How to maximize the profit for the given equation
It is a maximize profit question. Basically:
- There is a play which costs $180$
- Each attendee costs $0.4$
- Ticket price affects the overall attendance. When the ticket price is $5$, then there are $120$ attendees. If the ticket price is lowered by $0.1$, then there are $15$ more attendees.
How can I model this problem?
How can I maximize the profit? I believe it has something to do with derivatives, but I don't know how to approach it.
$\endgroup$ 11 Answer
$\begingroup$Fixed cost of the play is $180$, variable cost is $0.4$ by attendee.
Total cost is the fixed cost plus the variable cost multiplied by the number of attendees: $$c=180+0.4n$$
You can express the number of attendees as a function of ticket price: $$n=120+15\times\frac{5-p}{0.1}$$
Total revenue is the ticket price multiplied by number of attendees: $$r=pn$$
Profit is the total revenue minus the total cost: $$ \begin{align} \pi&=r-c\\ \pi&=pn-(180+0.4n)\\ \pi&=pn-180-0.4n\\ \pi&=(p-0.4)n-180\\ \pi&=(p-0.4)\left(120+15\times\frac{5-p}{0.1}\right)-180\\ \pi&=-150p^2+930p-528 \end{align} $$
You get a quadratic function of profit depending on the ticket price, whose maximum you can easily get by calculating the coordinates of the vertex of the quadratic parabola described by the function: $$ V\left(-\frac{b}{2a},\frac{b^2-4ac}{4a}\right) $$
In this case, the profit reaches its maximum of: $$\pi=\frac{b^2-4ac}{4a}=\frac{930^2-4(-150)(-528)}{4(-150)}=913.5$$ at the ticket price of: $$p=-\frac{b}{2a}=-\frac{930}{2(-150)}=3.1$$
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