# The relationship between gross profit and tax

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

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

$$y=y^*(t)$$

FOC with respect to y combine with above function

$$R’(y^*(t))-C’(y^*(t)) -t=0$$

Derivative wrt t

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

$$\frac{dy^*}{dt}=\frac{1}{R’’-C’’}<0$$

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.

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

Evaluating the profits at the optimum, you get $$\pi_G=R(y^*(t))-C(y^*(t)) \quad\text{and}\quad \pi_N=R(y^*(t))-C(y^*(t))-ty^*(t).$$ 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.