<s>[INST] Looking for an alternative approach to Find the minimum value of the function $f(x, y)=4 x^{2}+9 y^{2}-12 x-12 y+14$ is The minimum value of the function $f(x, y)=4 x^{2}+9 y^{2}-12 x-12 y+14$ is
My work
$$
\begin{aligned}
\text f(x, y) &=4 x^{2}+9 y^{2}-12 x-12 y+14 \\
&=\left(4 x^{2}-12 x+9\right)+\left(9 y^{2}-12 y+4\right)+1 \\
&=(2 x-3)^{2}+(3 y-2)^{2}+1 \geq 1
\end{aligned}
$$
So, minimum value of $\mathrm{f}(\mathrm{x}, \mathrm{y})$ is 1
My question is How can it be done by calculus?
 [/INST] $$f(x, y)=4 x^{2}+9 y^{2}-12 x-12 y+14$$
$$\frac{\partial f(x,y)}{\partial x}=8x-12$$
$$\frac{\partial f(x,y)}{\partial y}=18y-12$$
Set the partial derivatives equal to $0$; so $x=\frac{3}{2}$ and $y=\frac{2}{3}$
$$f\left(\frac{3}{2},\frac{2}{3}\right)=1$$
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