Question 6.39: Given the demand functions P = 100 − 2Q and Q = 80 − 10 ln(P......

(a) (i) Q only

P = 100 − 2Q

Step 1: \frac {dP}{dQ} = −2  → \frac {dQ}{dP} = \frac {1}{-2} = -\frac {1}{2}

Steps 2 and 3:

ε_d = \frac {dQ}{dP} \frac {P}{Q} = -\frac {1}{2} \frac {P}{Q} = \frac {-0.5}{1}  \frac {P}{Q}

ε_d = \frac {-0.5}{1} × \frac {100   −   2Q}{Q} = \frac {−50  +  Q}{Q}

ε_d = 1  –  \frac {50 }{Q}

This is ε_d expressed in terms of Q.

At Q = 20,

ε_d = 1  –  \frac {50 }{Q} = \frac {50 }{20} = −1.5

(a) (i) Q only

Q = 80 − 10 ln(P)

Step 1: \frac {dQ}{dP} = -\frac {10}{P}

Steps 2 and 3:

ε_d = \frac {dQ}{dP} \frac {P}{Q} = -\frac {10}{P} \frac {P}{Q} = -\frac {10}{Q}

ε_d = -\frac {10}{Q}

This is ε_d expressed in terms of Q.

At Q = 20,

ε_d = -\frac {10}{Q} = -\frac {10}{20} = -0.5

(a) (ii) P only

ε_d = \frac {dQ}{dP} \frac {P}{Q} = -\frac {1}{2} \frac {P}{Q}

= -\frac {1}{2} × \frac {P}{50  −  0.5P} = \frac {P}{−100  +  P}

= \frac {P}{P  −  100}

ε_d = \frac {P}{P − 100}

This is ε_d  expressed in terms of P.

At P = 60,

ε_d = \frac {P}{P  −  100} = \frac {60}{60  −  100} = −1.5  

(a) (ii) P only

ε_d = \frac {dQ}{dP} \frac {P}{Q} = \frac {-10}{P} \frac {P}{Q} = \frac {-10}{Q}

= \frac {-10}{80  −  10 ln(P)} × \frac {-0.1}{-0.1}

= \frac {1}{ln(P)  −  8}

ε_d = \frac {1}{ln(P)  −  8}

This is ε_d  expressed in terms of P.

At P = 60,

ε_d = \frac {1}{ln(P) −  8} = \frac {1}{ln(60)  −  8} = −0.256

(b) (i) To calculate the percentage change in Q when P increases by 5%, use the definition of elasticity given in equation (6.34)

ε_d = \frac {\text {\% change in quantity}}{\text {\% change in price}} → ε_d × \text {(\% change in P)} = \text {(\% change in Q)}

ε_d = × (5 \%) = \text {(\% change in Q)}

(−1.5)(5) = −7.5%

So, demand drops by 7.5% when price increases by 5%.

ε_d = × (5 \%) = \text {(\% change in Q)}

(−0.256)(5) = −1.28%

So, demand drops by 1.28% when price increases by 5%.

(ii) When price increases by 5%, the new price is P = \frac {105}{100} × 60 = 63

The corresponding value of Q for each function is

P = 100 − 2Q → Q = 50 − 0.5P

At P = 60, Q = 50 − 0.5(60) = 20.0

At P = 63, Q = 50 − 0.5(63) = 18.5

Therefore, the percentage change in Q

= \frac {\text {change in Q}}{\text {initial value of Q}} × 100

= \frac {18.5   −   20}{20} × 100 = −7.5%

This is exactly the same result as that calculated using elasticity.

Q = 80 − 10 ln(P)

At P = 60, Q = 80 − 10 ln(60) = 39.057

At P = 63, Q = 80 − 10 ln(63) = 38.569

Therefore, the percentage change in Q

= \frac {\text {change in Q}}{\text {initial value of Q}} × 100

= \frac {38.569   −   39.057}{39.057} × 100 

= −1.249%

This is not exactly the same result as that calculated using elasticity. The reason: for non-linear demand functions, the change over a small interval, P = 60 to 63, is approximately equal to the change at a point.

Therefore, for non-linear demand functions, the equation

ε_d = \frac {\text {\% change in quantity}}{\text {\% change in price}} → ε_d × \text {(\% change in P)} = \text {(\% change in Q)}

may be used to calculate the approximate percentage change in demand, Q, as a result of small percentage changes in P.