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Demand schedule: $Q_d=a_0-a_1P_d$
Supply schedule: $Q_s=b_0+b_1P_s$
$P_d$ and $P_s$ are prices faced by consumers and producers. $a_0,a_1,b_0,b_1$ are all positive constants, where $a_0>b_0$. The government imposes a tax $t$ per unit on consumers of this good where $0 \leq t\leq1$. That is, consumers face a price $P_d=P_s+t$. The government collects tax revenue $T(t)=tQ_d$. Find tax $t$* that maxmizes tax revenue. Verify that revenue attains a maximum at $t$*.

Attempt: First, I substituted $P_d=P_s+t$ into the demand function:

$Q_d=a_0-a_1P_d=a_0-a_1(P_s+t)=a_0-a_1P_s-a_1t$

Then, I found the new equilibrium:

$Q_d=Q_s$

$a_0-a_1P_s-a_1t=b_0+b_1P_s$

I solved for $P_s$:

$P_s=\frac{a_0-b_0-a_1t}{a_1+b_1}$

I substituted this into the tax revenue equation $T(t)$:

$T(t)=tQ_d=t(a_0-a_1P_s-a_1t)=a_0t-a_1tP_s-a_1t^2=a_0t-a_1t(\frac{a_0-b_0-a_1t}{a_1+b_1})-a_1t^2$

Then, I took the derivative and made it equal to zero:

$T'(t)=\frac{a_1 \left(b_0-2 b_1 t\right)+a_0 b_1}{a_1+b_1}=0$

What should my next steps be? Is my way of solving this problem correct at all? And in order to verify that there is a maximum at $t$* can I just take the second derivative?

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1 Answer 1

up vote 1 down vote accepted

Next, solve for $t$. Yes, all looks correct. Yes, the second derivative test will do the job.

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When I solve for $t$ I get: $$t\to \frac{a_1 b_0+a_0 b_1}{2 a_1 b_1}$$ Since $0\leq t\leq 1$, do i have to find a number value of $t$ or what I got above is enough? –  Koba May 13 '12 at 23:36
1  
If you aren't told the values of $a_0,a_1,b_0,b_1$, you can't give a numerical value for $t$. You can deduce that your formula satisfies $t\ge0$, but it may not satisfy $t\le1$; that depends on the values of $a_0,a_1,b_0,b_1$. So you can say there is no satisfactory value of $t$ if $a_1b_0+a_0b_1-2a_1b_1\gt0$. –  Gerry Myerson May 13 '12 at 23:45

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