Welcome to our community

Be a part of something great, join today!

emwhy's question at Yahoo! Answers regarding finding extrema in a given region

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Here is the question:

Find the extreme values of f, on a given region? Using LaGrange multipliers?


Find the extreme values of f(x,y) = 6x^2+3y^2-36x-1, on the region described by x^2+y^2 <=36.

I'm pretty sure you have to use LaGrange multipliers. I ended up with lamba=0, x=3, y= +/-sqrt(27) which doesn't seem right.


Additional Details

Actually I just got the absolute minimum value by taking partial derivatives of f with respect to x and y. How do I get the absolute max?
I have posted a link there to this thread so the OP can see my work.
 
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: emwhy's question at Yahoo! Questions regarding finding extrema in a given region

Hello emwhy,

First, let's find the critical values inside the given region. We do this by equating the partials to zero, and solving the resulting system:

We are given:

\(\displaystyle f(x,y)=6x^2+3y^2-36x-1\)

Hence:

\(\displaystyle f_x(x,y)=12x-36=12(x-3)=0\)

\(\displaystyle f_y(x,y)=6y=0\)

This gives us the critical point $(x,y)=(3,0)$. Since its distance from the origin is less than $6$, we know it is within the circular region.

Next, we may employ the second partials test to determine the nature of the critical point.

Let:

\(\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left(f_{xy}(x,y) \right)^2\)

Thus:

\(\displaystyle D(x,y)=12\cdot6-\left(0 \right)^2=72>0\)

Since \(\displaystyle f_{xx}(3,0)=12>0\) we conclude that the critical point $(3,0)$ is a relative minimum.

Now we want to check the boundary. In order to examine $f$ on the boundary of the region, we may represent the circle $x^2+y^2=6^2$ by means of the parametric equations $x=6\cos(t),\,y=6\sin(t),\,0\le t\le 2\pi$. Thus, on the boundary we can write $f$ as a function of a single variable $t$:

\(\displaystyle f(\cos(t),\sin(t))=f(t)=6\left(6\cos(t) \right)^2+3\left(6\sin(t) \right)^2-36\left(6\cos(t) \right)-1=108\cos^2(t)-216\cos(t)+107\)

Differentiating with respect to $t$ and equating to zero, we find:

\(\displaystyle f'(t)=-216\cos(t)\sin(t)+216\sin(t)=216\sin(t)\left(1-\cos(t) \right)=0\)

This gives us the critical values for $t$:

\(\displaystyle t=0,\,\pi,\,2\pi\)

Since the first and third critical values are boundary values for $t$, and because the parametrization of $x$ and $y$ have the same value for these two values, we need only consider:

\(\displaystyle t=0,\,\pi\)

This gives us the two critical points:

\(\displaystyle \left(x(0),y(0) \right)=(6,0)\)

\(\displaystyle \left(x(\pi),y(\pi) \right)=(-6,0)\)

Now, we may use the second derivative test to determine the nature of these extrema. Let's write the first derivative as:

\(\displaystyle f'(t)=-108\sin(2t)+216\sin(t)\)

Differentiating, we find:

\(\displaystyle f''(t)=-216\cos(2t)+216\cos(t)\)

Checking the critical values:

\(\displaystyle f''(0)=0\)

The second derivative test is inconclusive for this point, so using the first derivative test, we find:

\(\displaystyle f'\left(\frac{\pi}{2} \right)=216>0\)

\(\displaystyle f'\left(\frac{3\pi}{2} \right)=-216<0\)

Hence the point $(6,0)$ is a relative minimum.

\(\displaystyle f''(\pi)=-432<0\)

Hence the point $(-6,0)$ is a relative maximum.

So, we now know:

\(\displaystyle f_{\max}=f(-6,0)=6(-6)^2+3(0)^2-36(-6)-1=431\)

\(\displaystyle f_{\min}=f(3,0)=6(3)^2+3(0)^2-36(3)-1=-55\)