Question #221279

In the market for Fante Kenley, the supply and demand functions respectively are Qs = 0.25P+10 and  -0.5P+100

When there is excess demand, price adjusts according to the equation dp/dt = 0.5(Qd - Qs)

 

a) Find the long run equilibrium price, P* (that is, the price at which there is no excess demand or supply). 

b) Formulate and solve he first order differential equation giving P as a function of time, t. Is this market dynamically stable or unstable? 

c) If the initial price is P = 50, how close will the price be to its long run equilibrium value, when t = 10? 



1
Expert's answer
2021-07-29T12:57:01-0400

a)

Qs=0.25P+10Qd=0.5p+100Qs = 0.25P + 10\\ Qd = -0.5p + 100

If excess demand , price adjustment pt=0.5(QdQs)\frac{∂p}{∂t} = 0.5(Q^d − Q^s)

For Long run equilibrium put

QD=Qd0.25P+10=0.5p+1000.25P+0.5p=100100.30P=90P=900.30P=300QD = Qd\\ 0.25P + 10 = -0.5p + 100\\ 0.25P + 0.5p = 100 - 10 0.30P = 90\\ P = \frac{90}{0.30}\\ P = 300

Now put P = 300 in any equation to calculate the Equilibrium Quantity

Qs = 0.25P + 10

Qs = 0.25*300 + 10

Qs = 75 + 10

Qs = 80

Qs=0.25P+10Qs=0.25×300+10Qs=75+10Qs=80Qs = 0.25P + 10\\ Qs = 0.25\times300 + 10\\ Qs = 75 + 10\\ Qs = 80


b)

The Given Differentiation is pt=0.5(QdQs)\frac{∂p}{∂t} = 0.5(Q^d − Q^s)

pt=0.5(.5p+1000.25p10)\frac{∂p}{∂t} = 0.5(-.5p+100-0.25p-10)

which can be arranged as

pt=0.5(.5p+1000.25p10)\frac{∂p}{∂t} = 0.5(-.5p+100-0.25p-10)

pt=0.15p+45\frac{∂p}{∂t} = -0.15p+45 where b=-0.15 and c=45

p=Aebtcbp=Ae^{bt}-\frac{c}{b} where A=P(0)+cbA=P(0)+\frac{c}{b}

since b=-0.15 and c=45, we have cb=p\frac{c}{b}=-p^* (i.e equilibrium price)

P=(P(0)+P)ebt+pP = (P(0) + P*)e^{bt}+p^*

P=(P(0)+300)e0.15t+300P = (P(0) + 300)e^{0.15t} + 300

 Accordingly as b is < 0 this implies it will increase monotonically

 through time, therefore market is Dynamically unstable


c)

When P(0)=50,AP=(P(0)+300)e0.15t300P=(50+300)e0.15×10300P=(350)×4.4816300P=820.422When\space P(0) = 50, A\\ P= (P(0) + 300)_e^{0.15t}− 300\\P = (50 + 300)_e^{0.15\times10} − 300\\P = (350)\times4.4816 − 300\\P = 820.422



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