From Income and Prices to Three Representations
A two-good budget constraint can be shown as a table, an equation, or a graph. Each representation describes the same choices: which combinations of two goods use all of a consumer’s available income, and which combinations cost no more than that income. Translating between the three is useful because a table makes particular combinations easy to inspect, an equation gives a precise rule, and a graph displays the full set of combinations.
Assume a consumer has income \(m\), spends it only on goods \(x\) and \(y\), and faces constant per-unit prices \(p_x>0\) and \(p_y>0\). Quantities \(x\) and \(y\) are nonnegative and divisible. There are no taxes, discounts, or other costs in the examples unless stated. The consumer can leave some income unspent, so affordable bundles satisfy the inequality below; bundles on the budget line spend the entire income.
Worked Example: Building All Three Representations
Worked Example: Building All Three Representations
A consumer has $60 to spend. Good \(x\) costs $6 per unit and good \(y\) costs $3 per unit. Construct a budget table, write the equation in both forms, and describe how to graph it. Check a selected bundle against all three representations.
Method: start with the total-spending equation. Spending on \(x\) is \(6x\) dollars and spending on \(y\) is \(3y\) dollars. The boundary of the affordable set therefore satisfies \(6x+3y=60\).
Equation forms. Dividing by 3 gives \(2x+y=20\). Solving for \(y\) gives \(y=20-2x\). The coefficient \(-2\) means that, on the line, an increase of one unit of \(x\) corresponds to a decrease of two units of \(y\).
Budget table. For each selected \(x\), calculate \(y=(60-6x)/3\):
| Quantity of \(x\) | Spending on \(x\) | Remaining income | Quantity of \(y\) |
|---|---|---|---|
| 0 | $0 | $60 | 20 |
| 2 | $12 | $48 | 16 |
| 5 | $30 | $30 | 10 |
| 8 | $48 | $12 | 4 |
| 10 | $60 | $0 | 0 |
For example, at \(x=5\), the remaining income is \(60-6(5)=30\) dollars, which buys \(30/3=10\) units of \(y\). Substitution confirms the table entry: \(6(5)+3(10)=30+30=60\).
Graph. Put \(x\) on the horizontal axis and \(y\) on the vertical axis. When \(x=0\), the equation gives \(y=20\); when \(y=0\), it gives \(x=10\). Plot \((0,20)\) and \((10,0)\), then connect them with a straight line. The table points, such as \((5,10)\), should lie on that line.
Cross-check the bundle \((8,4)\). The table includes this point, and the equation gives \(2(8)+4=20\). Its cost is \(6(8)+3(4)=48+12=60\), so it lies on the budget line. All three representations agree. The common mistake avoided here is treating the remaining income as the quantity of \(y\); it must be divided by \(y\)’s price.
Why the Equation and Table Must Agree
The conversion is based on adding the two expenditures. If the consumer buys \(x\) units at \(p_x\) dollars each, the first expenditure is \(p_xx\). The second expenditure is \(p_yy\). Spending all income means their sum is \(m\). Spending no more than income means the sum is at most \(m\). This gives the budget line and the affordable set, respectively:
A table entry is found by choosing one quantity and solving this same equality for the other. Rearranging the boundary equation for \(y\) gives:
This rearrangement justifies the straight-line graph: \(m/p_y\) is the vertical intercept, and \(-p_x/p_y\) is the slope when \(x\) is horizontal and \(y\) is vertical. Setting \(y=0\) in the original equation gives the horizontal intercept \(x=m/p_x\). Thus, independently calculated intercepts, table entries, and plotted points should all satisfy the same equation.
Worked Example: A Fractional Slope and an Affordable Bundle
Worked Example: A Fractional Slope and an Affordable Bundle
A consumer has $84. Good \(x\) costs $7 per unit, and good \(y\) costs $4 per unit. Make a table using \(x=0,4,8,12\), derive an equation suitable for graphing, and determine whether \((8,8)\) is on the line, affordable but inside it, or unaffordable.
Method: calculate the remaining income for each selected quantity. The boundary equation is \(7x+4y=84\), so \(y=(84-7x)/4\).
| \(x\) | Spending on \(x\) | Remaining income | \(y\) on the line |
|---|---|---|---|
| 0 | $0 | $84 | 21 |
| 4 | $28 | $56 | 14 |
| 8 | $56 | $28 | 7 |
| 12 | $84 | $0 | 0 |
Solving the equation for \(y\) gives \(y=21-\frac{7}{4}x\), or \(y=21-1.75x\). The vertical intercept is \((0,21)\). Setting \(y=0\) gives \(7x=84\), hence the horizontal intercept is \((12,0)\). Plot these endpoints and connect them; for example, the table point \((8,7)\) is also on the line.
For the proposed bundle \((8,8)\), expenditure is \(7(8)+4(8)=56+32=88\) dollars. Since \(88>84\), it is unaffordable; it is not a point on or inside the budget set. As a check, the equation at \(x=8\) allows \(y=21-1.75(8)=7\), not 8. The common mistake avoided is rounding or ignoring the fractional slope and then plotting a point that does not satisfy the original spending equation.
