# How to get the minimizer for the following function?

For a function $$f(x_1,x_2)$$ that $$f(x_1,x_2)=x_1^2x_2^2-4x_1^2x_2+4x_1^2+2x_1x_2^2+x_2^2-8x_1x_2+8x_1-4x_2$$ , I try to get the minimizer point. It seems that the minimal values for $$f(x_1,x_2)$$ is $$-4$$.

I try to use the second-order derivative test, the gradient is $$\nabla f(x_1,x_2)=[2x_1x_2^2-8x_1x_2+8x_1+2x_2^2-8x_2+8, 2x_1^2x_2-4x_1^2+4x_1x_2+2x_2-8x_1-4]^T\\=[(2x_1+2)(x_2-2)^2, (x_1+1)^2(2x_2-4)]^T$$ Solve $$\nabla f(x_1,x_2)=0$$ I got the stationary points are two lines that $$(-1, x_2)$$ and $$(x_1,2)$$. For $$(-1, x_2)$$, its Hessian is $$\nabla^2f(-1,x_2)=\begin{bmatrix} 2(x_2-2)^2&0 \\ 0 & 0 \end{bmatrix}.$$ For $$(x_1, 2)$$, its Hessian is $$\nabla^2f(x_1, 2)=\begin{bmatrix} 0 & 0 \\ 0 & 2(x_1+1)^2 \end{bmatrix}.$$ This method is invalid. How to transform $$f(x_1,x_2)$$ into a new form to make this method work?

• Can you show how you calculated the stationary points and the Hessian? I think that's where the error is. Oct 6, 2020 at 23:44
• @Andrei I added it. But I do not think I have an error in computation. Oct 6, 2020 at 23:53

With the form of the gradient that you have, it seems like your original function is $$f(x_1,x_2)=(x_1+1)^2(x_2-2)^2+C$$ Expand the right hand side, and you get $$C=-4$$. The term containing squares is non-negative (it is $$0$$ if $$x_1=-1$$ or $$x_2=2$$). So the minimum value will be indeed $$-4$$.