$122,500
1) Since the revenue is given by this product: price x quantity, we can write it out and plug into that the given data:
[tex]\begin{gathered} R=P\cdot Q \\ R=(30+x)(4000-100x) \end{gathered}[/tex]
2) Now, let's expand those factors:
[tex]\begin{gathered} R=(30+x)(4000-100x) \\ R=120,000-3000x+4000x-100x^2 \\ R=-100x^2+1000x+120,000 \end{gathered}[/tex]
The maximum total revenue is given by the y-coordinate on the Vertex of that parabola. Let's use another formula for that:
[tex]Y_V=-\frac{\Delta}{4a}=\frac{-(1000^2-4(-100)(120000))}{4(-100)}=122500[/tex]
3) Hence, the maximum revenue would be $122,500
