Step 1. The equation for any budget constraint is the following:

[latex]\text{Budget }={P}_{1}\times{Q}_{1}+{P}_{2}\times{Q}_{2}+\dots+{P}_{n}\times{Q}_{n}\\[/latex]

where P and Q are the price and respective quantity of any number, n, of items purchased and Budget is the amount of income one has to spend.

Step 2. Apply the budget constraint equation to the scenario.

In Charlie’s case, this works out to be

[latex]\begin{array}{l}\text{Budget}={P}_{1}\times{Q}_{1}+{P}_{2}\times{Q}_{2}\\\text{Budget}=\$10\\\,\,\,\,\,\,\,\,\,\,\,\,{P}_{1}=\$2\left(\text{the price of a burger}\right)\\\,\,\,\,\,\,\,\,\,\,\,\,{Q}_{1}=\text{quantity of burgers}\left(\text{variable}\right)\\\,\,\,\,\,\,\,\,\,\,\,\,{P}_{2}=\$0.50\left(\text{the price of a bus ticket}\right)\\\,\,\,\,\,\,\,\,\,\,\,\,{Q}_{2}=\text{quantity of tickets}\left(\text{variable}\right)\end{array}[/latex]

For Charlie, this is

[latex]{\$10}={\$2}\times{Q}_{1}+{\$0.50}\times{Q}_{2}\\[/latex]

Step 3. Simplify the equation.

At this point we need to decide whether to solve for [latex]{Q}_{1} \\[/latex] or [latex]{Q}_{2} \\[/latex].

Remember, [latex]{Q}_{1} = \text{quantity of burgers} \\[/latex]. So, in this equation [latex]{Q}_{1} \\[/latex] represents the number of burgers Charlie can buy depending on how many bus tickets he wants to purchase in a given week. [latex]{Q}_{2}=\text{quantity of tickets} \\[/latex]. So, [latex]{Q}_{2} \\[/latex] represents the number of bus tickets Charlie can buy depending on how many burgers he wants to purchase in a given week.

We are going solve for [latex]{Q}_{1} \\[/latex].

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10=2Q_{1}+0.50Q_{2}\\\,\,\,10-2Q_{1}=0.50Q_{2}\\\,\,\,\,\,\,\,\,\,\,\,\,-2Q_{1}=-10+0.50Q_{2}\\\left(2\right)\left(-2Q_{1}\right)=\left(2\right)-10+\left(2\right)0.50Q_{2}\,\,\,\,\,\,\,\,\,\text{Clear decimal by multiplying everything by 2}\\\,\,\,\,\,\,\,\,\,\,\,\,-4Q_{1}=-20+Q_{2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Q_{1}=5-\frac{1}{4}Q_{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Divide both sides by}-4\end{array}\\[/latex]

Step 4. Use the equation.

Now we have an equation that helps us calculate the number of burgers Charlie can buy depending on how many bus tickets he wants to purchase in a given week.

For example, say he wants 8 bus tickets in a given week. [latex]{Q}_{2}\\[/latex] represents the number of bus tickets Charlie buys, so we plug in 8 for [latex]{Q}_{2}\\[/latex], which gives us

[latex]\begin{array}{l}{Q}_{1}={5}-\left(\frac{1}{4}\right)8\\{Q}_{1}={5}-2\\{Q}_{1}=3\end{array}\\[/latex]

This means Charlie can buy 3 burgers that week (point C on the graph, above).

Let’s try one more. Say Charlie has a week when he walks everywhere he goes so that he can splurge on burgers. He buys 0 bus tickets that week. [latex]{Q}_{2}\\[/latex] represents the number of bus tickets Charlie buys, so we plug in 0 for [latex]{Q}_{2}\\[/latex], giving us

[latex]{Q}_{1}={5}-\left(\frac{1}{4}\right)0\\{Q}_{1}={5}\\[/latex]

So, if Charlie doesn’t ride the bus, he can buy 5 burgers that week (point A on the graph).

If you plug other numbers of bus tickets into the equation, you get the results shown in Table 1, below, which are the points on Charlie’s budget constraint.

Step 4. Graph the results.

If we plot each point on a graph, we can see a line that shows us the number of burgers Charlie can buy depending on how many bus tickets he wants to purchase in a given week.

Figure 2. Charlie’s Budget Constraint

We can make two important observations about this graph. First, the slope of the line is negative (the line slopes downward). Remember in the last module when we discussed graphing, we noted that when when X and Y have a negative, or inverse, relationship, X and Y move in opposite directions—that is, as one rises, the other falls. This means that the only way to get more of one good is to give up some of the other.

Second, the slope is defined as the change in the number of burgers (shown on the vertical axis) Charlie can buy for every incremental change in the number of tickets (shown on the horizontal axis) he buys. If he buys one less burger, he can buy four more bus tickets. The slope of a budget constraint always shows the opportunity cost of the good that is on the horizontal axis. If Charlie has to give up lots of burgers to buy just one bus ticket, then the slope will be steeper, because the opportunity cost is greater.

Let’s look at this in action and see it on a graph. What if we change the price of the burger to $1? We will keep the price of bus tickets at 50 cents. Now, instead of buying 4 more tickets for every burger he gives up, Charlie can only buy 2 tickets for every burger he gives up. Figure 3, below, shows Charlie’s new budget constraint (and the change in slope).

Figure 3. Charlie’s New Budget Constraint