My Math Forum Inequality constraint NLP problem

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 February 15th, 2015, 04:28 PM #1 Newbie   Joined: Feb 2015 From: London Posts: 1 Thanks: 0 Inequality constraint NLP problem Hello, Regarding subject, I have a difficulty with following NLP problem. Max f(x) = X1 / (X2+1) s.t. X1 - X2 <= 2 X1>=0 X2>=0 Actually, I could get the answer with simple graphical way that optima is 2 at X = [2;0] However, I could not get the solution with Lagrange multiplier method. After partial differentiation of L(X1, X2, lamda), I could get 3 equations and 3 variables, but two equations are parallel so it does not give me a solution. Could anyone solve this problem? And also, could I get Kuhn-Tucker conditions (KKT) of this solution? In addition, can't we say this problem is convex? Thank you. Last edited by skipjack; February 16th, 2015 at 12:26 AM.
 February 15th, 2015, 07:27 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,559 Thanks: 2561 Math Focus: Mainly analysis and algebra To maximise a quotient, you minimise the denominator and maximise the numerator. Thus we want $x_2 = 0$ and then $x_1 - x_2 \le 2 \implies x_1 = 2$. Of course, you might suggest that we could add $a \gt 0$ to each of these values and still satisfy the inequalities. This would work if$${x_1 + a \over x_2 + 1 + a} \gt {x_1 \over x_2 + 1}$$ Since we have that $x_2 = x_1 - 2$ we then have \begin{aligned}{x_1 + a \over x_1 - 1 + a} &\gt {x_1 \over x_1 - 1} \\ (x_1 - 1)(x_1 + a) &\gt x_1(x_1 - 1 + a) \\ x_1^2 + (a-1)x_1 -a &\gt x_1^2 + (a - 1)x_1 \\ a \lt 0\end{aligned} But we have already stated the $a \gt 0$, and indeed $a \lt 0$ violates the minimum value of $x_2$. So $(x_1, x_2) = (2,0)$ is the only solution.
 February 16th, 2015, 12:51 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,104 Thanks: 1907 As 0 $\leqslant$ X1 $\leqslant$ 2 + X2, one must take X1 = 2 + X2 to maximize X1/(X2 + 1). That gives f(x) = (2 + X2)/(X2 + 1) = 2 - X2/(X2 + 1). As X2 $\geqslant$ 0, the maximum value of f(x) is 2, achieved when X2 = 0 (and X1 = 2). It would have been better to write f(X1, X2) instead of f(x).

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