Demand function (price function), market equilibrium

The price function tells you, for every quantity, the price at which that quantity is bought. A company sells apple juice. Its price function is

p_{N} (x) = 20 - 2 x

Here $x$ is the quantity in quantity units (QU) and $p$ is the price in monetary units (MU) per QU. For example:

p_{N} (6) = 20 - 2 \cdot 6 = 8

Six QU sell at 8 MU each.

Demand curve with maximum price and saturation quantity

$p_{N} (x)$ is the $y$ : the height of the graph above the position $x$ . The quantity runs horizontally, the price vertically.

Reading off prices

A larger quantity sells only at a lower price. Such a falling price function is called the demand function or price-demand function.

More examples:

	price function	wanted	calculation
a)	$p_{N} (x) = 20 - 2 x$	$p_{N} (3)$	$20 - 2 \cdot 3 = 14$
b)	$p_{N} (x) = 12 - 3 x$	$p_{N} (2)$	$12 - 3 \cdot 2 = 6$
c)	$p_{N} (x) = 30 - 0, 5 x$	$p_{N} (8)$	$30 - 0, 5 \cdot 8 = 26$
d)	$p_{N} (x) = 18 - 1 x$	$p_{N} (5)$	$18 - 1 \cdot 5 = 13$
e)	$p_{N} (x) = 24 - 4 x$	$p_{N} (4, 5)$	$24 - 4 \cdot 4, 5 = 6$

Remember: put in the quantity, get out the price.

Your turn

p_{N} (x) = 16 - 2 x

, wanted

p_{N} (3)

Solution

16 - 2 \cdot 3 = 10

Maximum price and saturation quantity

The curve has two edges. At the top the price at quantity $0$ , at the bottom the quantity at price $0$ .

p_{N} (0) = 20 - 2 \cdot 0 = 20

20 - 2 x = 0 \Rightarrow 2 x = 20 \Rightarrow x = \frac{20}{2} = 10

The upper edge is the maximum price, also called the prohibitive price: from 20 MU on, nobody buys. The lower edge is the saturation quantity: more than 10 QU the market does not take even for free.

Demand curve with maximum price 20 MU and saturation quantity 10 QU

More examples:

	price function	maximum price $p_{N} (0)$	saturation quantity ( $p_{N} (x) = 0$ )
a)	$p_{N} (x) = 20 - 2 x$	$20$	$x = \frac{20}{2} = 10$
b)	$p_{N} (x) = 12 - 3 x$	$12$	$x = \frac{12}{3} = 4$
c)	$p_{N} (x) = 30 - 0, 5 x$	$30$	$x = \frac{30}{0 , 5} = 60$
d)	$p_{N} (x) = 18 - 1 x$	$18$	$x = \frac{18}{1} = 18$
e)	$p_{N} (x) = 24 - 4 x$	$24$	$x = \frac{24}{4} = 6$

Remember: the maximum price is the number without $x$ . The saturation quantity is that number divided by the number in front of the $x$ .

Your turn

p_{N} (x) = 40 - 5 x

Solution

maximum price

40

MU, saturation quantity

x = \frac{40}{5} = 8

The slope of the price function

The derivative $p_{N}^{'} (x)$ tells you by how many MU the price changes when one more QU is sold.

p_{N} (x) = 20 - 2 x^{1} \to p_{N}^{'} (x) = - 2 \cdot 1 x^{0} = - 2

Double graph: the price function above, its derivative below

Every further QU pushes the price down by 2 MU. The minus sign shows the falling, and the $- 2$ is the same at every position: the line below runs flat.

