Finding the local extrema for a curve of intersection of two surfaces and Lagrange Multipliers

Question

Suppose that I'm trying to find a local extrema for $f\left(x,y\right)$ subject to the constraint $g\left(x,y\right)=c$, is this equivalent to finding the local extrema for the curve of intesection of the surfaces $f\left(x,y\right)$ & $h\left(x,y\right)=g\left(x,y\right)-c$?

To elaborate further, suppose that we're trying to locate an extrema for $f\left(x,y\right)=x^2y$ subject to $x^2+y^2=1$ is locating this equivalent to locating an extrema for the curve of intersections of the two surfaces $f\left(x,y\right)=x^2y$ & $h\left(x,y\right)=x^2+y^2-1$

Answer 1

It's really a 2D problem, not a 3D problem.

The task is to find the relative extrema of the function $f(x,y)=x^2y$ on the circle in the $xy$-plane whose equation is $x^2+y^2=1$.

The method of Lagrange multipliers will work, but isn't really needed here.

Given the constraint $x^2+y^2=1$, we can write $f(x,y)=(1-y^2)y$, which reduces the problem to that of finding the relative extrema of the function $g(y)=(1-y^2)y$ for $y\in [-1,1]$.

Can you finish it?

As regards your attempt to interpret the problem in 3D, note that in $\mathbb{R}^3$, the graph of the equation $x^2+y^2=1$ is a cylinder, which is not the same as the graph of the equation $z=x^2+y^2-1$ which is a paraboloid.

A correct 3D interpretation could be stated as:

Find the points $(x,y,z)\in\mathbb{R}^3$ on the curve of intersection of the surfaces with equations given by \begin{align*} &A:&&z=x^2y\\[4pt] &B:&&x^2+y^2=1\\[4pt] \end{align*} such that $z$ assumes a relative extreme value.

However the above 3D interpretation doesn't make the problem easier to solve.

Answer 2

We take $(x,y)$ to represent scalars, segment lengths as traditional and not coordinates of a surface.

Lets say we need to maximize volume $V =x^2 y$ of a brick of square section breadth and height x and length y subject to condition that extremities stay on unit cylinder:

$$ x^2+y^2= 1 \tag1 $$

To find optimum solution by application of Lagrange Multiplier

$$ \dfrac{2 x y}{x^2}=\dfrac{2x}{2y}\rightarrow 2 y^2= x^2$$

$$y=\pm \dfrac{1}{\sqrt2} \tag2x$$ is a pair of planes intersecting the given unit cylinder.

Solving (1) and (2)

$$ x=\pm\sqrt{\dfrac{2}{3}},\, y=\pm\sqrt{\dfrac{1}{3}};\,$$

This realizes a maximum volume ( that calculates to $\dfrac{16 \sqrt3}{9} )\;$ 3D brick in a rough hand-sketch: