Minimum point for $x^2 +2xy + 4y^2 + 6$ is to be found.
$\frac{\partial}{\partial x}$ and equated to $0$ to find $x=-y$
$\frac{\partial}{\partial y}$ and equated to $0$ to find $x = -4y$
I'm confused as to which is the correct solution
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1$(x,y)=(0,0)$ is the point where the minimum value (of $6$) is attained. – Kavi Rama Murthy Aug 25 '20 at 05:34
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$$x^2 +2xy + 4y^2 + 6=(x+y)^2+3y^2+6.$$
The first two terms are nonnegative and can simultaneously achieve zero. This is consistent with your method of taking partial derivatives. Solving your system also leads to $x=y=0$.
mjw
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