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

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Staff member
Feb 24, 2012
Here is the question:

How can I find the local maximum and minimum values and saddle points of the function f(x,y) = sin(x)sin(y)?

Where -π < x < π and -π < y < π
I have posed a link there to this thread so the OP can see my work.
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  • #2


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Feb 24, 2012
Hello Robin,

We are given the function:

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


\(\displaystyle -\pi<x<\pi\)

\(\displaystyle -\pi<y<\pi\)

Let's take a look at a plot of the function on the given domain:


Equating the first partials to zero, we obtain:

\(\displaystyle f_x(x,y)=\cos(x)\sin(y)=0\implies x=\pm\frac{\pi}{2},\,y=0\)

\(\displaystyle f_y(x,y)=\sin(x)\cos(y)=0\implies x=0,\,y=\pm\frac{\pi}{2}\)

Adding, we find:

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

Applying the angle-sum identity for sine, we find:

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

Observing that we require:

\(\displaystyle -2\pi<x+y<2\pi\)

We then have:

\(\displaystyle x+y=-\pi,\,0,\,\pi\)

Thus, we obtain the 5 critical points:

\(\displaystyle P_1(x,y)=\left(-\frac{\pi}{2},-\frac{\pi}{2} \right)\)

\(\displaystyle P_2(x,y)=\left(-\frac{\pi}{2},\frac{\pi}{2} \right)\)

\(\displaystyle P_3(x,y)=(0,0)\)

\(\displaystyle P_4(x,y)=\left(\frac{\pi}{2},-\frac{\pi}{2} \right)\)

\(\displaystyle P_5(x,y)=\left(\frac{\pi}{2},\frac{\pi}{2} \right)\)

To categorize these critical points, we may utilize the second partials test for relative extrema:

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

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

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


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

Critical point $(a,b)$$D(a,b)$$f_{xx}(a,b)$Conclusion
$\left(-\dfrac{\pi}{2},-\dfrac{\pi}{2} \right)$1-1relative maximum
$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2} \right)$11relative minimum
$(0,0)$-10saddle point
$\left(\dfrac{\pi}{2},-\dfrac{\pi}{2} \right)$11relative minimum
$\left(\dfrac{\pi}{2},\dfrac{\pi}{2} \right)$1-1relative maximum