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?
How to find stationary points for this function:
f = [6x (cosy)-2(x^2)(cosy)+(x^2)(sin2y)]
2 answers
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)
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)
1 answer