# 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?

• Your work is fine. You can also use derivatives to find the minimum value. But your approach is trickier and it is done by calculus. Jul 13, 2020 at 14:18
• callculus speaking to Integral calculus about calculus. Nice combination ! But, if I may ask, what is calculus ? Jul 13, 2020 at 14:22
• @Integralcalculus. Thank you ! Jul 13, 2020 at 14:36
• I never would have thought of this approach. Nice work Jul 13, 2020 at 14:39
• @K.defaoite thanks!
– user801681
Jul 13, 2020 at 14:40

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, $$f\left(\frac{3}{2},\frac{2}{3}\right) = 1.$$

• The sign of the det isn't useful. But rather, you should say that the Hessian is positive definite to conclud. e Jul 13, 2020 at 14:32

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.

• I think you made a mistake, changing $18y-12$ to $18y-9$ Jul 13, 2020 at 14:23

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 $$f(x,y)$$.

$$f(x, y)=4 x^{2}+9 y^{2}-12 x-12 y+14$$ $$\frac{\partial f(x,y)}{\partial x}=8x-12$$ $$\frac{\partial f(x,y)}{\partial y}=18y-12$$ Set the partial derivatives equal to $$0$$; so $$x=\frac{3}{2}$$ and $$y=\frac{2}{3}$$ $$f\left(\frac{3}{2},\frac{2}{3}\right)=1$$