{"id":5058,"date":"2016-07-20T21:05:55","date_gmt":"2016-07-20T21:05:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/microeconomics\/?post_type=chapter&p=5058"},"modified":"2016-08-02T16:37:15","modified_gmt":"2016-08-02T16:37:15","slug":"reading-calculating-opportunity-cost","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/atd-herkimer-microeconomics\/chapter\/reading-calculating-opportunity-cost\/","title":{"raw":"Reading: Calculating Opportunity Cost","rendered":"Reading: Calculating Opportunity Cost"},"content":{"raw":"\"Pencil<\/a>\r\n\r\nIt makes intuitive sense that Charlie\u00a0can 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\u00a0to determine what he can do with his very small budget, but when budgets and constraints are\u00a0more complex, it's important to know how to solve equations that demonstrate budget constraints and opportunity cost.\r\n\r\nVery simply, when Charlie\u00a0is spending his full budget on burgers and tickets, his\u00a0budget is equal to the total amount that he spends on burgers plus the total amount that he spends on bus tickets.\u00a0 For example, if Charlie\u00a0buys four bus tickets and four burgers with his $10 budget (point B on the graph below), the equation would be\r\n

[latex]\\$10=\\left(\\$2\\times4\\right)+\\left(\\$.50\\times4\\right)[\/latex]<\/p>\r\nYou can see this on the graph of Charlie's\u00a0budget constraint, Figure 1, below.\r\n\r\n[caption id=\"attachment_6580\" align=\"aligncenter\" width=\"500\"]\"Graph Figure 1. Charlie's Budget Constraint<\/strong>[\/caption]\r\n\r\nIf we want to answer the question \"How many burgers and bus tickets can Charlie\u00a0buy?\" then we need to use the budget constraint equation.\r\n

\r\n\r\nStep 1.<\/strong> The equation for any budget constraint is the following:\r\n

[latex]\\text{Budget }={P}_{1}\\times{Q}_{1}+{P}_{2}\\times{Q}_{2}+\\dots+{P}_{n}\\times{Q}_{n}[\/latex]<\/p>\r\nwhere 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.\r\n\r\nStep 2<\/strong>. Apply the budget constraint equation to the scenario.\r\n\r\nIn Charlie's\u00a0case, this works out to be\r\n

[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]<\/p>\r\nFor Charlie, this is\r\n

[latex]{\\$10}={\\$2}\\times{Q}_{1}+{\\$0.50}\\times{Q}_{2}[\/latex]<\/p>\r\nStep 3<\/strong>. Simplify the equation.\r\n\r\nAt this point we need to decide whether to solve for [latex]{Q}_{1}[\/latex] or [latex]{Q}_{2}[\/latex].\r\n\r\nRemember, [latex]{Q}_{1} = \\text{quantity of burgers} [\/latex]. So, in this equation [latex]{Q}_{1}[\/latex] represents the number of burgers Charlie\u00a0can buy depending on how many bus tickets he wants to purchase in a given week.\u00a0[latex]{Q}_{2}=\\\\text{quantity of tickets}\\[\/latex]. So, [latex]{Q}_{2}[\/latex] represents the number of bus tickets Charlie\u00a0can buy depending on how many burgers he wants to purchase in a given week.\r\n\r\nWe are going solve for [latex]{Q}_{1}[\/latex].\r\n

[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]<\/p>\r\nStep 4<\/strong>. Use the equation.\r\n\r\nNow we have an equation that helps us calculate the number of burgers Charlie\u00a0can buy depending on how many bus tickets he wants to purchase in a given week.\r\n\r\nFor example, say he wants 8 bus tickets in a given week. [latex]{Q}_{2}[\/latex] represents the number of bus tickets Charlie\u00a0buys, so we plug in 8 for [latex]{Q}_{2}[\/latex], which gives us\r\n

[latex]\\begin{array}{l}{Q}_{1}={5}-\\left(\\frac{1}{4}\\right)8\\\\{Q}_{1}={5}-2\\\\{Q}_{1}=3\\end{array}[\/latex]<\/p>\r\nThis\u00a0means Charlie\u00a0can buy 3 burgers that week (point C on the graph, above).\r\n\r\nLet's try one more. Say Charlie\u00a0has a week when\u00a0he 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\u00a0buys, so we plug in 0 for [latex]{Q}_{2}[\/latex], giving us\r\n

