I was studying for a quiz when a wild question appears. It goes like this:
How can I determine the stationary points of the function $f(x,y) = 2x^2 + 2xy – y^3$?
My Work
Stationary points are points where the slope of the function becomes zero. With that in mind, I got to differentiate the above function and
set its derivative to zero to get its stationary point.
Now implicitly differentiating the above equation:
$$z = 2x^2 + 2xy – y^3$$
$$dz = 4x + 2d(xy) – d(y^3)$$
$$dz = 4x + 2(xy' + y') – 3y^2y'$$
$$dz = 4x + 2xy' + 2y' – 3y^2 y'$$
$$dz = 4x + y'(2x + 2 – 3y^2)$$
Finally…..
$$ y' = \frac{dz – 4x}{2x + 2 – 3y^2} $$
With the form I got above, I see two derivatives, the $dz$ and $y'$. Setting the $dz$ and $y'$ to zero, I will get
$x = 0$. Something's wrong.
How can I properly get the stationary points of the function $f(x,y) = 2x^2 + 2xy – y^3$?
Best Answer
The stationary points of a real valued function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ are those points $x_0$ where the derivative in every direction equals zero, or equivalently, the gradient is zero.
So, you have the gradient of $f$ is
$$\nabla f = (4x+y)\mathbf{i} + (2x-3y^2)\mathbf{j} = \mathbf{0}.$$