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Suppose I wish to maximize the function $f(x,y)$ subject to the equality constraint $g(x,y)=c$ as well as the non-negativity constraints $x\geq0$, $y\geq0$.

If I first solve it ignoring the non-negativity constraints and find that $x<0$ and $y>0$ is it valid to say that the optimum value of $x=0$?

The reason I ask is that it seems that a negative value for $x$ suggests that you would want the lowest value of $x$ possible. However, in the textbooks I have referred to the only way they suggest to deal with such problems is to use the Karush-Kuhn-Tucker conditions so I wanted to know whether my reasoning is valid or flawed.

K.K.McDonald
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hk39
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1 Answers1

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This is a broad question, and depends a lot on the kind of function that you want to optimize.

For instance consider the following cases:

  1. $f(x,y)=1-2x$ with $g(x,y)=x-y=0$:

This problem has no maximum if the non-negativity constraint is not considered, and when $x\geq 0$ the maximum is attained in $x=0$.

  1. $f(x,y)=(x-1/4)^2$ with $x^2+y^2=1$:

Again no maximum if the non-negativity constraint is not considered, but when $x\geq 0$ the maximum is attained in $x=1$.

I agree that there are several problems for which Kuhn-Tucker conditions are not necessary, and sometimes it is easier to solve these problems by analyzing increasing or decreasing of the functions, or restricting to specific values of $x$ and $y$, or even using a substitution of $y$ in terms of $x$ from the equation $g(x,y)=c$.

However, the importance of K-T conditions is that they provide a systematic way to deal with optimization problems when some restrictions of inequalities are given.

In my opinion, the ideal is that a student knows the possible short-cuts that might appear on the different problems, together with the general method (Kuhn-Tucker) in case no "bright idea" come to the mind.

Darío G
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  • Thank you for your answer especially clarifying the importance of the K-T conditions. I have two further questions. Firstly, in your second example I am struggling to see why you cannot have $(-1,0)$ as a maximum without the non-negativity constraint. Secondly, now I realize that my question was broad, does my reasoning described above work for when the level curves of the function are strictly convex? – hk39 Apr 26 '16 at 17:59
  • You are right! Actually the maximums of the problem (2) were in $(1,0)$ and $(-1,0)$. (When I constructed the example I didn't thought much about the restriction). Now, regarding your question, consider the function $f(x,y)=\dfrac{1}{4}x^2-y$ with the restriction $g(x,y)=x^3-x+y=1$ (the graph of the restriction is the graph of the polynomial $y=-x^3+x=-x(x+1)(x-1)$ but moving up one unit!). You will find that without the restrictions there are no maximums (the cubic will always intersect higher parabolas on the negative side of $x$'s) but the maximum is neither in $x=0$ nor in $y=0$. – Darío G Apr 26 '16 at 18:31