Constrained Optimisation Problem: Confusion with Algebra/Contradiction

Question

I completed the following constrained optimisation problem:

Maximum and minimum values of $f(x,y) = x^2 + 2xy + y^2$ on the ellipse $g(x,y) = 2x^2 + y^2 - xy - 4x = 0$.

$\nabla f = \lambda \nabla g$

$\therefore (2x + 2y, 2x + 2y) = \lambda(4x - y - 4, 2y - x)$

We have a system of 3 equations in 3 unknowns:

$2x + 2y = \lambda(4x - y - 4)$,

$2x + 2y = \lambda(2y - x)$,

$2x^2 + y^2 - xy - 4x = 0$.

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$2x + 2y = \lambda(4x - y - 4)$

$\implies \lambda = \dfrac{2x + 2y}{4x - y - 4} \forall \ 4x - y - 4 \not= 0$

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$2x + 2y = \lambda(2y - x)$

$\implies \dfrac{2x + 2y}{2y - x} \forall \ 2y - x \not= 0$

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$\therefore \dfrac{2x + 2y}{4x - y - 4} = \dfrac{2x + 2y}{2y - x} \forall \ y \not= \dfrac{x}{2}, y \not= 4x - 4$ Remember this part: This is where my confusion lies.

But if we continue with our reasoning, we get the following (correct) points of potential maxima/minima:

$(x,y) = (0,0)$, $(x,y) = (1, -1)$, $(x,y) = (2,2)$, and $(x,y) = \left(\dfrac{2}{7}, \dfrac{-6}{-7}\right)$.

And, finally, using these points we find that $f_{max} = 16$ at $(x,y) = (2,2)$, and $f_{min} = 0$ at $(x,y) = (0,0)$ and $(x,y) = (1,-1)$.

But here's the problem: We had above that $y \not= \dfrac{x}{2}$; therefore, $(x,y) = (0,0)$ would get us $0 \not= \dfrac{0}{2} \implies 0 \not= 0$, which is obviously false. Also, as we just saw, $(0,0)$ is a solution for $f_{min}$!

The algebra here is confusing me, since we cannot have $y = \dfrac{x}{2}$, but $(0,0)$ is apparently the minima. Is this not a contradiction? Did I do something wrong? Or am I supposed to discard the solution $(0,0)$ and just have $f_{min} = 0$ at $(1, -1)$? What am I misunderstanding?

I would greatly appreciate it if people could please take the time to clarify this.

Answer 1

I think the confusion is this:

$y \neq \frac x 2$ is not a result of your calculations at any point (and therefore not to be understood as a restriction on the set of solutions), but an assumption you make to get from

$2x + 2y = \lambda(2y - x)$

to

$\lambda = \dfrac{2x + 2y}{2y - x} $

This means that you still have to consider the possibility that $2y-x = 0$, and solve the set of equations under this assumption - which eventually yields the $x=y=0$ solution.

An analogy: Suppose we have

$3x = x^2$

What you are doing is:

$\implies x=3 \,$ if $\, x\neq0$

Which is correct, but does not imply that $ x$ is not $0$!

Rather, we still need to check whether the other possibility, $x=0$, also leads to viable solutions (which here, it does.)

Answer 2

$(x+y)^2\geq0$ for all real variables.

The equality occurs for $x=y=0$ for example.

Thus, $0$ is a minimal value.

For $x=y=2$ we'll get a value $16$.

We'll prove that it's a maximal value.

Indeed, we need to prove that $$(x+y)^2\leq16$$ or $$16-(x+y)^2\geq0$$ or $$16-(x+y)^2+4(2x^2+y^2-xy-4x)\geq0$$ or $$3(x-y)^2+4(x-2)^2\geq0,$$ which is obvious.

Done!

Answer 3

You don't have to solve like that and exclude $x=2y$ or other limitations.

The system

$ \left\{ {\begin{array}{*{20}{l}} {2x + 2y = \lambda \left( {4x - y - 4} \right)} \\ {2x + 2y = \lambda \left( {2y - x} \right)} \\ {2{x^2} + {y^2} - xy - 4x = 0} \end{array}} \right. $

can be solved looking at the first two equations which say

$\lambda \left( {4x - y - 4} \right)= \lambda \left( {2y - x} \right) $

and then

$4x-y-4=2y-x\to y=\frac{1}{3} (5 x-4)$

substitute in the third

$2 x^2-\frac{1}{3} (5 x-4) x-4 x+\frac{1}{9} (5 x-4)^2=0$

expand and simplify

$7 x^2-16 x+4=0\to x_1=\dfrac{2}{7};\;x_2=2$

$y_1=-\dfrac{6}{7};\;y_2=2$

Plug these values in the first equation to get

$4+4=\lambda(8-2-4)\to \lambda=4$

Substitute in the second

$2x+2y=4(2y-x) \to 6(x-y)=0 \to x=y$

which can be plugged in the third and give

$2x^2+x^2-x^2-4x^2=0\to x_3=0;\;y_3=0$

this last $(0,0)$ is the minimum for the given function, while according to my calculations $(2,2)$ gives the maximum

Answer 4

First note that the feasible set is compact, so the $\max,\min$ exist. Since $Dg((x,y)) \neq 0$ for all $Ix,y)$ satisfying the constraints, we know that the Lagrange conditions apply (in particular, the solution is not determined completely by $g$). Hence all solutions (and some more) can be determined by solving the Lagrange equations.

Second, note that you don't need to solve for $\lambda$.

If $\lambda = 0$ then you have $x=-y$ and substituting this into the constraint gives $y \in \{ -1,0 \} $ which correspond to a value of $0$.

If $\lambda \neq 0$ then you have $5x-3y-4=0$, from which you can solve for $x={ 1\over 5}(3x+4)$, and substituting this into the constraint gives $y \in \{ -{6 \over 7},2 \} $ which correspond to values of $({4 \over 7})^2, 4^2$.

Hence the $\min$ is $0$ and the $\max$ is $4^2$.