[latex]{Q}_{1}={5}-\\left(\\frac{1}{4}\\right)0\\\\{Q}_{1}={5}[\/latex]<\/p>\r\nSo, if Charlie\u00a0doesn't ride the bus, he can buy 5 burgers that week (point A on the graph).\r\n\r\nIf 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\u00a0budget constraint.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Table 1.<\/th>\r\n<\/tr>\r\n
Point<\/th>\r\nQuantity of Burgers (at $2)<\/th>\r\nQuantity of Bus Tickets (at 50 cents)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
A<\/td>\r\n5<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
B<\/td>\r\n4<\/td>\r\n4<\/td>\r\n<\/tr>\r\n
C<\/td>\r\n3<\/td>\r\n8<\/td>\r\n<\/tr>\r\n
D<\/td>\r\n2<\/td>\r\n12<\/td>\r\n<\/tr>\r\n
E<\/td>\r\n1<\/td>\r\n16<\/td>\r\n<\/tr>\r\n
F<\/td>\r\n0<\/td>\r\n20<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nStep 4<\/strong>. <\/strong>Graph the results.\r\n\r\nIf we plot each point on a graph, we can see a line that shows us the number of burgers Charlie\u00a0can buy depending on how many bus tickets he wants to purchase in a given week.\r\n\r\n[caption id=\"attachment_6580\" align=\"aligncenter\" width=\"500\"]\"Graph Figure 2. Charlie's Budget Constraint<\/strong>[\/caption]\r\n\r\nWe 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\u00a0negative, or inverse, relationship, X and Y move in opposite directions\u2014that is, as one rises, the other\u00a0falls.\u00a0This means that the only way to get more of one good is to give up some of the other.\r\n\r\nSecond, the slope is defined as the change in the number of burgers (shown on the vertical axis) Charlie\u00a0can 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\u00a0is on the horizontal axis. If Charlie has to give up lots of\u00a0burgers to buy just one bus ticket, then the slope will be steeper, because the opportunity cost is greater.\r\n\r\nLet'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).\r\n\r\n[caption id=\"attachment_6581\" align=\"aligncenter\" width=\"501\"]\"Graph Figure 3. Charlie's New Budget Constraint<\/strong>[\/caption]\r\n\r\n<\/div>\r\n

Self Check: The Cost of Choices<\/h2>\r\nAnswer the question(s) below to see how well you understand the topics covered in the previous section. This short quiz does not\u00a0<\/strong>count toward your grade in the class, and you can retake it an unlimited number of times.\r\n

You\u2019ll have more success on the Self Check if you\u2019ve completed the two Readings in this section.<\/span><\/p>\r\nUse this quiz to check your understanding and decide whether to (1) study the previous section further or (2) move on to the next section.\r\n\r\nhttps:\/\/assessments.lumenlearning.com\/assessments\/1551","rendered":"

\"Pencil<\/a><\/p>\n

It makes intuitive sense that Charlie\u00a0can 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\u00a0to determine what he can do with his very small budget, but when budgets and constraints are\u00a0more complex, it’s important to know how to solve equations that demonstrate budget constraints and opportunity cost.<\/p>\n

Very simply, when Charlie\u00a0is spending his full budget on burgers and tickets, his\u00a0budget is equal to the total amount that he spends on burgers plus the total amount that he spends on bus tickets.\u00a0 For example, if Charlie\u00a0buys four bus tickets and four burgers with his $10 budget (point B on the graph below), the equation would be<\/p>\n

[latex]\\$10=\\left(\\$2\\times4\\right)+\\left(\\$.50\\times4\\right)[\/latex]<\/p>\n

You can see this on the graph of Charlie’s\u00a0budget constraint, Figure 1, below.<\/p>\n

\"Graph<\/p>\n

Figure 1. Charlie’s Budget Constraint<\/strong><\/p>\n<\/div>\n

If we want to answer the question “How many burgers and bus tickets can Charlie\u00a0buy?” then we need to use the budget constraint equation.<\/p>\n

