Find the extreme values of the function, subject to constraints
$$f(x,y) = y^2 - x^2 ; x^2 + y^2 = 36$$
Here is what I have done:
Using the substitution method we can say:
$$y = \sqrt{-x^2 + 36}$$
Then plugging that into our original equation:
$$f(x,y) = (\sqrt{-x^2 + 36})^2 - x^2$$ which simplifies to $$-2x^2+36$$
now $$f_x(x,y) = 2(-2x) + 0$$
set that $=0$ and get $x=0$
Now find if min/max:
$$f_{xx} = -4 $$
$$-4 < 0; maximum$$
Answer: There is a maximum point at $x=0$?
Is this right?
$\endgroup$ 62 Answers
$\begingroup$Let $x = 6\cos \theta, y = 6\sin \theta \implies F(x,y) = F(\theta) = 36\cos (2\theta)\le 36$, and the max is $36$. The min is $-36$ and this occurs when $\cos (2\theta) = -1$ which corresponds to $x^2 - y^2 = -36$. Since $x^2+y^2 = 36$, we have: $x = 0, y = \pm 6$.
$\endgroup$ 4 $\begingroup$Your method will find every extreme point except for those where the domain (i.e. the points on $x^2 + y^2 = 36$) move parallel to the $y$-axis. The reason for that is probably best explained if we have a simple drawing to help:
Now, what does this drawing represent? There is the black circle, which is the constraint $x^2 + y^2 = 36$. Then there are red level lines of the function $36-2x^2$, which, as long as you only care about points on the circle, is the same function as $f(x) = y^2 - x^2$. Following these level lines, you will find that there are extremal points when $x = 0$.
But you miss the extremal points that might happen where the red lines are tangent to the circle. That's because $36-2x^2$ just passes on by while the circle itself "doubles back" and changes which direction it goes. For instance, moving clockwise from the top point of the circle, where the function value is $36$ and down towards the rightmost point where the function value is $-36$, the value of $x$ increases, which means that the value of $36-2x^2$ decreases. However, the moment you get to the rightmost point of the circle, the circle starts going back towards lower $x$-values. That means that if you walk along the circle, the value of $x$ starts decreasing again, which means the function value starts increasing. So we must have hit a minimum point.
The reason that your partial derivatives miss this is that they do not take into account that the circle we follow turns around. They only care about what happens if we increase the value of the $x$-coordinate. At most of the points on the circle, you can move along it by the $x$-coordinate, and therefore the partial derivative with respect to $x$ can tell you what happens (here it is important that the new function expression $36-2x^2$ doesn't have any $y$ in it). But not at those two points where the circle crosses the $x$-axis. At those points, moving along the circle means changing from increasing $x$ to decreasing $x$ (or vice versa).
The solution is to also do the other substitution, into the function $2y^2-36$. This function will let you find any extreme values where the circle is not horizontal. Together these two methods will let you cover any point on the circle, and you can rest easy knowing you've found all extreme values.
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