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 $\displaystyle s^2+t^2$ on the curve

$\displaystyle s^2+2t^2-2t=4$

by the method of Lagrange Multipliers.

I already have this half answered, and got the following:

Max: $\displaystyle (s,t)=(\pm 2, 1),s^2+t^2=5$

Min: $\displaystyle (s,t)=(0,2),s^2+t^2=4$, $\displaystyle (s,t)=(0,-1),s^2+t^2=1$

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

If $\displaystyle s^2=4-2t^2+2t$ is substituted into $\displaystyle s^2+t^2$, we get a function $\displaystyle h(t)=4+2t-t^2$, which has only a maximum value on $\displaystyle R$. Explain how the extreme values obtained in the first part can be obtained from $\displaystyle h(t)$.

See, the issue is that $\displaystyle h(t)$ is a parabola, which has a maximum at 5, but no minimum (or at least no absolute minimum). So how would you use $\displaystyle h(t)$ 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.