More examples:

	price function	derivative	meaning
a)	$p_{N} (x) = 20 - 2 x^{1}$	$- 2 \cdot 1 x^{0} = - 2$	2 MU less per QU
b)	$p_{N} (x) = 12 - 3 x^{1}$	$- 3 \cdot 1 x^{0} = - 3$	3 MU less per QU
c)	$p_{N} (x) = 30 - 0, 5 x^{1}$	$- 0, 5 \cdot 1 x^{0} = - 0, 5$	0,50 MU less per QU
d)	$p_{N} (x) = 18 - 1 x^{1}$	$- 1 \cdot 1 x^{0} = - 1$	1 MU less per QU
e)	$p_{A} (x) = 2 + 1 x^{1}$	$1 \cdot 1 x^{0} = 1$	1 MU more per QU

Remember: the derivative of the price function is the price change per additional QU. Minus means falling, plus means rising.

Your turn

p_{N} (x) = 25 - 4 x

, wanted

p_{N}^{'} (x)

and its meaning.

Solution

- 4 \cdot 1 x^{0} = - 4

, 4 MU less per QU

Setting up the price function

Usually you have no formula, only two measurements. To get the price function $p_{N} (x) = m x + b$ from them, you take the $m$ first.

At 2 QU the customers pay 16 MU, at 5 QU only 10 MU.

Derivation:

m = \frac{10 - 16}{5 - 2} = \frac{- 6}{3} = - 2

The $b$ comes from either of the two points. Putting in $P_{1} (2 ∣ 16)$ :

16 = - 2 \cdot 2 + b \Rightarrow 16 = - 4 + b \Rightarrow b = 20

p_{N} (x) = - 2 x + 20

Check with the second point:

p_{N} (5) = - 2 \cdot 5 + 20 = 10 ✓

Scheme:

1. slope from both points: $m = \frac{p _{2} - p _{1}}{x _{2} - x _{1}}$ 2. put one point into $p_{N} (x) = m x + b$ and solve for $b$ 3. write down the price function: $p_{N} (x) = m x + b$ 4. check with the other point

More examples:

	points	$m$	$b$	price function
a)	$P_{1} (2 ∣ 16)$ , $P_{2} (5 ∣ 10)$	$\frac{10 - 16}{5 - 2} = - 2$	$16 = - 2 \cdot 2 + b \Rightarrow b = 20$	$p_{N} (x) = - 2 x + 20$
b)	$P_{1} (2 ∣ 25)$ , $P_{2} (6 ∣ 15)$	$\frac{15 - 25}{6 - 2} = - 2, 5$	$25 = - 2, 5 \cdot 2 + b \Rightarrow b = 30$	$p_{N} (x) = - 2, 5 x + 30$
c)	$P_{1} (5 ∣ 20)$ , $P_{2} (15 ∣ 10)$	$\frac{10 - 20}{15 - 5} = - 1$	$20 = - 1 \cdot 5 + b \Rightarrow b = 25$	$p_{N} (x) = - 1 x + 25$

Your turn

P_{1} (1 ∣ 18)

P_{2} (4 ∣ 12)

Solution

m = \frac{12 - 18}{4 - 1} = - 2

18 = - 2 \cdot 1 + b \Rightarrow b = 20

, so

p_{N} (x) = - 2 x + 20

The economically meaningful domain

Negative quantities do not exist, and neither do negative prices. Of the whole line only the piece between the two edges is left.

Only the piece between 0 and the saturation quantity makes economic sense

For $p_{N} (x) = 20 - 2 x$ :

0 \leq x \leq 10

Remember: the economically meaningful domain runs from $0$ to the saturation quantity.

Your turn

p_{N} (x) = 12 - 3 x

Solution

0 \leq x \leq 4

The supply function

The supply function tells you at which price the producers deliver a given quantity. It rises: larger quantities they deliver only for more money. For example:

p_{A} (x) = 2 + x

p_{A} (6) = 2 + 6 = 8

The edge at the bottom is the minimum supply price: below 2 MU nobody delivers.

