Find Extrema of f(x,y)=sin(x)sin(y) | Yahoo Answers

In summary: Therefore, in summary, to find the local maximum and minimum values and saddle points of the function f(x,y) = sin(x)sin(y) on the given domain, we can equate the first partial derivatives to zero and use the second partials test to classify the critical points as either relative extrema or saddle points.
  • #1
MarkFL
Gold Member
MHB
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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
Hello Robin,

We are given the function:

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

where:

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

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

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

View attachment 1676

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:

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

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

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

Hence:

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

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
 

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FAQ: Find Extrema of f(x,y)=sin(x)sin(y) | Yahoo Answers

What is the purpose of finding extrema of a function?

Finding extrema allows us to identify the maximum and minimum values of a function, which can provide important insights into the behavior and characteristics of the function.

How do you find the extrema of a two-variable function like f(x,y)=sin(x)sin(y)?

To find the extrema of a two-variable function, we use partial derivatives. First, we take the partial derivative with respect to x, set it equal to 0, and solve for x. Then, we take the partial derivative with respect to y, set it equal to 0, and solve for y. The resulting values of x and y will give us the coordinates of the extrema.

Can a two-variable function have more than one extrema?

Yes, a two-variable function can have multiple extrema. These can include local extrema, which are the maximum and minimum values in a specific region, and global extrema, which are the overall maximum and minimum values of the function.

How can we determine whether an extrema is a maximum or minimum?

We can determine whether an extrema is a maximum or minimum by using the second derivative test. If the second derivative is positive at the extrema, it is a minimum, and if the second derivative is negative, it is a maximum.

Are there any applications of finding extrema in real-life situations?

Yes, finding extrema has various applications in fields such as economics, physics, and engineering. For example, in economics, finding the maximum profit of a company can involve finding the extrema of a profit function. In physics, finding the maximum height of a projectile involves finding the extrema of a height function. In engineering, finding the minimum cost of production can involve finding the extrema of a cost function.

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