Optimization problem - Graphically

Question

How can we draw the cobb-douglas optimization problem with constraint and solve it graphically?

We have the function $U(x,y)=x^{0.4}y^{0.3}$.

Do we have to draw the constraint and some level curves of $U(x,y)$ ?

Answer 1

Do we have to draw the constraint and some level curves of U(x,y) ?

That is right. I would recommend to use a function plotter since a single utility function only is hard enough to draw. You solve the utility function for y:

$$y=\left( \frac{\overline U}{x^{0.4}} \right)^{10/3}$$

For different levels of $\overline U$ the function plotter can draw different curves.

For different values of $\overline U$ the array of curve looks like below

The different levels of $\overline U$ are

$4.070905-3; \quad 4.070905-2.5; \quad4.070905-2;\quad 4.070905-1.5\quad \ldots \quad 4.070905+2$

From the graph you can read off that $(x^*,y^*)=(10,5)$

Answer 2

Let $p_1, p_2, m > 0$, $\alpha \in [0,1]$, $X = \{(x,y) \in \mathbb R^2 \text{ s.t. } p_1 x + p_2 y = m\}$ and let $$ (x^*, y^*) = \underset{(x,y)\in X}{\arg \max} \,\, x^\alpha y^{1-\alpha} $$

Solve this analytically and you get $$ x^*= \alpha \frac{m}{p_1} $$ $$ y^* = (1-\alpha) \frac{m}{p_2}. $$

Economically, this means that the agent spends fraction $\alpha$ of his wealth $m$ on good $x$ and fraction $1-\alpha$ of his wealth on good $y$.

Graphically, it means that $(x^*, y^*) = \alpha r + (1-\alpha)t$, where $r$ is the point at which the constraint intersects the $x$-axis and $t$ is the point at which the constraint intersects the $y$-axis.

For example, if $\alpha = \frac{1}{2}$ then $(x^*, y^*)$ will lie half-way between $r$ and $t$.

If you have exponents that add up to a value different from $1$, simply divide your exponents by their sum (this does not change the solution to the optimization problem).

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Applied to your case:

The exponents $0.4$ and $0.3$ don't add up to $1$, but as I said, this is not a problem. You get $$ \alpha = \frac{0.4}{0.3+0.4} = \frac{4}{7}. $$ Therefore the solution will lie $\frac{4}{7}$ the way from $t$ to $r$.

What are the intersections of the constraint and the axes? Plugging $y=0$ into the constraint yields $x=\frac{m}{p_1}$, plugging in $x=0$ yields $y=\frac{m}{p_2}$. Therefore $$ r=\left(\frac{35}{2}, 0\right) $$ and $$ t=\left(0, \frac{35}{3}\right) $$ so that $$ (x^*, y^*) = \frac{4}{7} \left(\frac{35}{2}, 0\right) + \frac{3}{7}\left(0, \frac{35}{3}\right) = \left(10, 5\right). $$

Note that this geometric approach allows you to find the exact solution graphically with just pen and paper and without needing to approximate a solution via indifference curves.