More examples:

	supply function	minimum supply price $p_{A} (0)$	derivative
a)	$p_{A} (x) = 2 + 1 x^{1}$	$2$	$1 \cdot 1 x^{0} = 1$
b)	$p_{A} (x) = 4 + 2 x^{1}$	$4$	$2 \cdot 1 x^{0} = 2$
c)	$p_{A} (x) = 6 + 3 x^{1}$	$6$	$3 \cdot 1 x^{0} = 3$
d)	$p_{A} (x) = 3 + 0, 5 x^{1}$	$3$	$0, 5 \cdot 1 x^{0} = 0, 5$

Remember: the supply function rises, its derivative is positive. The minimum supply price is the number without $x$ .

Your turn

p_{A} (x) = 5 + 2 x

, wanted the minimum supply price and

p_{A} (4)

Solution

5

MU and

5 + 2 \cdot 4 = 13

The market equilibrium

A trade happens when customers and producers mean the same price at the same quantity. The same price means the same height of the two functions:

p_{N} (x) = p_{A} (x)

20 - 2 x = 2 + x

20 = 2 + 3 x

18 = 3 x

x = 6

The price you take from either of the two equations. Both work, because at this quantity both carry the same price:

p_{N} (6) = 20 - 2 \cdot 6 = 8

Check in the other one:

p_{A} (6) = 2 + 6 = 8 ✓

Equilibrium quantity $x = 6$ QU, equilibrium price $p = 8$ MU, so $S (6 ∣ 8)$ .

Scheme:

1. set both price functions equal: $p_{N} (x) = p_{A} (x)$ 2. solve the equation for $x$ , that is the equilibrium quantity 3. put this $x$ into either function, that is the equilibrium price 4. check in the other function 5. answer as a point: $S (x ∣ p)$

More examples:

	demand	supply	set equal	quantity	price	answer
a)	$p_{N} (x) = 20 - 2 x$	$p_{A} (x) = 2 + x$	$20 - 2 x = 2 + x$	$x = \frac{18}{3} = 6$	$2 + 6 = 8$	$S (6 ∣ 8)$
b)	$p_{N} (x) = 12 - x$	$p_{A} (x) = 4 + x$	$12 - x = 4 + x$	$x = \frac{8}{2} = 4$	$4 + 4 = 8$	$S (4 ∣ 8)$
c)	$p_{N} (x) = 30 - 3 x$	$p_{A} (x) = 6 + 3 x$	$30 - 3 x = 6 + 3 x$	$x = \frac{24}{6} = 4$	$6 + 3 \cdot 4 = 18$	$S (4 ∣ 18)$
d)	$p_{N} (x) = 15 - 0, 5 x$	$p_{A} (x) = 3 + x$	$15 - 0, 5 x = 3 + x$	$x = \frac{12}{1 , 5} = 8$	$3 + 8 = 11$	$S (8 ∣ 11)$
e)	$p_{N} (x) = 24 - 2 x$	$p_{A} (x) = 4 + 2 x$	$24 - 2 x = 4 + 2 x$	$x = \frac{20}{4} = 5$	$4 + 2 \cdot 5 = 14$	$S (5 ∣ 14)$

Remember: set equal, solve for $x$ , put $x$ in, answer $S (x ∣ p)$ .

Your turn

p_{N} (x) = 18 - 2 x

and

p_{A} (x) = 6 + x

Solution

18 - 2 x = 6 + x

x = \frac{12}{3} = 4

p = 6 + 4 = 10

, so

S (4 ∣ 10)

Prices above and below the equilibrium

Excess supply at 12 MU, excess demand at 6 MU

	price	demanded $x_{N}$	supplied $x_{A}$	difference	name
a)	$12$ MU	$\frac{20 - 12}{2} = 4$ QU	$12 - 2 = 10$ QU	$6$ QU too many	excess supply (surplus)
b)	$8$ MU	$\frac{20 - 8}{2} = 6$ QU	$8 - 2 = 6$ QU	$0$ QU	market equilibrium
c)	$6$ MU	$\frac{20 - 6}{2} = 7$ QU	$6 - 2 = 4$ QU	$3$ QU too few	excess demand (shortage)

Remember: above the equilibrium price goods are left over, below it goods are missing. Only at the intersection do both sides fit together.

