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I am trying to solve a quadratic equation with 2 variables to obtain minima or maxima points.
The equation I have is:

$$f(x,y)=15.794x^2+0.0004y^2+0.015xy-75.325x-0.149y+96.954$$

But when I solve this for global extrema points, using 2 partial derivative equation(keeping one variable constant at a time), I'm getting single value for $x$ but 2 values for $y$.
i.e. $x=2.3$ and $y=149.75$ and $162.15$. I'm not being able to understand if it is possible or I am making a mistake.

A little help with this matter will be highly appreciated.
Thank you.

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$$\begin{cases}\frac{\partial f}{\partial x} = 31.588 x+0.015y-75.325 = 0\\ \frac{\partial f}{\partial y} = 0.0008y+0.015x-0.149 = 0\\ \end{cases}$$

The previous is a linear system, then it can have no solution, 1 solution, infinite solution. It is not possible to have 2 solution!

In this case you have only 1 solution:

$$\begin{cases} x = 2.3139\\y = 142.864\end{cases}$$

This is the only stationary point of your equation.

To establish if it is a maximum or a minimum, you need the hessian matrix:

$$H = \left[\begin{array}{cc}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial x\partial y} & \frac{\partial^2 f}{\partial y^2 } \end{array}\right] = \left[\begin{array}{cc}31.588 & 0.015 \\ 0.015 & 0.0008 \end{array}\right]$$

The hessian has two positive eigenvalues ($\lambda_1 \simeq 0.0008, \lambda_2 \simeq 31.588$), and then the stationary point is a local minimum. Furthermore, since $H$ is constant for every point, then this point is a global minimum.

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  • $\begingroup$ Yes, I probably made a mistake. I plugged the value of x in eq.2 i.e. df/dy. It was giving me a value ~162. $\endgroup$ – diffracteD Sep 18 '15 at 15:52
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Bivariate quadratic functions can't have more than one extremum. The partial derivatives are linear in the variables and form a 2x2 linear system, that has a single solution. [Except degenerate cases with no solution or an infinity of them, but we are not in such a case.]

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