# Thread: Find the max and min

1. ## Find the max and min

Find the maximum and minumum values of $\displaystyle s^2+t^2$ on the curve
$\displaystyle s^2+2t^2-2t=4$
by the method of Lagrange Multipliers.
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 max value on R. Explain how the extreme values you obtained in the first part can be obtained from h(t).

I only know how to find the max and min from the orgin to to a surface, how do i deal with it when is from $\displaystyle s^2+t^2$

2. Originally Posted by wopashui
Find the maximum and minumum values of $\displaystyle s^2+t^2$ on the curve
$\displaystyle s^2+2t^2-2t=4$
by the method of Lagrange Multipliers.
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 max value on R. Explain how the extreme values you obtained in the first part can be obtained from h(t).

I only know how to find the max and min from the orgin to to a surface, how do i deal with it when is from $\displaystyle s^2+t^2$
You want to find the extrema of

$\displaystyle f(s,t)=s^2+t^2$

subject to the constraint:

$\displaystyle g(s,t)=s^2+2t^2-2t-4=0.$

So you form the Lagrangian:

$\displaystyle L(s,t,\lambda)=f(s,t)+\lambda g(s,t)$

and the maximum and minimum are amoung the stationary points of $\displaystyle L(s,t,\lambda)$

CB

3. Originally Posted by CaptainBlack
You want to find the extrema of

$\displaystyle f(s,t)=s^2+t^2$

subject to the constraint:

$\displaystyle g(s,t)=s^2+2t^2-2t-4=0.$

So you form the Lagrangian:

$\displaystyle L(s,t,\lambda)=f(s,t)+\lambdag(s,t)$
You mean $\displaystyle L(s,t,\lambda)=f(s,t)+\lambda g(s,t)$.
(you don't have a space between "\lambda" and "g".)

and the maximum and minimum are amoung the stationary points of $\displaystyle L(s,t,\lambda)$

CB

4. so to find this, I need to find the values of x y and lamda, do I let $\displaystyle L_x =0$, $\displaystyle L_y=0$, and $\displaystyle L_{lamba} =0$ to do this

5. Running through this one's first half, I already found the maximum and minimum values. I'll post them up (but not the steps).

Max: $\displaystyle (s,t)=(\pm 2, 1),s^2+t^2=5$
Min: $\displaystyle (s,t)=(0,2),s^2+t^2=4$ and $\displaystyle (s,t)=(0,-1),s^2+t^2=1$

The second half of this question, however, has me a bit puzzled, as it seems to be a bit of a trick question. The issue I'm having is that $\displaystyle h(t)$ is a parabola that has a maximum at 5 (which matches the first half), though it doesn't have a minimum (or at least no absolute minimum). So how would one find the minimum extrema using $\displaystyle h(t)$?