Graphing indifference curves to visualize solutions?

Question

I am having trouble with being able to graph indifference curves. This is a particularly important skill to have especially when trying to visualize corner solutions, and when the Lagrangian method doesn't necessarily give us the solution i.e. perfect substitutes.

I particularly struggle when the Utility function in question has a mixture of different types of curves.

For example:
$U(x, y) = min$ {$x, y$} + $max$ {$\frac{x}{2}, \frac{y}{2}$}, or
$U(x, y) = min$ {$ax + by, cx + dy$}

My approach for the first one: I individually plotted the two curves by setting them equal to a constant. I got a L shaped curve for the min part after evaluating various cases (x = y, x > y, x < y), and the opposite of that for the max part again after evaluating cases.

However, I am having a hard time visualizing the combination of both. Like how would the shifts effect the shape? In general, what strategy do you recommend when trying to plot different types of indifference curves such as ones mentioned above?

Answer 1

You can easily plot these using Desmos.

On the left side, define the utility function with the equation $$ U\left(x,y\right)=\min\left(x,y\right)+\max\left(\frac{x}{2},\frac{y}{2}\right) $$ Then ask for the set of points $(x,y)$ which satisfy the equation for a utility level, e.g. 2, $$ 2 = U\left(x,y\right). $$ It is important to place the function on the left hand side when defining the function (first equation), and to place the function on the right hand side when asking for the set of points (second equation).

Here is the implementation.

Answer 2

For your second example $ax + by = cx + dy$ is the line where the kink will be. However, this will not be L shaped (Assuming a,b,c,d > 0) the indifference curves may have an outward or inward kink depending on the values of a,b,c,d values. Try it out in the Desmos environment Giskard linked.

Answer 3

Your first example is relatively easy, since the two terms have the same condition. That is, they both depend on whether $x>y$ (note $x>y$ iff $\frac x 2> \frac y2$), and if you go through the two cases, they simplify to $\frac{x+y+min\{x,y\}}2$. For an indifference curve, set that equal to a constant: $\frac{x+y+\min\{x,y\}}2=c_1$. We can then introduce another constant $c_2 = 2c_1$ to get $x+y+\min\{x,y\}=c_2$. Now just draw two curves, one for $x>y$, and one for $x<y$. The dividing line will be the diagonal line $x=y$. To the bottom-right, we have $x>y$, so $x+2y=c_2$, so we have a bunch of lines with slope $-\frac 12$. And to the top-left, we have lines with slope $-2$.

In the second example, the dividing line is $ax+by=cx+dy$, which simplifies to $y=\frac{b-d}{c-a}x$. On one side of that line, we have lines with slope $=-\frac ab$, and on the other side we have lines with slope $-\frac cd$.