Answer:
The production level that maximizes Silky's profit is [tex]5000[/tex] ties.
Explanation:
Hi
First of all, as we have [tex]Q(P)=150-0.2P[/tex], we need to transcript it as price in function of the quantity so
[tex]P(Q)=\frac{150-Q}{0.2}=750-5Q[/tex]
Then we need to find income function that is [tex]I(Q)=Q*P(Q)=750Q-5Q^{2}[/tex]. After derivate it [tex]I'(Q)=750-10Q[/tex].
The optimum level is when we have [tex]MC=I'(Q)[/tex], therefore,
[tex]5Q=750-10Q[/tex], as we clear it for [tex]Q[/tex] we find that
[tex]Q=\frac{750}{15}=50[/tex], finally as we have that [tex]Q[/tex] is measured in hundreds of ties, the production level that maximizes Silky's profit is [tex]5000[/tex] ties.