Find the absolute maxima and minima of the function f (x, y) = (4x-x ^ 2) cos⁡ (y) in the rectangular plate 1≤x≤3, -π / 4≤y≤π / 4

Question

I need to find that problem for my math class

Answer 1

f(x,y)=(4x-x²)cos(y).

∂f/∂x=(4-2x)cos(y) and ∂f/∂y=-(4x-x²)sin(y).

So there is a maximum when 4-2x=0, x=2. The next derivative is negative so this confirms a maximum.

There is a maximum when -sin(y)=0 so y=0 in the given interval.

Therefore there is a maximum when x=2 and y=0, f(x,y)=4.

To find the minimum we need to look at the intervals. When x=1, 4x-x²=3 and when x=3, 4x-x²=3. We know there is a maximum at x=2.

When y=-π/4, cos(y)=√2/2=cos(-y). This is less than cos(0), so is a minimum for y.

We need a minimum for f, so we need x to be 1 or 3 and y to be ±π/4.

There are 4 absolute minima where f=3√2/2 for (x,y)=(1,-π/4), (1,π/4), (3,-π/4), (3,π/4).