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Find the stationary points of w = $ −3x^2 − 4xy − y^2− 12y + 16x $ which reside at 1st quadrant. I did this problem with traditional method : like find $f_x$ = 0 $f_y$ = 0 and checking $f_{xx} f_{yy} - f_{xy}^2 $ which I got negative. I got the critical point as = <-20,34>. Now how can I get a critical point on first quadrant? They've given the answer as <8/3,0>.

user187604
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  • If you found the only critical point, and it's not in the first quadrant, then your job is easy: There are no critical points in the first quadrant. – quasi Jun 28 '18 at 02:45
  • @quasi sorry I had to add the answer. Please check again – user187604 Jun 28 '18 at 02:49
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    Books have typos. Presumably the problem itself has a typo (e.g., the coefficients of $w$ are different). – quasi Jun 28 '18 at 02:51

1 Answers1

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According to the needed results the question should be asked as:

What the stationary points for $f(x,y)=-3 x^2 - 4 x y - y^2 - 12 y + 16x$
subjected to $x \ge 0$ and $y \ge 0$ ?

In this case with the help of some slack variables $\epsilon_1,\epsilon_2$ to cope with the inequalities, the Lagrangian could be represented as

$$ L(x,y,\lambda,\epsilon) = f(x,y) + \lambda_1(x-\epsilon_1^2)+\lambda_2(y-\epsilon_2^2) $$

given the stationary conditions

$$ \nabla L = \left\{ \begin{array}{rcl} \lambda_1-6 x-4 y+16=0 \\ \lambda_2-4 x-2 y-12=0 \\ x-\epsilon_1^2=0 \\ y-\epsilon_2^2=0 \\ -2 \epsilon_1 \lambda_1=0 \\ -2 \epsilon_2 \lambda_2=0 \\ \end{array} \right. $$

Solving those conditions we obtain the points

$$ (0,0) \;\;\; \mbox{and}\;\;\; (\frac 83,0) $$

Attached a plot showing the stationary points

enter image description here

NOTE

The simple stationary points can be classified as: relative minimum, relative maximum and saddle points.

This qualification is make with the contribution of the function Hessian, considering the active restrictions, at each point. Analyzing the solutions we can say if the restrictions are active. For instance in our case

$$ \left[ \begin{array}{cccccc} x & y & \lambda_1 & \lambda_2 & \epsilon_1 & \epsilon_2 \\ 0 & 0 & -16 & 12 & 0 & 0 \\ \frac{8}{3} & 0 & 0 & \frac{68}{3} & \pm2 \sqrt{\frac{2}{3}} & 0 \\ \end{array} \right] $$

The first point $(0,0)$ activates the two restrictions as can be observed from $\epsilon_1=\epsilon_2 = 0$. The second point has only the $y = 0$ restriction active $\epsilon_2 = 0$

Cesareo
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