Gradient Of $f(x,y)=2x^3+xy^2+5x^2+y^2$

Question

If I am told to take the gradient of the function:$$f(x,y)=2x^3+xy^2+5x^2+y^2$$

So it is $F_{x}+F_{y}=0$?

Or in this case:

$$\{6x^2+y^2+10x\}+[2xy+2y]=0$$

And to find the critical points we take:

$F_{x}=0$ and $F_{y}=0$

Or in this case:

$6x^2+y^2+10x=0$ and $2xy+2y=0$

And the critical point are: $(1,\pm 2),(-\frac{5}{3},0),(0,0)$

Answer 1

So it is $F_{x}+F_{y}=0$?

No, the gradient of a function $f:ℝ^n→ℝ$ is defined as: $$∇f(x_1,…,x_n)=\pmatrix{\frac{∂}{∂_{x_1}}f \\\vdots \\ \frac{∂}{∂_{x_n}}f}.$$

With $\frac{∂}{∂_{x_i}}$ denoting the partial derivative to $x_i$.

So in your case, for $f:ℝ^2→ℝ$, it is $$∇f(x,y) = \pmatrix{\frac{∂}{∂_x}f(x,y) \\ \frac{∂}{∂_y}f(x,y)}.$$

So now you just need to calculate the derivatives. \begin{align*} \frac{∂}{∂_x}f(x,y) &= \frac{∂}{∂_x}[2x^3+xy^2+5x^2+y^2] \\ &= 6x^2+y^2+10x \\ \\ \frac{∂}{∂_y}f(x,y) &= 2xy+2y \end{align*}

edit: Since you want to calculate maxima and minima you are correct, that the necessary condition (just like in 1D) is: $$∇f=0.$$ And that is short for $$\frac{∂}{∂_{x_i}} f = 0, \quad i=1,…,n$$ since the gradient is simply a vector.

But you also have to check that the second derivative is "$>0$" or "$<0$", just like in 1D. And now that we are talking about multivariate functions, the second derivative is a matrix, the so-called Hessian, and "$\lessgtr 0$" is called definiteness.

Wikipedia illustrates that nice for the special 2D case and for arbitrary dimension.

Answer 2

I do not understand why you consider the equation $f_x+f_y=0$.

But your critical points are correct.

Edit: it should read $(-1,\pm 2)$.