finding maxima and minima of a quadratic equation

Question

I'm dealing with a quadratic equation(with 2 independent variable) which looks like:
$$f(x,y) = 15.390x^2 - 0.001y^2 - 0.003xy - 69.985x + 0.263y + 58.740 $$ But I'm not being able to determine the maxima and minima for this equation.

So, a bit idea or reference regarding this issue would be nice.
Thank you.

Answer 1

You need to find the derivatives in terms of both x and y i.e. $$\frac {\partial f} {\partial x} = 30.78x - 0.003y -69.985$$ $$\frac {\partial f} {\partial y} =-0.002y-0.003x+0.263$$

If you want to find the minimal/maximal points, then you should put the above 2 derivatives equal to $0$ and solve simultaneously.

If you would like to go further and determine whether that point is a minimum or a maximum, then create the Hessian which is just the matrix of 2nd order partial derivatives, $$H(x,y)=\begin{pmatrix}\frac {\partial ^2f} {\partial x^2} & \frac {\partial ^2f} {\partial x \partial y} \\ \frac {\partial ^2f} {\partial y \partial x} & \frac {\partial ^2f} {\partial y^2}\\\end{pmatrix}$$ Which is $$H(x,y)=\begin{pmatrix} 30.78 & -0.003 \\ -0.003 & -0.002 \end{pmatrix}$$

Now since the Hessian is symmetric, then you have real eigenvalues. From here, you calculate the eigenvalues for this function. If both are positive then it is convex i.e. it is a minimum, if both are negative then it is concave i.e. it is a maximum.

Answer 2

First find the partial derivatives and put them equal to zero....

$$f_x(x,y) = 2(15.39)x -0.003y - 69.985 = 0$$ $$f_y(x,y) = -2(0.001)y -0.003x - 0.263 = 0$$

Solve these simultaneously to find the critical point $(x_c, y_c)$

Determine if critical point is maximum or minimum by using $$D(x_c,y_c) = f_{xx}(x_c,y_c)f_{yy}(x_c,y_c) - [f_{xy}(x_c,y_c)]^2$$

If $D > 0$ then we have Max/Min.

Distinguish between these as follows

• $f_{xx} < 0$ and $f_{yy} < 0$ at $(x_c,y_c)$ point is Max

• $f_{xx} > 0$ and $f_{yy} > 0$ at $(x_c,y_c)$ point is Min