Question

In: Economics

Let the demand and supply functions for a commodity be Q =D(P) ∂D < 0 ∂P...

Let the demand and supply functions for a commodity be Q =D(P) ∂D < 0
∂P
Q=S(P,t) ∂S>0, ∂S<0 ∂P ∂t
where t is the tax rate on the commodity.
(a) What are the endogenous and exogenous variables? (b) Derive the total differential of each equation.
(c) Use Cramer’s rule to compute dQ/dt and dP/dt. (d) Determine the sign of dQ/dt and dP/dt.
(e) Use the Q − P diagram to explain your results.
Find the Taylor series with n = 4 and x0 = −2 for the function
φ(x) = 1 − x 1+x

Solutions

Expert Solution

Here,

Demand Fn: Q=D(P)

Supply function: Q=S(P,t)

Here, P and t are the exogeneous variables

Q is the endogenous variable

Q=D(P)

Taking total derivative: dQ/dP = ∂D/∂P -------------- (1)

and dQ/dP = (∂S/ ∂P) + (∂S/∂t)(dt/dP) -------------(2)

dQ/dt = (∂S/∂P)( dP/dt) + ∂S/∂t -------------(3)

Put (1) in (2) and rearranging

∂D/∂P = (∂S/ ∂P) + (∂S/∂t)(dt/dP)

(∂D/∂P)(dP/dt) = (∂S/ ∂P)(dP/dt) + (∂S/∂t)

(∂S/∂t) = [(∂D/∂P) - (∂S/ ∂P)](dP/dt) ------------(4)

Now, let (dP/dt) = x and dQ/dt=y

Rewriting (3) and (4)

(3) => y -(∂S/∂P)x = ∂S/∂t

(4) => (∂S/∂t) = [(∂D/∂P) - (∂S/ ∂P)]x

Solving for x and y (Use Cramer’s matrix to simplify)

x= [(∂D/∂P) - (∂S/ ∂P)]/ (∂S/∂t)

y= ∂S/∂t +(∂S/∂P)* [(∂D/∂P) - (∂S/ ∂P)]/ (∂S/∂t)

Sign of x positive because (∂D/∂P) is negative, (∂S/ ∂P) is positive => Numerator is negative

But denominator is also negative

Sign of y ambiguous

∂S/∂t is negative

(∂S/∂P) is positive

[(∂D/∂P) - (∂S/ ∂P)]/ (∂S/∂t) is positive


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