# Cannot find Maxima using KKT method

I am learning KKT method and failing to find the maxima for following $f(x)$

$$f(x_1,x_2) = 7x_1^2 + 6x_1 + 5x_2^2$$ subject to $$x_1 + x_2 \leq 10$$ $$x_1-3x_2\leq 9$$ $$x_1,x_2 \geq 0$$

My Solution: Using Lagrange multipliers $\lambda_1, \lambda_2$ for $g_1, g_2$, respectively. I studied following 4 cases (all of which does not give the maxima)

Case I : $\lambda_1 =0,\lambda_2 =0$ $\to$ This yields $x_1 =-3/7, x_2 =0$. Since $x$ must be greater than or equal to Zero, this case does not hold for KKT.

Case II: $\lambda_1 \neq 0,\lambda_2 =0$, $(x_1,x_2) = (47/12,73/12$). But $\lambda_1 < 0$. Hence this case also does not give maxima

Case III: $\lambda_1 = 0,\lambda_2 \neq 0$ , $x_2 < 0$ and also not valid case

Case IV : $\lambda_1 \neq 0,\lambda_2 \neq 0$, ($x_1,x_2$) = ($21/4,19/4$), but $\lambda_2 < 0$. Hence it seems it is also not working.

Where am I going wrong? Any help will be much appreciated

• I know that this isn't what you asked for, but it may be worthwhile to note that a maximum of a continuous convex function over a polytope is attained at one of its vertices. Commented Apr 18, 2018 at 5:28
• In case IV you need $x_1-3x_2=9$, which your solution does not satisfy. It would help us if you provided the KKT conditions. Commented Apr 18, 2018 at 11:41

I learned that Lagrange Multipliers were to be used only for a 'path' on a function, like $f(x,y)=2x^2-3xy+4y^3$ restricted to $x+y=10$. So, in my ignorance, I went with the brute force method involving $f_x, f_y, f_{xx}, f_{yy}, and f_{xy}$. I also changed the variables (since $x_1$ and $x_2$ are confusing) to $x$ and $y$.

$f(x,y) = yx^2 + 6x + 5y^2$
Restrictions: $x+y≤10$, $x-3y≤9$, $x≥0, y≥0$ So, first you try to find the mini-max of the entire function. If they appear inside your bounds, you're done. Let's try it...

$f_x = 14x+6, f_y = 10y$
$f_{xx} = 14, f_{yy} = 10$
$f_{xy} = 0$
$D = f_{xx}f_{yy}-(f_{xy})^2 = 140$

Darn it. We have to do more work, by which I mean: 1. check boundaries, 2. check corners. I did the work for checking the boundaries, so in order for me to type less (and it really is effortless in this case to check the boundaries if you want to yourself), I'll just announce that the boundaries are as helpful as rotten tomatoes.
We have corners where boundaries intersect: $(0,10),(0,0),(9,0),(\frac{39}4,\frac14)$. Plug em in to check:
(0,10) --> 500
(0,0) --> 0
(9,0) --> 621
(39/4,1/4) --> 724.25

And there you have it! (0,0) is the min, (39/4,1/4) is the max.
Unless I goofed (which is highly possible). Hope this helps, even if it completely discards your preferred method!

• Although you solved the problem, Lagrange multipliers can definitely be used in higher dimensions, which is the KKT method. Commented Apr 18, 2018 at 11:41