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The problem statement is find the maximum and minimum of $f(x,y) = \frac 3 2 x^2y^2+5x + 18y + 25$ under the constraints $ -5 \le 2x - y \le 5 $ and $ 0 \le x + 2y \le 2.5$. I graphed the restricted domain here.

For unconstrained extrema (aka local extrema) $f(x,y)$ has a critical point $\large (\frac{-3 ~10^{2/3}}{5},\frac{-~10^{2/3}}{6}) $ which turns out to be a saddle point using the second derivative test. I don't know if you would include that in the test points, but in any case this point is not in the domain.

Next, I tested the boundaries, and I tried to use Lagrange multipliers. But note that the domain is a parallelogram bounded by $4$ line segments. Thus we have $4$ restrictions. $$ 2x -y -5= 0 \\ 2x -y +5= 0 \\ x+2y = 0 \\ x + 2y - 2.5 = 0 $$ Then I defined $$ L:= f(x,y) - \lambda_1 (2x-y-5) - \lambda_2( 2x-y+5 ) - \lambda_3(x+2y) - \lambda_4(x+2y-2.5)$$ Solving $\nabla L = \bf 0$, I obtain no solutions. Note that I realize I can plug in the lines into the original function and use one variable calculus approach but I was wondering if I could use Lagrange multipliers here. According to Wolfram the maximum is 67 at the critical point $(-1.5, -2)$. My question, is the Lagrange multiplier approach doomed from the start? What is wrong with my hypotheses.

john
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    You can't have these conditions simultaneously. There are, for instance, no points satisfying $2x-y-5=0$ and $2x-y+5=0$. – lulu Apr 22 '22 at 17:22
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    You need to consider the 4 pieces of the boundary separately. – SolubleFish Apr 22 '22 at 17:23
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    The point $\left(-1.5,-2\right)$ does not belong to the given region. – José Carlos Santos Apr 22 '22 at 17:25
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    The maximum is at $(-1.5,2)$. Note the sign difference for $y$ – Andrei Apr 22 '22 at 18:33
  • @lulu I can only do Lagrange multipliers on one constraint equation at a time? Also how do you account for the fact that it is not a restriction of type $g(x,y) = c $ because it is line segment, specifically in the original problem $-5 \le 2x - y \le 5$ – john Apr 23 '22 at 03:55
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    Each boundary segment is indeed a constraint of that type, but the entire boundary is not. The multiple constraint version of multipliers which you use is applicable to simultaneous constraints, that is, a set of constraints all of which must be satisfied by the admissible points. So you could go boundary segment by boundary segment, or you could use general principles to look at the intersections (barring error, the point you want is the intersection of the second and fourth segments). – lulu Apr 23 '22 at 10:22
  • @lulu Sorry, I'm confused on this sentence you wrote "Each boundary segment is indeed a constraint of that type, but the entire boundary is not" . When you say full boundary do you mean the full line $2x-y = 5 $ versus a line segment $-5 \le 2x - y \le 5$. Thanks for explaining! – john Apr 24 '22 at 18:16
  • You have defined the boundary in piecewise fashion. The boundary as a whole is not defined by a nice, differentiable function. It has corners. As a result, you will need to consider the corners, regardless of what you do with the smooth parts of the boundary. – lulu Apr 24 '22 at 18:21

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