28 Reading: Calculating Opportunity Cost
It makes intuitive sense that Charlie can buy only a limited number of bus tickets and burgers with a limited budget. Also, the more burgers he buys, the fewer bus tickets he can buy. With a simple example like this, it isn’t too hard to determine what he can do with his very small budget, but when budgets and constraints are more complex, it’s important to know how to solve equations that demonstrate budget constraints and opportunity cost.
For example, if Charlie buys four bus tickets and four burgers with his $10 budget (point B on the graph below), the equation would be
[latex]\$10=\left(\$2\times4\right)+\left(\$.50\times4\right)[/latex]
You can see this on the graph of Charlie’s budget constraint, Figure 1, below.
If we want to answer the question “How many burgers and bus tickets can Charlie buy?” then we need to use the budget constraint equation.
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}[