The terms at a glance

term	symbol	how to get it	in the picture
quantity	$x$ in QU	given	horizontal axis
price	$p$ in MU	$p = p (x)$	vertical axis
demand function (price-demand function)	$p_{N} (x)$	falling line	blue curve
supply function	$p_{A} (x)$	rising line	orange curve
maximum price (prohibitive price)	$p_{N} (0)$	number without $x$	blue curve meets the price axis
saturation quantity	$x$ with $p_{N} (x) = 0$	number without $x$ divided by the number in front of $x$	blue curve meets the quantity axis
minimum supply price	$p_{A} (0)$	number without $x$	orange curve meets the price axis
price change per QU	$p^{'} (x)$	differentiate	slope triangle on the curve
equilibrium quantity	$x$	from $p_{N} (x) = p_{A} (x)$	intersection, read horizontally
equilibrium price	$p$	$x$ put in	intersection, read vertically
economically meaningful domain	$0 \leq x \leq x_{s a t}$	from $0$ to the saturation quantity	blue piece of the curve

Exercises

Wanted are the equilibrium quantity and the equilibrium price.

a) $p_{N} (x) = 12 - x$ and $p_{A} (x) = 4 + x$

b) $p_{N} (x) = 30 - 3 x$ and $p_{A} (x) = 6 + 3 x$

c) $p_{N} (x) = 15 - 0, 5 x$ and $p_{A} (x) = 3 + x$

d) $p_{N} (x) = 24 - 2 x$ and $p_{A} (x) = 4 + 2 x$

e) $p_{N} (x) = 40 - 4 x$ and $p_{A} (x) = 10 + x$

f) $p_{N} (x) = 18 - 1, 5 x$ and $p_{A} (x) = 3 + 1, 5 x$

Solutions step by step

a) $12 - x = 4 + x$ ; $12 = 4 + 2 x$ ; $8 = 2 x$ ; $x = 4$ ; $p_{A} (4) = 4 + 4 = 8$ ; check $p_{N} (4) = 12 - 4 = 8$ ; $S (4 ∣ 8)$

b) $30 - 3 x = 6 + 3 x$ ; $30 = 6 + 6 x$ ; $24 = 6 x$ ; $x = 4$ ; $p_{A} (4) = 6 + 3 \cdot 4 = 18$ ; check $p_{N} (4) = 30 - 3 \cdot 4 = 18$ ; $S (4 ∣ 18)$

c) $15 - 0, 5 x = 3 + x$ ; $15 = 3 + 1, 5 x$ ; $12 = 1, 5 x$ ; $x = 8$ ; $p_{A} (8) = 3 + 8 = 11$ ; check $p_{N} (8) = 15 - 0, 5 \cdot 8 = 11$ ; $S (8 ∣ 11)$

d) $24 - 2 x = 4 + 2 x$ ; $24 = 4 + 4 x$ ; $20 = 4 x$ ; $x = 5$ ; $p_{A} (5) = 4 + 2 \cdot 5 = 14$ ; check $p_{N} (5) = 24 - 2 \cdot 5 = 14$ ; $S (5 ∣ 14)$

e) $40 - 4 x = 10 + x$ ; $40 = 10 + 5 x$ ; $30 = 5 x$ ; $x = 6$ ; $p_{A} (6) = 10 + 6 = 16$ ; check $p_{N} (6) = 40 - 4 \cdot 6 = 16$ ; $S (6 ∣ 16)$

f) $18 - 1, 5 x = 3 + 1, 5 x$ ; $18 = 3 + 3 x$ ; $15 = 3 x$ ; $x = 5$ ; $p_{A} (5) = 3 + 1, 5 \cdot 5 = 10, 5$ ; check $p_{N} (5) = 18 - 1, 5 \cdot 5 = 10, 5$ ; $S (5 ∣ 10, 5)$