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Consider the following Constrained Optimization Problem:

$$max: -2(x+y)^2 +5x-y$$ Subject to:
$$x≤B$$ $$-x+2y≤3$$ $$y≥0$$
with$$ beta = 0$$
I have found the solution that solves the problem:$$x=beta$$ $$y=0$$ $$lambda_1 = 5-4beta$$ $$lambda_2 =0$$ $$lambda_3=4beta +1$$I found several other sets of solutions by not making any prior assumptions about the constraints, i.e., whether or not they were binding or slack.
I can see that the original function is decreasing in positive values of $y$, would it be reasonable to assume $y=0$ and then solve the problem? Or is working the entire problem out the only way to be sure that the solution is correct?

Solving with the help of Lagrange Multipliers and introducing some slack variables $(epsilon_i)$ to reduce the inequalities to equalities we have with $f(x,y) = -2(x+y)^2+5x-y$

$$
L(x,y,lambda,epsilon) = f(x,y)+lambda_1(x-beta+epsilon_1^2)+lambda_2(-x+2y-3-epsilon_2^2)+lambda_3(y-epsilon_3^2)
$$

The stationary conditions are

$$
L_x = 5+lambda_1-lambda_2-4(x+y)=0\
L_y = -1+2lambda_2+lambda_3-4(x+y) = 0\
L_{lambda_1} = -beta+x+epsilon_1^2 = 0\
L_{lambda_2} = 2y-3-x-epsilon_2^2 = 0\
L_{lambda_3} = y-epsilon_3^2 = 0\
L_{epsilon_1} = lambda_1epsilon_1 = 0\
L_{epsilon_2} = lambda_2epsilon_2 = 0\
L_{epsilon_3} = lambda_3epsilon_3 = 0
$$

Solving those equations we have the possible stationary points as well as the $f(x,y)$ value according to the table

$$
begin{array}{ccccccccc}
x & y & lambda_1 & lambda_2 &lambda_3 &epsilon_1 &epsilon_2 &epsilon_3 & f\
beta & frac{beta +3}{2} & 9 beta +frac{9}{2} & 3 beta +frac{7}{2} & 0 & 0 & 0 & -frac{sqrt{beta +3}}{sqrt{2}} & -frac{3}{2} (3 beta
(beta +1)+4) \
beta & frac{beta +3}{2} & 9 beta +frac{9}{2} & 3 beta +frac{7}{2} & 0 & 0 & 0 & frac{sqrt{beta +3}}{sqrt{2}} & -frac{3}{2} (3 beta
(beta +1)+4) \
beta & 0 & 4 beta -5 & 0 & 4 beta +1 & 0 & -sqrt{-beta -3} & 0 & (5-2 beta ) beta \
beta & 0 & 4 beta -5 & 0 & 4 beta +1 & 0 & sqrt{-beta -3} & 0 & (5-2 beta ) beta \
beta & -beta -frac{1}{4} & -6 & 0 & 0 & 0 & -sqrt{-3 beta -frac{7}{2}} & -frac{1}{2} sqrt{-4 beta -1} & 6 beta +frac{1}{8} \
beta & -beta -frac{1}{4} & -6 & 0 & 0 & 0 & -sqrt{-3 beta -frac{7}{2}} & frac{1}{2} sqrt{-4 beta -1} & 6 beta +frac{1}{8} \
beta & -beta -frac{1}{4} & -6 & 0 & 0 & 0 & sqrt{-3 beta -frac{7}{2}} & -frac{1}{2} sqrt{-4 beta -1} & 6 beta +frac{1}{8} \
beta & -beta -frac{1}{4} & -6 & 0 & 0 & 0 & sqrt{-3 beta -frac{7}{2}} & frac{1}{2} sqrt{-4 beta -1} & 6 beta +frac{1}{8} \
-3 & 0 & 0 & 17 & -45 & -sqrt{beta +3} & 0 & 0 & -33 \
-3 & 0 & 0 & 17 & -45 & sqrt{beta +3} & 0 & 0 & -33 \
-frac{1}{2} & frac{5}{4} & 0 & 2 & 0 & -sqrt{beta +frac{1}{2}} & 0 & -frac{sqrt{5}}{2} & -frac{39}{8} \
-frac{1}{2} & frac{5}{4} & 0 & 2 & 0 & -sqrt{beta +frac{1}{2}} & 0 & frac{sqrt{5}}{2} & -frac{39}{8} \
-frac{1}{2} & frac{5}{4} & 0 & 2 & 0 & sqrt{beta +frac{1}{2}} & 0 & -frac{sqrt{5}}{2} & -frac{39}{8} \
-frac{1}{2} & frac{5}{4} & 0 & 2 & 0 & sqrt{beta +frac{1}{2}} & 0 & frac{sqrt{5}}{2} & -frac{39}{8} \
end{array}
$$

NOTES

1-Some values are duplicate due to the adoption of $epsilon_i$ squared.

2-Solutions with at least one $epsilon_i = 0$ are located at the boundary. As we can observe, all solutions are at the feasible region boundary.

3-Once the $beta$ value is fixed, the maximum and minimum can be chosen.