The Budget Set

Draw the region of bundles you can afford, set by prices and your money.

If the two goods together cost more than your money, that bundle is out. The borderline is the budget line. When one good gets pricier, the line rotates inward. With more money, the line shifts outward in parallel. The budget set and slope formulas are below.

For example, a 6,000-won allowance buys 🍎 apples at 1,000 won each and 🍌 bananas at 2,000 won each. Apples alone, you can get six; bananas alone, three. If apples rise to 2,000 won, only three apples fit the budget.

Bp,w={x∈R+2:p1x1+p2x2≤w}B_{p,w} = \{x \in \mathbb{R}^2_+ : p_1 x_1 + p_2 x_2 \le w\}
slope=−p1p2\text{slope} = -\frac{p_1}{p_2}

Part of these learning paths

Examples & Exercises

With p1=1p_1 = 1, p2=2p_2 = 2, w=60w = 60, find the two intercepts and the slope of the budget line, and check whether the probe bundle b=(30,15)b = (30, 15) exhausts the budget exactly.

x1-intercept=wp1=60,x2-intercept=wp2=30x_1\text{-intercept} = \frac{w}{p_1} = 60, \quad x_2\text{-intercept} = \frac{w}{p_2} = 30
slope=−p1p2=−12\text{slope} = -\frac{p_1}{p_2} = -\frac{1}{2}
p⋅b=1⋅30+2⋅15=60=wp \cdot b = 1 \cdot 30 + 2 \cdot 15 = 60 = w
p=(1,2)p = (1, 2), w=60w = 60, probe b=(30,15)b = (30, 15) — exactly on the budget line.
Show proof

  1. On the budget line p1x1+p2x2=wp_1 x_1 + p_2 x_2 = w, setting x2=0x_2 = 0 gives x1=w/p1=60x_1 = w/p_1 = 60, and setting x1=0x_1 = 0 gives x2=w/p2=30x_2 = w/p_2 = 30.

  2. The slope is −p1p2=−12-\frac{p_1}{p_2} = -\frac{1}{2} — one more unit of good 1 costs 1/21/2 unit of good 2.

  3. The probe bundle costs p⋅b=1⋅30+2⋅15=60=wp \cdot b = 1 \cdot 30 + 2 \cdot 15 = 60 = w. The budget is fully spent, so Walras' law holds and the point bb lies on the budget line.

With p2=1p_2 = 1, w=60w = 60, how does the budget line move when p1p_1 rises from 11 to 22?

Is the bundle (30,30)(30, 30) still affordable at the new prices?

(60,60)→p1:1→2(30,60)(intercepts)(60, 60) \xrightarrow{p_1: 1 \to 2} (30, 60) \quad \text{(intercepts)}
p′⋅(30,30)=2⋅30+30=90>60p' \cdot (30, 30) = 2 \cdot 30 + 30 = 90 > 60
p1:1→2p_1: 1 \to 2, p2=1p_2 = 1, w=60w = 60 — the probe (30,30)(30, 30) is no longer affordable.
Show proof

  1. The original intercepts were (w/p1,w/p2)=(60,60)(w/p_1, w/p_2) = (60, 60). At p1=2p_1 = 2 the x1x_1-intercept shrinks to 60/2=3060/2 = 30 while the x2x_2-intercept stays at 60/1=6060/1 = 60.

  2. So the budget line rotates inward around the x2x_2-intercept — the relative price p1/p2p_1/p_2 rose, steepening the slope from −1-1 to −2-2.

  3. The bundle (30,30)(30, 30) now costs 2⋅30+1⋅30=90>602 \cdot 30 + 1 \cdot 30 = 90 > 60, so it drops out of the budget set — which is why the probe dot lands outside the shaded region.

3. Consumer Choice