Worked Example: Inferring Prices From Intercepts
Worked Example: Inferring Prices From Intercepts
A consumer’s income is $105. A graph of the budget line has horizontal intercept 15 units of \(x\) and vertical intercept 21 units of \(y\). Infer each price, construct a short table, and verify that the bundle \((10,7)\) lies on the line.
Method: use the meaning of each intercept. At the horizontal intercept, all income is spent on \(x\), so \(p_x(15)=105\). Therefore \(p_x=105/15=\$7\) per unit. At the vertical intercept, all income is spent on \(y\), so \(p_y(21)=105\), giving \(p_y=105/21=\$5\) per unit.
The spending equation is \(7x+5y=105\). Solving for \(y\) gives \(y=(105-7x)/5=21-1.4x\). A table can be constructed by selecting \(x\) values and calculating the remaining amount for \(y\):
| \(x\) | Spending on \(x\) | Remaining income | \(y\) on the line |
|---|---|---|---|
| 0 | $0 | $105 | 21 |
| 5 | $35 | $70 | 14 |
| 10 | $70 | $35 | 7 |
| 15 | $105 | $0 | 0 |
The graph therefore joins \((0,21)\) to \((15,0)\), and includes table points such as \((10,7)\). To check the requested bundle directly, its cost is \(7(10)+5(7)=70+35=105\) dollars. It is exactly on the line. The inferred prices, equation, table, and graph are consistent. The common mistake avoided is reversing the intercept-price relationship: the price is income divided by the maximum affordable quantity, not the quantity divided by income.
Reading and Checking the Graph
For positive prices and income, the budget line has a negative slope in the usual graph with \(x\) horizontal and \(y\) vertical. Its endpoints are the maximum quantities available when all income is spent on just one good: \(m/p_x\) units of \(x\), or \(m/p_y\) units of \(y\). Connecting those intercepts gives the boundary. Points below it in the first quadrant are affordable because their total spending is less than income; points above it require more than the available income.
A plotted table point need not be an intercept. It can be used as an additional accuracy check: substitute its coordinates into \(p_xx+p_yy=m\). If the equality fails, check the arithmetic, the axis labels, and whether the correct price was used for each good. A table row with a negative remaining income is not a feasible point on the line in the nonnegative quadrant.
Write down income \(m\), both prices, and which quantity belongs to each good.
Use \(p_xx+p_yy=m\) for the line and \(p_xx+p_yy\leq m\) for affordable bundles.
Choose quantities for one good and divide the remaining income by the other good’s price.
Find both intercepts, plot them, connect them, and check at least one table point in the original equation.
A Common Representation Trap
The equation \(p_xx+p_yy=m\) is a spending equation, so each price must multiply the quantity of the good it prices. If \(p_x=6\), \(p_y=3\), and \(m=60\), the correct equation is \(6x+3y=60\). Reversing the prices gives \(3x+6y=60\), which describes a different line: it would imply a maximum of 20 units of \(x\) and 10 units of \(y\), rather than the correct intercepts of 10 and 20. Check the proposed intercepts directly: at \(x=10,y=0\), the correct spending is \(6(10)=60\); at \(x=0,y=20\), it is \(3(20)=60\).
Check Your Understanding
For each problem, use the income and prices to connect the equation, selected table values, and graph or bundle check.
- A consumer has income $48; \(p_x=\$4\) and \(p_y=\$2\). Write the budget-line equation solved for \(y\), give both intercepts, and find the line’s \(y\)-value at \(x=6\). (Answer: \(y=24-2x\); intercepts \((12,0)\) and \((0,24)\); at \(x=6\), \(y=12\).)
- Income is $90, \(p_x=\$6\), and \(p_y=\$5\). Complete the budget-line table entries for \(x=0,5,10,15\), and state the endpoints to plot. (Answer: corresponding \(y\)-values \(18,12,6,0\); endpoints \((0,18)\) and \((15,0)\).)
- A consumer has $72, with \(p_x=\$8\) and \(p_y=\$4\). Is \((5,8)\) on the line, affordable but inside it, or unaffordable? Show the expenditure check. (Answer: on the line; \(8(5)+4(8)=40+32=\$72\).)
- Income is $100 and the budget-line intercepts are 20 units of \(x\) and 25 units of \(y\). Infer both prices, write the budget equation, and find the \(y\)-quantity paired with \(x=10\) on the line. (Answer: \(p_x=\$5\), \(p_y=\$4\); \(5x+4y=100\); \(y=12.5\).)
- Income is $63, \(p_x=\$7\), and \(p_y=\$3\). A student writes \(y=21-\frac{3}{7}x\). Identify the error, give the correct equation solved for \(y\), and state the maximum affordable quantity of \(x\). (Answer: the price ratio was reversed; \(y=21-\frac{7}{3}x\); maximum \(x=9\).)