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Consider a utility function on the form $u(q_{1},q_{2},q_{3}) = min\{\alpha ln(q_{1}) + (1 - \alpha) ln(q_{2}), ln(q_{3})\}$

I know that optimal behaviour requires $\alpha ln (q_{1}) + (1 - \alpha) ln(q_{2}) = ln(q_{3})$

I tried substituting this into the budget constraint (after raising both sides to the power of $e$)

$p_{1}q_{1} + p_{2}q_{2} = y$

But I am unsure how to proceed in finding the Marshallian demand, since I have two variables $(q_{1}, q_{2})$ and only one equation. I have tried different transformations but my main problem is that it feels like I'm missing an equation.

Any help is greatly appreciated.

EDIT: I just re-read the question and I misspecified the budget constraint! The consumer only spends on $q_{1}$ and $q_{2}$.

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  • $\begingroup$ So since $\ln(0)$ is not defined but $\lim_{q_3 \to 0} \ln(q_3) = -\infty$ the task is to maximize w.r.t. $q_1,q_2$ the formula $$ \min\{\alpha \ln(q_{1}) + (1 - \alpha) \ln(q_{2}), -\infty\} = -\infty? $$ $\endgroup$ – Giskard Oct 28 '18 at 11:36
  • $\begingroup$ I believe we just have to assume some exogenous non-zero $q_{3}$. In the question I am given that $q_{1}$ and $q_{2}$ represent tea and coffee and $q_{3}$ represents sugar. I tried a transformation such that the UMP became min$\{\phi_{1} + \phi_{2}, \phi_{3}\}$ and thus at optimum $\phi_{1} + \phi_{2} = \phi_{3}$. Wouldn't this mean that the agent chooses the cheaper one of $\phi_{1}$ and $\phi_{2}$? $\endgroup$ – Gensys Oct 28 '18 at 11:45
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So, I have not really worked out the maths behind solving this. I prefer to take a shortcut and just compute it numerically. Here is what I have.

These are the contour surfaces and the budget constraint. Notice how the solution will be at the "kinked" curve in the middle of each surface. enter image description here

The levels 1 and 2 are only for example. The third level, u = 3.92762 is related to the next part of my answer. Also, BC is the budget curve with income = 300. I have chosen $\alpha = 0.5$ and $P_1 = 2, P_2=3, P_3=1$.

Now, using the parameters mentioned in the last paragraph, the maximum is found to be at $q_1 = 65.8896, q_2 = 39.1448, q_3 = 50.7863$ and the maximized utility is $u = 3.92762$. I assumed that all goods are strictly consumed at positive amounts.

I am attaching my Mathematica code below, so that you can replicate and modify these results as you want

u = Min[a*Log[q1] + ((1 - a)*Log[q2]), Log[q3]]

BC = p1*q1 + p2*q2 + p3*q3
a = 0.5; p1 = 2; p2 = 3; p3 = 1;

ContourPlot3D[{u == 1, u == 2, u == 3.92762, BC == 300}, {q1, 0,
100}, {q2, 0, 100} , {q3, 0, 100}, AxesLabel -> Automatic,
ContourStyle -> {Opacity[0.5], Opacity[0.5], Opacity[0.5], Opacity [1]}, Mesh -> None, PlotLegends -> "Expressions"]

NMaximize[{u, BC <= 300, q1 > 0, q2 > 0, q3 > 0}, {q1, q2, q3}, Reals]

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