Qno: (a) obtain a solution $(x^{*}, y^{*})$ by graphical method.
(b) Formulate the Lagrangean.
(c) Obtain all the K-T necessary conditions
(d) Using values for $(x^{*}, y^{*})$, obtain the values for the Lagrange multipliers in the problem.
(e) obtain the dual problem.
Solution: (a) I have done it.
(b) I have done it.
(c) I have done it.
(d) I got stuck in here.
The K-T means Kuhn Tucker necessary conditions were
$$L_x=2+4 \times \lambda, x \geq 0, x \times L_x=0 \implies L_x=0 $$
$$L_y=3+ \lambda, y \geq 0, y \times L_y=0 \implies L_y=0$$
$$L_{\lambda}=-5+4x+y, \lambda \geq 0, \lambda \times L_{\lambda}=0 \implies L_{\lambda}=0$$
From these, I have
$$2+4 \times \lambda = 0$$
$$3+\lambda=0$$
$$-5+4x+y=0$$
Case1: Assume $\lambda ≠0$
Case2: Assume $\lambda=0$
I don't know how to tackle these two cases in order to find the values of $x, y, \lambda$ to maximize the utility function. I have tried my best but unable to get the true answer.
Also help me to get dual problem or at least give me hint for that part. Thanks
The function $f(x,y)=2x+3y$ is linear non-constant, therefore it has no critical points. This is also justified by the system \begin{cases} f_x=2\ne 0\\ f_y=3\ne 0 \end{cases} Similarly, the function $L(x,y,\lambda)=2x+3y+\lambda(4x+y-5)$ has no critical point (this follows from your result).
However, $f$ is continuous and the considered domain is compact, therefore $f(x,y)$ achieves maximal and minimal value on the domain. The candidates for "where" it happens are the intersection points of the lines $x=0,\; y=0,\; 4x+y-5=0.$
Find the three points, put their coordinates in $f(x,y).$ The greatest is the maximum, the lowest ist minimum of $f(x,y)$ on the domain.