Question about solving of the equation using derivatives.

Question

What I want to do is to find a certain x,y that minimizes a function f(x,y) which is given below.

$x, y \in R$, (R is real set)

$f(x,y) = x^{2}+y^{2}$

Yes, the answer is (x,y) = (0,0).

But, assume that I have no idea what the (x,y) is.

What I only know is there exist a minimum value of f(x,y).

So, I can use derivative to find the (x,y).

$\frac{\partial f(x,y)}{\partial x} = 2x + 2y\frac{\partial y}{\partial x}$

$\frac{\partial }{\partial y}\frac{\partial f(x,y)}{\partial x} = \frac{\partial (2x + 2y\frac{\partial y}{\partial x})}{\partial y}$

$\frac{\partial }{\partial y}\frac{\partial f(x,y)}{\partial x} = 2\frac{\partial x}{\partial y} + 2\frac{\partial y}{\partial x} + 2y\frac{\partial }{\partial y}\frac{\partial y}{\partial x}$

Because certain (x,y) makes f(x,y) = 0, it is obvious that ...

$\frac{\partial }{\partial y}\frac{\partial f(x,y)}{\partial x} = 0$

$2\frac{\partial x}{\partial y} + 2\frac{\partial y}{\partial x} + 2y\frac{\partial }{\partial y}\frac{\partial y}{\partial x} = 0$

From here, I have a question.

Using the last equation, how can I find the (x,y) that I tried to find is (0,0)?

Because of lack of my calculus, I have no idea what to do with this equation.

Thanks for reading.

-min soo kim-

Answer 1

The correct way is to find the critical points such that

• $f_x=2x=0$

• $f_y=2y=0$

and the only solution is $(x,y)=(0,0)$ and that's a minimum since $f(x,y)\ge 0$ and $f(x,y)=0$ if and only if $(x,y)=(0,0)$.

Answer 2

At critical points such as minima and maxima, the Jacobian matrix will be the zero matrix, i.e. $$Jf(x,y)=\pmatrix{\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}}=(0,0).$$ So you need only to solve $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$.