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?
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Your solution is nice, but to do it with calculus set $\dfrac{\partial f}{\partial x}=8x-12=0$ and $\dfrac{\partial f}{\partial y}=18y-12=0$ to find $(x,y)=(\frac32,\frac23)$, and then evaluate $f(x,y)$.
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If a function attains a minimum at point $(x_0, y_0)$ then both partial derivatives at this point are zero. In this case, derivative with respect to $x$ is $$ f(x, y)=8 x -12 $$ and the derivative with respect to $y$ is $$ f(x, y)= 18y - 12 $$ This means that $0 = 8x - 12 \Rightarrow x = 3/2 $ and $0 = 18 y - 12 \Rightarrow y =2/3$, which coincides with your result.
Work out the partial derivatives of the function, namely $$ \frac{\partial f}{\partial x} = 8x - 12, \qquad \frac{\partial f}{\partial y} = 18y - 12 $$ Now you set both derivatives equal to zero to find the stationary points: $8x - 12 = 0$ and $18x - 12 = 0$ and you get a single point with co-ordinates $(3/2,2/3)$. It remains to prove that it is a minimum. This can be done with the Hessian matrix, i.e. by finding the second order partial derivatives: $$ \frac{\partial^2 f}{\partial x^2} = 8, \qquad \frac{\partial^2 f}{\partial x\partial y} = 0, \qquad \frac{\partial^2 f}{\partial y^2} = 18. $$ The hessian is therefore $$ \begin{pmatrix} 8 & 0 \\ 0 & 18 \end{pmatrix} $$ Since both eigenvalues are positive (the matrix is positive), the point is a minimum.
Finally, $$ f\left(\frac{3}{2},\frac{2}{3}\right) = 1. $$