Corner solutions and demand for special utilities

The tangency condition assumes three things: the indifference curve has a slope at the best bundle, it is convex, and the best bundle buys some of both goods. Take any one away and the recipe can give nonsense.

• No curvature: straight-line curves (perfect substitutes) have the same slope everywhere. Either they match the budget line everywhere or nowhere.
• A kink: L-shaped curves and the min-of-two-lines have no slope at their corner.
• Wrong curvature: curves that bow away from the origin (max{x, y}) make a tangency the worst point on the line.
• A bad: a good that lowers utility is bought not at all.
• Too expensive an interior: the tangency exists but needs more income than she has, or gives a negative quantity.

A corner solution buys only one good. The correct rule replaces the equality with an inequality. Compare marginal utility per rupee:

At a corner on the x-axis, her MRS is at least the price ratio: she would happily swap more y for x, but she has no y to swap. So the working rule is the one the exams reward: list the candidates and compare utilities. The candidates are the two ends of the budget line, the kink if there is one, and any tangency inside the budget. Evaluate the utility at each and take the largest.

Choose the tastes and slide the price of x. The left picture shows the budget line, the best curve she reaches and the chosen bundle; the right shows her demand for x as its price changes. Where the choice jumps from one kind to another, the demand curve jumps too. The note under the plots says where.

Perfect substitutes, u = ax + by: compare a/p_x with b/p_y. Whichever is larger gets her whole income; if they are equal, every point on the budget line is as good as every other (JNU 2018 sets this as a written question, with u = x + y and equal prices). For u = 2X + Y with X + 2Y = 10, X gives 2/1 = 2 per rupee and Y gives 1/2: she buys 10 of X and none of Y (CUET 2024). For u = 3(x₁ + x₂) with prices Re 1 and Rs 3 and income 300, good 1 is three times as cheap and she buys x₁ = 300, x₂ = 0 (JNU 2011). U = min{x₁, x₂} + max{x₁, x₂} equals x₁ + x₂, so with prices 1 and 2 and income 150 she buys 150 of good 1 (JNU 2015).

A bad is the same rule with a negative weight: U = 5M − 10B has a negative value per rupee on butter, so she buys no butter at all and spends the whole Rs 5 on milk (CUET 2025). She would not even buy butter if it were free.

For U = max{x, y}, a bundle is worth its larger entry. A mixture is worth less than the better of its ends, so the budget line’s middle is its worst part. With prices of Rs 2 and income Rs 100, all x (50 units, utility 50) or all y (50 units, utility 50) is optimal and the mix (25, 25) has utility 25 (JNU 2010). If one good is cheaper she buys only that one.

JAM 2023 gives u = max{0.5x₁, 0.5x₂} + min{x₁, x₂}, with p₂ = 1 and asks for the lowest p₁ above which she buys only good 2. Compare three candidates, with income y: only good 2 gives 0.5y; only good 1 gives 0.5y/p₁; the balanced bundle x₁ = x₂ = y/(p₁ + 1) gives 1.5y/(p₁ + 1). The balanced bundle beats good 2 alone while 1.5/(p₁ + 1) > 0.5, that is p₁ < 2. At p₁ = 2 they tie; above 2 only good 2 is bought. The threshold is 2.

With u = min{x₁, x₂} she buys the same amount of each. Let that amount be t. Then (p₁ + p₂)t = m, so t = m/(p₁ + p₂). With prices 1 and 2 and income 150, she buys t = 50 of each (JNU 2012). One more rupee of income buys 1/3 more of each, since each set of one of each costs Rs 3. The income expansion path is the line x₁ = x₂.

For u = g(x) + y the MRS is g′(x), which depends on x only. Set it equal to p_x/p_y and income never enters:

For ln x + y with prices 2 and 1, x* = ½ at every income that can afford it. Spending on x is Rs 1, so at income 10 she buys y = 9 and at income 20, y = 19 (JNU 2016, 2017). The extra Rs 10 all went on y.

The corner appears when the interior x* costs more than her income. For u = √x + y with P_y = 2 and P_x = 1/(4M), the interior x* is (2/(2 × 1/4M))² = 16M², costing 16M² × 1/(4M) = 4M. That exceeds her income M, so she spends everything on x and buys no y (JNU 2016). Whether she buys both goods depends on prices and income, so questions that ask “does the bundle include both?” with no numbers have the answer “cannot say”. With P_x = P_y the same holds: small income buys only x, larger income buys x = ¼ and the rest in y, so y can be less or more than x (JNU 2016).

JNU 2016 gives u = min{2x + y, x + 2y}. The first line is smaller when x < y and the second when x > y, so the corner is on x = y, and utility rises in both goods. A concave kinked function like this has its best bundle at the kink or at an end of the budget line. With income m and prices p_x, p_y:

• all on y: utility m/p_y (here x = 0, y = m/p_y, so min{y, 2y} = y);
• at the kink x = y = m/(p_x + p_y): utility 3m/(p_x + p_y);
• all on x: utility m/p_x, worse than y alone when x is dearer.

The kink beats all-y when 3/(p_x + p_y) > 1/p_y, which is p_x < 2p_y. So she buys x = 0 exactly when p_x > 2p_y, which is the keyed answer.

A utility such as (x − γ_x)^a(y − γ_y)^1−a is a Cobb–Douglas function with its origin moved. Replace x − γ_x by X and y − γ_y by Y, and the problem is Lesson 5 with a different budget. If S = m − p_xγ_x − p_yγ_y is the income left after paying for the shifts:

For U = x₂(x₁ + 1) with income 142 and prices 2 and 3: the shift in x₁ is γ = −1, so S = 142 + 2 = 144, and with a = ½, x₁ = −1 + 72/2 = 35 and x₂ = 72/3 = 24 (GATE 2023). For (2x₁ − 1)^0.25(x₂ − 4)^0.75, which is proportional to (x₁ − ½)^0.25(x₂ − 4)^0.75, with prices 2 and 1 and budget 73: S = 73 − 1 − 4 = 68, x₁ = ½ + 0.25 × 68/2 = 9 and x₂ = 4 + 0.75 × 68 = 55, so x₁ + x₂ = 64 (JAM 2024).