Find the max/min values of s^2+t^2 on a curve by the method of Lagrange Multipliers

Quote:

Originally Posted by Runty

I've gotten an answer for the first half of this, but not the second one, which I believe is some sort of trick question.

Find the maximum and minimum values of $s^2+t^2$ on the curve
$s^2+2t^2-2t=4$
by the method of Lagrange Multipliers.
I already have this half answered, and got the following:
Max: $(s,t)=(\pm 2, 1),s^2+t^2=5$
Min: $(s,t)=(0,2),s^2+t^2=4$ , $(s,t)=(0,-1),s^2+t^2=1$

Now for the second half, which I believe is some sort of trick question.

If $s^2=4-2t^2+2t$ is substituted into $s^2+t^2$ , we get a function $h(t)=4+2t-t^2$ , which has only a maximum value on $R$ . Explain how the extreme values obtained in the first part can be obtained from $h(t)$ .
See, the issue is that $h(t)$ is a parabola, which has a maximum at 5, but no minimum (or at least no absolute minimum). So how would you use $h(t)$ to find the minimum values? That's what I feel is the "trick" part of the question.

If anyone could clarify the second half of the question, I'd appreciate it.

The point is that h(t) is not defined for all values of t. The curve $s^2+2t^2-2t=4$ represents an ellipse with centre at (0,1/2) and minor axis 3/2. So t has to lie between –1 and +2. If you look for the minimum value of h(t) in that interval, you will find that it occurs at an endpoint.