f(x,y) = (y^2 -3y) sinx in the open rectangle 0 < x < 2pi and 1 < y < 5.
i have found (pi, 3), (pi/2, 3/2) and (3pi/2, 3/2) which are saddle point, local min and local max respectively.
Am i missing any other points? Also is my answer correct?
How do i find the global maximum value for this function over the open rectangle 0 <= x <= 2pi and 1 <= y <= 5?
With $f(x, y)= (y^2- 3y)sin(x)$ then $f_x= (y^2- 3y)cos(x)$ and $f_y= (2y- 3)sin(x)$. At a stationary point both of those must be 0. sin(x)= 0 for x a multiple of $\pi$, $n= k\pi$ for k an integer. Since here you are requiring that $0< x< \pi$, that means $x= \pi$. In that case $cos(x)= -1$ so we must have $y^2- 3y= y(y- 3)= 0$, y= 0 or y= 3. That is, two stationary points are $(\pi, 0)$ and $(\pi, 3)$. cos(x)= 0 for $x= \pi/2$ and $x= 3\pi/2$. Since sine is not 0 for those values, we must have $2y- 3= 0$ so $y= \frac{3}{2}$. Two other points are $\left(\pi/2, \frac{3}{2}\right)$ and $\left(3\pi/2, \frac{3}{2}\right)$.
You appear to have missed $(\pi, 0)$.
As for the last part of the question, finding the global maximum in that region, first evaluate the given function at each of the stationary points inside that region. However, it might happen that a global maximum occurs on the boundary. The boundary consists of the four line segments- x= 0 for y between 1 and 5, $x= 2\pi$ for y between 1 and 5, $y= 1$ for x between 0 and $2\pi$, y= 5 for x between 0 and $2\pi$. On the first of those, the function to be maximized is $f(0, y)= (y^2- 3y)sin(0)= 0$ for all y. On the second, the function to be maximized is $f(2\pi, y)= 0$ for all y. On the third, the function to be maximized is $f(x, 1)= -2sin(x)$. On the last, the function to be maximized is $f(x, 5)= 10sin(x)$. Find the points that maximize those by setting the derivative equal to 0 and determine the value of f at each such point.
Finally, the boundary also consist of the boundaries of those lines- that is, the four vertices of that rectangle, $(0, 1)$, $(2\pi, 1)$, $(2\pi, 5)$, and $(0, 5)$. Find the value of f at those four points. Now select the largest of all those values as the global maximum.