I tried to do following question but I am not sure about my solution. Please tell me your opinions

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Since gross profit is strictly concave, I can say that $R’’(y)-C’’(y) <0$

Now I maximize the net profit and

FOC is $R’(y)-C’(y) -t=0$

SOC is hold because of strict concavity in y.

That is $MR(y)=MC(y)+t$ from FOC.

Now show that the implicit relationship btw y and t can be solved for the explicit choice function


FOC with respect to y combine with above function

$R’(y^*(t))-C’(y^*(t)) -t=0$

Derivative wrt t

$$R’’(y) (dy^*/dt)-C’’(y) (dy^*/dt)-1=0$$


which means that the output will decrease as the tax goes up.

Think again gross profit

$G\pi =R(y^*(t))-C(y^*(t))$

I am not sure about my solution after this point especially.

Suppose the firm is price taker then $R(y)=py$

So $$G\pi =py^*(t)-C(y^*(t))$$

That is, since the output will decrease as tax rises, revenue $R(y) $ will decrease as well. Accordingly gross profit will decrease.


I am not sure about the last part especially. Thank you.


Most of what you have done was correct. Here are the steps to finish the argument.

Evaluating the profits at the optimum, you get \begin{equation} \pi_G=R(y^*(t))-C(y^*(t)) \quad\text{and}\quad \pi_N=R(y^*(t))-C(y^*(t))-ty^*(t). \end{equation} Differentiate the net profits with respect to $t$: \begin{align} \frac{\mathrm d\pi_N}{\mathrm dt}&=R'(y^*)\frac{\mathrm dy^*}{\mathrm dt}-C'(y^*)\frac{\mathrm dy^*}{\mathrm dt}-t\frac{\mathrm dy^*}{\mathrm dt}-y^*\\&=\underbrace{[R'(y^*)-C'(y^*)-t]}_{=0 \text{ due to FOC}}\frac{\mathrm dy^*}{\mathrm dt}-y^*<0, \end{align} which is true assuming $y^*>0$. Differentiate the gross profits with respect to $t$: \begin{align} \frac{\mathrm d\pi_G}{\mathrm dt}&=R'(y^*)\frac{\mathrm dy^*}{\mathrm dt}-C'(y^*)\frac{\mathrm dy^*}{\mathrm dt}=[R'(y^*)-C'(y^*)]\frac{\mathrm dy^*}{\mathrm dt}<0. \end{align} This is true because from the FOC, we know that for any positive tax, $R'(y^*)-C'(y^*)>0$, and since you've correctly derived $\frac{\mathrm dy^*}{\mathrm dt}<0$, their product must be negative.


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