How to find stationary points for this function:
f = [6x (cosy)-2(x^2)(cosy)+(x^2)(sin2y)]
f(x,y) = 6x (cosy) - 2(x^2)(cosy) + (x^2)(sin2y)
= (6x - 2x^2) cosy + x^2 sin2y
∂f/∂x = (6 - 4x)cosy + 2x sin2y
∂f/∂y = (2x^2 - 6x) siny + 2x^2 cos2y
stationary points are where ∂f/∂x = 0 and ∂f/∂y = 0
so, what do you get when trying to satisfy that condition?
I have almost solved this problem now. I'm still wondering what is meant by "single stationary point within range"?
In actual problem, it was asked to find single stationary point in range 0<x<6 and the corresponding value of y.
But after solving equations, I got six stationary points and the corresponding values at them, which are:
f(0,π/2) = 0 (saddle point)
f(0,π/2) = 0 (saddle point)
f(2,π/2)= 0 (saddle point)
f(2,5π/6) = -3 (local minimum)
f(2,π/6) = 3 (local maximum)
f(6,π/2) = 0 (saddle point)
To find the stationary points of a function, we need to locate the points where the slope (or gradient) of the function is equal to zero. In order to do that, we will partially differentiate the function with respect to each variable and then set the resulting equations to zero.
Here's a step-by-step guide on how to find the stationary points for the given function:
Step 1: Partially differentiate the function with respect to x.
To do this, treat y as a constant and differentiate each term one by one with respect to x:
∂f/∂x = 6(cos(y)) - 4x(cos(y)) + 2x(sin(2y))
Step 2: Partially differentiate the function with respect to y.
To do this, treat x as a constant and differentiate each term one by one with respect to y:
∂f/∂y = -6x(sin(y)) - 2x^2(sin(y)) + 2x^2(cos(2y))
Step 3: Set both partial derivatives to zero and solve for x and y.
Set ∂f/∂x = 0 and ∂f/∂y = 0:
6(cos(y)) - 4x(cos(y)) + 2x(sin(2y)) = 0 --- (Equation A)
-6x(sin(y)) - 2x^2(sin(y)) + 2x^2(cos(2y)) = 0 --- (Equation B)
To solve these equations, you can either use algebraic techniques or numerical methods, such as graphing or employing a computer software.
Please note that the given function is somewhat complex and finding explicit solutions may not be straightforward or even possible. In such cases, numerical approximations or graphical methods might be necessary.
Once you find the values of x and y that satisfy both equations (A and B), those points will correspond to the stationary points of the function f.