)in mathbb R^2$ on the circle only must be considered for the objective function $f(x,y)=yx^2$ to be maximized.

Let's see the solutions to understand it further.

Method 1. Use the contour curves $y=frac f{x^2}$, where $f$ is considered constant. Draw the contour curves (for various positive values of $f$ for maximum) and the constraint on the same graph:

Note that, if you look at the first quadrant, the red contour line implies the value of $f_1=1$ (it does not intersect the circle, so does not satisfy the constaint), the green $f_2=frac12$ (again, it does not satisfy the constaint), the solid black $f_3=frac2{3sqrt{3}}$ (it touches the circle and the touching point is the optimal), the blue $f_4=frac15$ (it crosses the circle at two points and at those two points the constaint is satisfied, however, those two points are not optimal, because the value of $f_4=frac15$ is less than $f_3$.

How to find the touching point? You need to make sure the contour curve $y=frac f{x^2}$ and the circle $x^2+y^2=1$ have a common tangent line. Let $(x_0,y_0)$ be the tangent point. Then:
$$begin{cases}y=frac f{x_0^2}-frac{x_0}{sqrt{1-x_0^2}}(x-x_0) \ y=frac f{x_0^2}-frac{2f}{x_0^3}(x-x_0) end{cases} Rightarrow x_0=sqrt{frac 23}; f_{text{max}}=frac{2}{3sqrt{3}}.$$

Method 2. Just for reference. Use AM-GM:
$$x^2+y^2=1 Rightarrow 1=frac{x^2}{2}+frac{x^2}{2}+y^2ge 3sqrt[3]{frac14x^4y^2} Rightarrow yx^2le frac{2}{sqrt{27}}=frac{2sqrt{3}}{9},$$
equality occurs for $left|frac x{sqrt{2}}right|=y=frac1{sqrt{3}}$. Hence: $f(pmsqrt{frac{2}{3}}, frac{1}{sqrt{3}})=frac{2sqrt{3}}{9}.$