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Question:

Given that

$W = W_A + W_B,$ where $$W_A = 3/4 \ln(x_a) + 1/4 \ln(y_a),$$ and $$W_B = 2/3 \ln(x_b) + 1/3 \ln(y_b).$$ How can we maximise $W$ with given constraints $$1-) \quad e_x^a + e_x^b = x_a + x_b,$$ $$2-) \quad e_y^a + e_y^b = y_a + y_b,$$ $$3-) \quad p_x x_a + p_y y_a = p_x e_x^a + p_y e_y^a,$$ $$4-) \quad p_X x_b + p_y y_b = p_x e_x^b + p_y e_y^b,$$ $$5-) \quad p_x > 0, p_y > 0,$$ where $(e_x^a, e_y^a) = (2,1)$ and $(e_x^b, e_y^b) = (1,2)$ ?

Attempt:

We have the unknowns (x_a, y_a, x_b, y_b, p_x, p_y), and we need to maximise $W$, so I tried to maximise each $W_i$ to maximise $W$. To do that, I have used the Lagrange's multiplier method, but after doing lots of algebra, I couldn't find any solution to this maximisation problem.

So, first of all, how can we solve this problem ? Secondly, is there any quick way of solving this kind of "complex" maximisation problems ?

Any help or hint is appreciated.

Further thoughts:

If we sum equation $3$ and $4$, using and $1$, we can derive the equation $2$, so we have $6$ unknowns and only 5 equation if we also consider the two equations coming from Lagrange's method, hence the system should not be solvable uniquely. However, this is a problem coming from economics, and I would expect it to have a solution that maximises this function $W$.

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  • $\begingroup$ did you appreciate my answer? $\endgroup$
    – LinAlg
    Jul 5 '18 at 20:46
  • $\begingroup$ @LinAlg Of course. I have probably forgot to accept it along the line, sorry for that. $\endgroup$
    – Our
    Jul 6 '18 at 4:31
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Let me write the problem is a more readable format: $$ \begin{align} \max \quad & 0.75 \log(x_1) + \frac{2}{3} \ln(x_2) + 0.25 \log(x_3) + \frac{1}{3} \ln(x_4) \\ \text{s.t.}\quad& x_1 + x_2 = 3 \\ & x_3 + x_4 = 3 \\ & p_x (x_1-2) + p_y (x_3-1) = 0 \\ & p_x (x_2-1) + p_y (x_4-2) = 0 \\ &x \in \mathbb{R}^4, \; p \in \mathbb{R}^2_{++} \end{align} $$ Let me then eliminate $x_2$ and $x_4$: $$ \begin{align} \max \quad & 0.75 \log(x_1) + \frac{2}{3} \ln(3-x_1) + 0.25 \log(x_3) + \frac{1}{3} \ln(3-x_3) \\ \text{s.t.}\quad& p_x (x_1-2) + p_y (x_3-1) = 0 \\ & p_x (2-x_1) + p_y (1-x_3) = 0 \\ &x \in \mathbb{R}^4, \; p \in \mathbb{R}^2_{++} \end{align} $$ The last two constraints are the same and reveal that you need either $x_1<2$ and $x_3>1$ or $x_1>2$ and $x_3<1$. The unconstrained optimum is at $x_1 = 27/17 \approx 1.59$ and $x_2 =9/7\approx 1.29$. This satisfies the constraint as well, so it is feasible and optimal.

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  • $\begingroup$ $p_x$ and $p_y$ has to have a positive value. I'm sorry that I did not mentioned that. $\endgroup$
    – Our
    Apr 27 '18 at 11:59

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