# Solving for global Maximum and minimum on a interval

We need to determine the global maximum and minimum of:

$$f(x,y)=y^2-16x^2$$ on the interval of: $$\{(x,y) : y ≤ 1−x^2,y ≥ 0\}$$

My initial thought was that I could use extreme value theorem, later realising that was wrong. Then I started to use Lagrange multiplier because these were constraints. I did it and go the answers to be $$(-1,0)$$ and $$(1,0)$$ as well as $$(-3,-8)$$ and $$(3,-8)$$. Although when you evaluate it comes up as all these values are equal to zero, when plugged into $$f(x,y$$). Some help is really appreciated!!!

• $f(\pm1,0)=1$ and $(\pm3,8)$ do not belong to the domain. – AugSB Apr 4 at 18:27
• $f(\pm1,0)=-16\ne1$ – Peter Foreman Apr 4 at 18:28

The set $$\{(x,y) : y \leq 1-x^2, y \geq 0\}$$ is not an interval, it is a region in the $$xy$$-plane.

First you need to find the critical points of $$f(x,y)$$, i.e., points where the gradient is zero. If any of these lie within the region above, they are candidates for the locations of the max/min.

You also need to find critical points of $$f(x,y)$$ along the boundaries of the region above, i.e., along the curve $$y=1-x^2$$ and the line $$y=0$$. For the first curve you should use Lagrange multipliers, and for the second you can just plug in $$y=0$$ and treat it as a function of $$x$$.

Finally, you also need to test the points where the boundary curves of the region intersect, i.e., where $$y=1-x^2$$ and $$y=0$$ intersect.

Once you have the list of all these critical points, plug them all into $$f$$ to see where the min and max occur.

$$y≤1−x^2,y≥0$$ or,

$$x^2+y ≤ 1$$, or, $$0≤y≤1$$ and $$0≤x^2≤1$$ or $$0≤|x|≤1$$.

To maximize $$|x|$$ has to be minimum which is zero(x=0), in that case $$y=1$$ (maximum) , to minimize similarly $$|x|=1$$ or $$x=±1$$ and $$y=0$$.