Curve of intersection problem using lagrange multipliers

Question

Let C be the curve of intersection of the following two surfaces

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

$$2x^2+4y^4+z=3\tag2$$

Find points on C which are closest to and furthest from the origin

Given that we are maximizing/minimizing distance, we can define $$f(x,y,z)=d^2=(x-0)^2+(y-0)^2+(z-0)^2=x^2+y^2+z^2$$

My approach included that since we know that $x^2+y^2=1$, we can change $f(x,y,z)$ to $1+z^2$, and therefore only have the 2nd constraint instead of two constraints.

Taking the gradients of both, we get the following system of equations:

$$0=2λx$$

$$0=8λy$$

$$2z=λ$$

$$2x^2+4y^4+z=3$$

I am honestly at a complete loss of where to go from here. For the record, yes I could do this question without eliminating the first constraint, but I'm looking for how to do it this way. Any help would be great!

Answer 1

In fact your curve may be proyected to the plane $OYZ$ and this simplify the problem. This is so because the intersection of $(1)$ and $(2)$ is $$R(y,z)=4y^4-2y^2+z-1=0$$ Hence the functional to minimise is $$f(y,z) =y^2+z^2$$

it results $$\vec{grad}\,f(x,y,z) = \lambda\,\vec{grad}\,R(x,y,z)$$ Resulting $$2\left[\begin{array}{c} y\\ z \end{array}\right]=\lambda\left[ \begin{array}{c} 4y(4y^2-1)\\ 1\end{array}\right] \qquad with \qquad 4y^4-2y^2+z-1=0 $$ Clearly (if I have not committed any mistake) $\lambda=2z$ and the following system wil determine the values for $y$ and $z$ $$y\left[1-4z(4y^2-1)\right]=0$$ $$4y^4-2y^2+z-1=0$$ Since $x$ is not solved, we obtain it from the restriction $x^2+y^2=1$, and the global optimum will be the minimum of $|(x,y,z)|^2$ of the computed points

Answer 2

We can use the third equation to substitute for $\lambda$ in the first two:

$$ 4xz = 0 $$ $$ 16yz = 0 $$

From here, we can consider two cases: (i) $z=0$ and (ii) $z \neq 0$.

In (i), the fourth equation reduces to $$ 2x^2 + 4y^4 = 3 $$

In this solution, the objective is always $1$, so we only need to find the values of $x$ and $y$ that satisfy the displayed equation above, along with $x^2 + y^2 = 1$. It should be straightforward to solve this.

In (ii), we know that $x=y=0$, and $z=3$. However, this violates the first constraint, so this cannot be a solution.

It's easy to intuit from the objective that the solution found above has to be a minimum. I'll need to think more carefully about maxima -- will edit this when I figure that part out.

Answer 3

You want to have your point on the constraints. It means that total differential of the constraints must be zero. From the first constraint $$2xdx+2ydy=0 \Rightarrow dy=-\frac{x}{y}dx$$ From the second constraint $$4xdx+16y^3dy+dz=0$$ $$\Rightarrow 4xdx-16y^2xdx+dz=0$$ $$\Rightarrow dz=16y^2xdx+4xdx$$ To maximize the distance equation wrt x,y,z $$x^2+y^2+z^2\Rightarrow 2xdx+2ydy+2zdz=0$$ Replacing dy and dz $$2xdx-2xdx+2z(16y^2xdx+4xdx)=0$$ $$\Rightarrow 2zx(16y^2+4)dx=0$$ This equation holds for 3 case

1)$z=0$ Using the constraint equations you will find 4 points. Calculating the distance with these points you will get the global minima

2) $x=0$ Using the constraint equations you will find 2 points. Calculating the distance with these points you will see that these points are not global extremum

3) $16y^2+4=0$ Using the constraint equations you will find 4 points. Calculating the distance with these points you will get the global maxima