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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?

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

1 Answer 1

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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 $-4$.

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