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pheobee's question at Yahoo! Answers regarding extrema for a function of two variables

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MarkFL

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Feb 24, 2012
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Here is the question:

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional g?

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

f(x, y) = 3 sin x sin y, −π < x < π, −π < y < π
I have posted a link there to this thread so the OP can view my work.
 
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MarkFL

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Feb 24, 2012
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Hello pheobee,

We are given the function:

\(\displaystyle f(x,y)=3\sin(x)\sin(y)\) where $-\pi<x,y<\pi$.

First, to find the critical points, we must solve the simultaneous system:

\(\displaystyle f_x(x,y)=3\cos(x)\sin(y)=0\)

\(\displaystyle f_y(x,y)=3\sin(x)\cos(y)=0\)

This implies by addition:

\(\displaystyle \sin(x)\cos(y)+\cos(x)\sin(y)=\sin(x+y)=0\)

And by subtraction:

\(\displaystyle \sin(x)\cos(y)-\cos(x)\sin(y)=\sin(x-y)=0\)

Hence, we may state:

\(\displaystyle x\pm y=k\pi\)

Now, given the first partials, and the stated domains, we obtain from this the following critical points:

\(\displaystyle (x,y)=\left(-\frac{\pi}{2},-\frac{\pi}{2} \right),\,\left(-\frac{\pi}{2},\frac{\pi}{2} \right),\,(0,0),\,\left(\frac{\pi}{2},-\frac{\pi}{2} \right),\,\left(\frac{\pi}{2},\frac{\pi}{2} \right)\)

Next, we may use the second partials test to determine the nature of the potential extrema associated with these 5 critical points.

First, we compute:

\(\displaystyle f_{xx}(x,y)=-3\sin(x)\sin(y)\)

\(\displaystyle f_{yy}(x,y)=-3\sin(x)\sin(y)\)

\(\displaystyle f_{xy}(x,y)=3\cos(x)\cos(y)\)

And we define:

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

Using our second partials, we then find:

\(\displaystyle D(x,y)=9\left(\sin^2(x)\sin^2(y)-\cos^2(x)\cos^2(y) \right)\)

So now, we analyze the critical points:

1.) \(\displaystyle (x,y)=\left(-\frac{\pi}{2},-\frac{\pi}{2} \right)\)

\(\displaystyle D(x,y)=9>0\)

\(\displaystyle f_{xx}(x,y)=-3<0\)

We conclude this point is at a relative maximum.

The value of the function at this point is:

\(\displaystyle f(x,y)=3\)

2.) \(\displaystyle (x,y)=\left(-\frac{\pi}{2},\frac{\pi}{2} \right)\)

\(\displaystyle D(x,y)=9>0\)

\(\displaystyle f_{xx}(x,y)=3>0\)

We conclude this point is at a relative minimum.

The value of the function at this point is:

\(\displaystyle f(x,y)=-3\)

3.) \(\displaystyle (x,y)=(0,0)\)

\(\displaystyle D(x,y)=-9<0\)

We conclude this point is not at an extremum (saddle point).

The value of the function at this point is:

\(\displaystyle f(x,y)=0\)

4.) \(\displaystyle (x,y)=\left(\frac{\pi}{2},-\frac{\pi}{2} \right)\)

\(\displaystyle D(x,y)=9>0\)

\(\displaystyle f_{xx}(x,y)=3>0\)

We conclude this point is at a relative minimum.

The value of the function at this point is:

\(\displaystyle f(x,y)=-3\)

5.) \(\displaystyle (x,y)=\left(\frac{\pi}{2},\frac{\pi}{2} \right)\)

\(\displaystyle D(x,y)=9>0\)

\(\displaystyle f_{xx}(x,y)=-3<0\)

We conclude this point is at a relative maximum.

The value of the function at this point is:

\(\displaystyle f(x,y)=3\)

And so we may conclude:

\(\displaystyle f_{\min}=-3\)

\(\displaystyle f_{\max}=3\)

Here is a plot of the given function on the stated domain, showing all 5 points:

pheobee.jpg

Here is a link to the program I used to plot the function:

z=3sin(x)sin(y) where x=-pi to pi,y=-pi to pi - Wolfram|Alpha