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Problem

Given demand $D(p)=A-ap$, and $A,a>0$ and a fixed price $0<p_1<A/a$ by some company.

My solution so far

CS is $CS=\int_{p}^{A/a}D(p)dp=\int_{p}^{A/a}(A-ap)dp=\frac{1}{2a}(A-ap)^2=\frac{(A-ap)^2}{2a}$

Problem

Find the CS now with a tax $p_1+t>A/a$

My solution so far

Just let $p_2=p_1+t>A/a$. Insert $p_2$ into the previous result since the integration would be the same and we get $\frac{(A-a(p+t))^2}{2a}$

Main Problem

Find the DWL

My solution so far

I get the general formulas $(p_2-p_1)p_1-\int_{p_1}^{p_2}A-ap\;dp$ and $1/2 \cdot (p_2-p_1)\cdot (Q_1-Q_2)$ And then the answer with the second formula that

$1/2*((p+t)-p)*((A-ap)-(A-a*(p+t)))=1/2\cdot t(a(p+t)-ap)$

Is this approach correct throughout the taxation influence on CS and the DWL?

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I think you are making it unnecessarily hard for yourself. You can calculate the deadweight loss (DWL) simply as the difference between consumer surplus pre-tax and post tax, and also by adding tax revenue as it is not necessarily wasted if government uses it, so that would be:

$$DWL = \frac{(A−ap)^2}{2a} - \frac{(A−a(p+t))^2}{2a} +t(A-a(p+t))= A t - \frac{1}{2} \left( a t (2 p + t) \right) + t(A-a(p+t)) $$

In addition if you have further equations that explicitly model supply you should also add to the deadweight loss the difference between producer surplus pre-tax and post tax but if that is not specified it is common to assume that supply is just flat line (so there is no producer surplus).

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  • $\begingroup$ Thanks. I am trying to learn the integral way of doing this for more complex functions in the future $\endgroup$ – user31331 Dec 3 '20 at 7:46
  • $\begingroup$ @bymathformath for the integral way just subtract integral before and after $\endgroup$ – 1muflon1 Dec 3 '20 at 7:47
  • $\begingroup$ Ah okay - I see. The thing that is confusing me abit this approach is that; the previous was to find the loss in $CS$ as a cause of the tax; the next one is to find DWL. By this approach they seem the same? (I got the loss cause of tax to $-At-atp$) $\endgroup$ – user31331 Dec 3 '20 at 7:49
  • $\begingroup$ @bymathformath they should be same but in the above I think you made some small mistake somewhere (it is close to being correct but not quite - it is easier to calculate difference in two integrals than actually calculate the area of DWL $\endgroup$ – 1muflon1 Dec 3 '20 at 7:53
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    $\begingroup$ @bymathformath okay I will have look at it tomorrow, also comments should not be for this sort of question unrelated requests I edited my answer above to reflect what was discussed here in the comments and then afterwards once you have read this comment move them to the chat- if you want to bring some question to my attention you can just tag me under that specific question with @ 1muflon1 (without the space) or you can open chat and tag me discretely there so it does not create unnecessary comment $\endgroup$ – 1muflon1 Dec 16 '20 at 20:08

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