MHB Finding the maxima and minima of a function

[brunette15]

I have the following function:
f(x,y) = xye^(-x2-y2)

I am trying to find all the maximum and minimum points.

I have started off by finding the partial derivatives of the function and equation them to zero.

df/dx = ye^(-x2-y2)(1-2x²) = 0

df/dy = xe^(-x2-y2)(1-2y²) = 0

However, i am stuck for actually how to solve the coordinates. Do i solve for the x values in df/dx and y values df/dy? And how do i find the corresponding coordinate after that?

If someone could work me through from here I would really appreciate it.

Thanks in advance!

 

[I like Serena]

I have the following function:
f(x,y) = xye^(-x2-y2)

I am trying to find all the maximum and minimum points.

I have started off by finding the partial derivatives of the function and equation them to zero.

df/dx = ye^(-x2-y2)(1-2x²) = 0

df/dy = xe^(-x2-y2)(1-2y²) = 0

However, i am stuck for actually how to solve the coordinates. Do i solve for the x values in df/dx and y values df/dy? And how do i find the corresponding coordinate after that?

If someone could work me through from here I would really appreciate it.

Thanks in advance!

Hey brunette15! (Smile)

Both equations should be true at the same time. Solving the set of equations will give you the coordinates.
Since $e^{-x^2-y^2}$ can never be equal to zero, we can divide it out, yielding:
$$\begin{cases}
y(1-2x^2) = 0 \\
x(1-2y^2) = 0
\end{cases} \Rightarrow \begin{cases}
y = 0 \vee 1-2x^2 = 0 \\
x = 0 \vee 1-2y^2 = 0
\end{cases}
$$
If $y = 0$ then the first equation is satisfied. From the second equation it then follows that $x=0$.
So $(0,0)$ is a solution.
Can you find the other 4 solutions? (Wondering)
Which of those solutions will be maximums or minimums?

 

[brunette15]

Hey brunette15! (Smile)

Both equations should be true at the same time. Solving the set of equations will give you the coordinates.
Since $e^{-x^2=y^2}$ can never be equal to zero, we can divide it out, yielding:
$$\begin{cases}
y(1-2x^2) = 0 \\
x(1-2y^2) = 0
\end{cases} \Rightarrow \begin{cases}
y = 0 \vee 1-2x^2 = 0 \\
x = 0 \vee 1-2y^2 = 0
\end{cases}
$$
If $y = 0$ then the first equation is satisfied. From the second equation it then follows that $x=0$.
So $(0,0)$ is a solution.
Can you find the other 4 solutions? (Wondering)
Which of those solutions will be maximums or minimums?

I see! Thankyou!