Finding Extremes of Multivariable Function $f(x,y,z)=xz−yz $

Question

I am seeking assistance in finding the extrema of the function $f(x,y,z)=xz−yz$ when evaluated at points on the curve of intersection of two given surfaces. The surfaces of interest are defined by the following equations:

$$yz=2$$

$$ x^2+z^2=2$$

My goal is to determine the points at which the function $f$ reaches extrema (either maxima or minima) on the curve of intersection of these two surfaces. However, I need guidance on how to approach this problem.

So far, I have managed to parametrize the intersection curve and have concluded that the extrema of the function $f(x,y,z)$ on this curve correspond to points where the parameter of the parametrization, $t$, is constant. I have also derived the parametric equations for the curve:

$x(t) = \pm\sqrt{\frac{2(t^2 - 2)}{t^2}}$

$y(t)=t$

$z(t) = \frac{2}{t}$

At this point, I find myself in need of determining the specific values of $t$ for which $f$ attains extrema on the curve. How can I proceed from here to ascertain whether these points are maxima or minima? Any further assistance or guidance would be greatly appreciated. Thank you!

Answer 1

The Lagrange multiplier condition is that to find constrained extrema of $f$ subject to $g_1, g_2 = 0$, we solve the system $\nabla f = \lambda g_1 + \mu g_2$ (or some equivalent condition).

In this case, this reduces to $(z, -z, x-y) = \lambda (0, z, y) + \mu (2x, 0, 2z)$. For now, let's assume that both $\lambda$ and $\mu$ are non-zero. Then we get

$$z = 2\mu x$$ $$-z = \lambda z$$ $$x-y = \lambda y + 2\mu z$$

From inspection, the second equation gives us $(\lambda +1)z = 0$. So let's try both cases (i.e. $z = 0$ or $\lambda = -1$)

If $\lambda = -1$, we get $z = 2\mu x$ and the bottom equation becomes $x = 2\mu z$. Add em up and we have $x+z = 2\mu (x+z)$. Rearranging, we have $(2\mu-1)(x+z) = 0$. So either $\mu = \frac{1}{2}$ or $x = -z$. Let's try both. With $\mu = 1/2$, we have $z = x$. Great so essentially we have $x = \pm z$, sub this into $g_2$ and we get $2x^2 = 2$ or $x = \pm 1$. Following carefully, this gives us the following points to test at $(1,2,1), (1,-2,-1), (-1,-2,-1), (-1,2,1)$.

I'm not going to complete the answer to the question but you need to go back and consider what if $z = 0$, and what if $\lambda = 0$ and what if $\mu = 0$ (why?). Spoiler alert though, our maxima and minima are actually all at those $4$ points we just found above!

Answer 2

No Lagrange multipliers are needed. We have $$f(x,y,z)=xz-yz=xz-2$$ Moreover $$-1=-{1\over 2}(x^2+z^2)\le xz\le{1\over 2}(x^2+z^2)=1$$ The first equality holds iff $x=-z=\pm 1,$ while the second iff $x=z=\pm 1.$ Therefore the maximal value is equal $-1$ and the minimal value is equal $-3.$ By analyzing the above the maximal value is attained at two points $\,\pm (1,2,1)$ and the minimal value at two points $\,\pm (-1,2,1).$
The methods shows that the values $-1$ and $-3$ are global extremes.