\n

Step 1.<\/strong> The equation for any budget constraint is the following:<\/p>\n

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

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.<\/p>\n

Step 2<\/strong>. Apply the budget constraint equation to the scenario.<\/p>\n

In Charlie’s\u00a0case, this works out to be<\/p>\n

[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]<\/p>\n

For Charlie, this is<\/p>\n

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

Step 3<\/strong>. Simplify the equation.<\/p>\n

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

Remember, [latex]{Q}_{1} = \\text{quantity of burgers}[\/latex]. So, in this equation [latex]{Q}_{1}[\/latex] represents the number of burgers Charlie\u00a0can buy depending on how many bus tickets he wants to purchase in a given week.\u00a0[latex]{Q}_{2}=\\\\text{quantity of tickets}\\[\/latex]. So, [latex]{Q}_{2}[\/latex] represents the number of bus tickets Charlie\u00a0can buy depending on how many burgers he wants to purchase in a given week.<\/p>\n

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

[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]<\/p>\n

Step 4<\/strong>. Use the equation.<\/p>\n

Now we have an equation that helps us calculate the number of burgers Charlie\u00a0can buy depending on how many bus tickets he wants to purchase in a given week.<\/p>\n

For example, say he wants 8 bus tickets in a given week. [latex]{Q}_{2}[\/latex] represents the number of bus tickets Charlie\u00a0buys, so we plug in 8 for [latex]{Q}_{2}[\/latex], which gives us<\/p>\n

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

This\u00a0means Charlie\u00a0can buy 3 burgers that week (point C on the graph, above).<\/p>\n

Let’s try one more. Say Charlie\u00a0has a week when\u00a0he 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\u00a0buys, so we plug in 0 for [latex]{Q}_{2}[\/latex], giving us<\/p>\n

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

So, if Charlie\u00a0doesn’t ride the bus, he can buy 5 burgers that week (point A on the graph).<\/p>\n

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\u00a0budget constraint.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
Table 1.<\/th>\n<\/tr>\n
Point<\/th>\nQuantity of Burgers (at $2)<\/th>\nQuantity of Bus Tickets (at 50 cents)<\/th>\n<\/tr>\n<\/thead>\n
A<\/td>\n5<\/td>\n0<\/td>\n<\/tr>\n
B<\/td>\n4<\/td>\n4<\/td>\n<\/tr>\n
C<\/td>\n3<\/td>\n8<\/td>\n<\/tr>\n
D<\/td>\n2<\/td>\n12<\/td>\n<\/tr>\n
E<\/td>\n1<\/td>\n16<\/td>\n<\/tr>\n
F<\/td>\n0<\/td>\n20<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Step 4<\/strong>. <\/strong>Graph the results.<\/p>\n

If we plot each point on a graph, we can see a line that shows us the number of burgers Charlie\u00a0can buy depending on how many bus tickets he wants to purchase in a given week.<\/p>\n

\"Graph<\/p>\n

Figure 2. Charlie’s Budget Constraint<\/strong><\/p>\n<\/div>\n

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\u00a0negative, or inverse, relationship, X and Y move in opposite directions\u2014that is, as one rises, the other\u00a0falls.\u00a0This means that the only way to get more of one good is to give up some of the other.<\/p>\n

Second, the slope is defined as the change in the number of burgers (shown on the vertical axis) Charlie\u00a0can 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\u00a0is on the horizontal axis. If Charlie has to give up lots of\u00a0burgers to buy just one bus ticket, then the slope will be steeper, because the opportunity cost is greater.<\/p>\n

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).<\/p>\n

\"Graph<\/p>\n

Figure 3. Charlie’s New Budget Constraint<\/strong><\/p>\n<\/div>\n<\/div>\n

Self Check: The Cost of Choices<\/h2>\n

Answer the question(s) below to see how well you understand the topics covered in the previous section. This short quiz does not\u00a0<\/strong>count toward your grade in the class, and you can retake it an unlimited number of times.<\/p>\n

You\u2019ll have more success on the Self Check if you\u2019ve completed the two Readings in this section.<\/span><\/p>\n

Use this quiz to check your understanding and decide whether to (1) study the previous section further or (2) move on to the next section.<\/p>\n

\t