Extrema of a function f(x,y,z)

edumat

New member
Joined
Aug 14, 2013
Messages
13
I need to find the extrema of \(\displaystyle f(x,y,z) = x(y+z)\) on the intersection curve of \(\displaystyle xz=1\) (hyperbolic cylinder) and \(\displaystyle x^2+y^2=1\) (circular cylinder)

Can anyone help me to get started? How can I proceed with the intersection of both cylinders?
 
Let \(\displaystyle x=\cos t, y=\sin t\) then \(\displaystyle xz=1\) implies \(\displaystyle z=\sec t\).

So the intersection of the cylinders is the curve given by \(\displaystyle \{(\cos t, \sin t, \sec t); 0\le t < 2\pi, t\neq \pi/2, 3\pi/2\}\).

Now you can sub these into \(\displaystyle f\) and find those special values of \(\displaystyle t\)
 
Last edited:
I need to find the extrema of \(\displaystyle f(x,y,z) = x(y+z)\) on the intersection curve of \(\displaystyle xz=1\) (hyperbolic cylinder) and \(\displaystyle x^2+y^2=1\) (circular cylinder)

Can anyone help me to get started? How can I proceed with the intersection of both cylinders?
I note that \(\displaystyle f(x,y,z) = xy + xz = xy + 1\), which makes me wonder about allowing \(\displaystyle x = 0\), or in the parametric form allowing \(\displaystyle t = \pi /2\) [see response by daon2].

If we show that \(\displaystyle \displaystyle \lim_{x \to 0+}(xz) = 1\text{ and }\lim_{x \to 0-}(xz) = 1\),

can we then treat \(\displaystyle f(x,y,z)\) as being continuous at \(\displaystyle x=0\) ??
 
I note that \(\displaystyle f(x,y,z) = xy + xz = xy + 1\), which makes me wonder about allowing \(\displaystyle x = 0\), or in the parametric form allowing \(\displaystyle t = \pi /2\) [see response by daon2].

If we show that \(\displaystyle \displaystyle \lim_{x \to 0+}(xz) = 1\text{ and }\lim_{x \to 0-}(xz) = 1\),

can we then treat \(\displaystyle f(x,y,z)\) as being continuous at \(\displaystyle x=0\) ??

Hmm.. the function is when considered by itself yes. But the surface xz=1 is not defined on the yz or xy planes. So the constraint restricts the domain of f, changing the function. As you point out, the limit will exist, but continuity is out the window.
 
Thanks guys. Using the trigonometric substitution suggested by daon2 I was able to solve the problem.
 
Top