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i have this function $f (x, y) = - 2x^2 - 3y^2 + 4xy + 4x - 10y$. in the compact set D bounded by the lines $x = 0$, $y = 0$ and $x + y = - 6$. I have to find global maximum and minimum. My problem is that I can't understand what to do with these $x = 0$ and $y = 0$.

I found the global maximum and minimum with the equation $x + y= -6$. I found y = -6-x and substituted into the main function $f (x, -6-x ) = - 2x^2 - 3y^2 + 4xy + 4x - 10y$. The result is $-9x^2 - 36^x -48$. I found the derivative $-18y-36 = 0$ and got that the global extremums of the function are $f (-2, 4) = -136$ and $f (4, -2)= - 40$.

But I don't know how to use that $x = 0$ and $y = 0$. Can you give me some advice?

Crystal
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  • What you have calculated (I did not check your calculations) are not the extrema of $f$ (on $D$) but those of its restriction to one side of the triangle $D.$ This will be useful but is not sufficient. Apply this method. I.e. calculate similarly the extrema of the restriction to the 2 other sides, and the value of $f$ at critical points in the interior of the triangle. A critical point is a point where $\partial f/\partial x=\partial f/\partial y=0.$ – Anne Bauval Apr 13 '23 at 17:20
  • Please format your formulas. – Anne Bauval Apr 13 '23 at 17:23

1 Answers1

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Assuming $D = \{x\le 0 \cap y\le 0 \cap x + y + 6 \ge 0 \} $ the lagrangian reads:

$$ L(x,y,\lambda,s) = -2x^2 - 3 y^2 + 4 x y + 4 x - 10 y + \lambda_1(x + y + 6 -s_1^2)+\lambda_2(x+s_2^2)+\lambda_3(y+s_3^2) $$

The stationary points of this lagrangian are the solutions for

$$ \nabla L = 0 = \cases{ \lambda_1+\lambda_2-4 x+4 y+4\\ \lambda_1+\lambda_3+4 x-6 y-10\\ x+y+6-s_1^2\\ x+s_2^2\\ y+s_3^2\\ \lambda_1s_1\\ \lambda_2 s_2\\ \lambda_3s_3 } $$

Using the fact $\lambda_is_i = 0$ and after solving the $2^3$ possible linear systems, we arrive at

$$ \left[ \begin{array}{ccccccccc} f & x & y & \lambda_1 & \lambda_2 & \lambda_3 & s_1^2&s_2^2&s_3^2\\ -96 & -6 & 0 & -28 & 0 & 62 & 0 & 6 & 0 \\ -48 & 0 & -6 & -26 & 46 & 0 & 0 & 0 & 6 \\ 0 & 0 & 0 & 0 & -4 & 10 & 6 & 0 & 0 \\ \frac{25}{3} & 0 & -\frac{5}{3} & 0 & \frac{8}{3} & 0 & \frac{13}{3} & 0 & \frac{5}{3} \\ \frac{97}{9} & -\frac{23}{9} & -\frac{31}{9} & -\frac{4}{9} & 0 & 0 & 0 & \frac{23}{9} & \frac{31}{9} \\ 11 & -2 & -3 & 0 & 0 & 0 & 1 & 2 & 3 \\ \end{array} \right] $$

Here when $s_i = 0$ indicates that the corresponding restriction is actuating otherwise, the solution is interior to this restriction. Now the last step is the qualification of those points...

Cesareo
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  • Thank you for your answer, but the problem is that I haven't study this method yet. We have to resolve this task with derivatives. – Crystal Apr 14 '23 at 05:50