Answer with Step-by-step explanation:
We are given that a demand equation is given by
[tex]q=8000-100p[/tex]
We have to find the maximum marginal revenue for the given production levels.
[tex]R(q)=qp[/tex]
[tex]100p=8000-q[/tex]
[tex]p=80-\frac{q}{100}[/tex]
Substitute the value then we get
[tex]R(q)=q(80-\frac{q}{100})[/tex]
[tex]R(q)=80q-\frac{q^2}{100}[/tex]
Differentiate w.r.t q
Then, we get
Marginal revenue,[tex]R'(q)=80-\frac{q}{50}[/tex]
[tex]R'(q)=0[/tex]
[tex]80-\frac{q}{50}=0[/tex]
[tex]\frac{q}{50}=80[/tex]
[tex]q=4000[/tex]
Again differentiate
[tex]R''(q)=-\frac{1}{50} <0[/tex]
Hence , the marginal revenue is maximum at q=4000
Now, the maximum marginal revenue is given by
[tex]R'(4000)=80-\frac{4000}{50}=0[/tex]
