Maximizing $f(x, y, z) = xyz$ on the intersection of a cylinder and a plane

Question

What are the extrema of $f(x, y, z) = xyz$ on the intersection of the cylinder $x^2 + y^2 = r^2$ and the plane through the origin $ax + by + cz = 0$ in $\mathbf{R^3}$? It is sensible to assume that $c \neq 0$, so that the plane and the cylinder intersect to form an ellipse. Note that this function has extrema in this case, since the intersection is compact.

I tried to use Lagrange multipliers, which leads to the system of equations

\begin{align*} yz &= 2 \lambda_1 x + \lambda_2 a \\ xz &= 2 \lambda_1 y + \lambda_2 b \\ xy &= \lambda_2 c \\ ax + by + cz &= 0 \\ x^2 + y^2 &= r^2 \end{align*} where $\lambda_1$ and $\lambda_2$ are the Lagrange multipliers. Giving these to a computer algebra system gives extraordinarily complicated results.

Is there a better way to approach this problem?

Answer 1

HINT

Let parametrize as

• $x=r\cos \theta$
• $y=r\sin \theta$
• $z=-\frac {ra}c \cos \theta-\frac {rb}c \sin \theta$

then

$$f(x,y,z)=g(\theta)=\frac{r^3}{c}\cos\theta\sin \theta (-a\cos \theta-b\sin \theta)$$

Answer 2

From the above answer, its better to consider min/max of the function $f(u,v) = uv(au+bv)= au^2v+buv^2, u^2+v^2=1, ab \neq 0$. Lagrange Multiplier seems a favorite candidate for this: $f_u = \lambda (2u), f_v = \lambda (2v)\implies 2auv+bv^2=2u\lambda, au^2+2buv=2v\lambda$. We will look at the cases $u = 0, v = 0, \lambda = 0$ later. In the mean time assume $uv\lambda \neq 0\implies \dfrac{2auv+bv^2}{2buv+au^2}=\dfrac{u}{v}$. Put $t= \dfrac{u}{v}\implies \dfrac{2at+b}{2bt+at^2}=t\implies at^3+2bt^2-2at-b=0\implies t^3+2dt^2-2t-d=0, d = \dfrac{b}{a}$. From here you can solve for $t$ and thus $\dfrac{u}{v}$ and together with $u^2+v^2=1$ you can solve for $u$ and $v$ individually. You can google the internet about the solution of a full cubic equation. Since you don't know $a,b$ the problem is more complicated than it should be.