S N's question at Yahoo Answers regarding revenue maximization

[MarkFL]

Here is the question:

Quadratic functions homework help?

A hall charges \$30 per person for a sports banquet when 120 people attend. For every 10 extra people that attend, the hall will decrease the price by \$1.50 per person. What number of people will maximize the revenue for the hall ?

Please answer this question with step by step instructions and explanations :) thank you

I have posted a link to this topic so the OP can see my work.

 

[MarkFL]

Hello S N,

Let's let $C$ be the amount charged in dollars and $P$ be the number of people that attend.

We are told that we $P$ increases by 10, then $C$ decreases by 1.5, so we may state the slope of the line is:

$$m=\frac{\Delta P}{\Delta C}=\frac{10}{-1.5}=-\frac{20}{3}$$

we are given the point on the line $(30,120)$, and so using the point-slope formula, we may determine the linear relationship between $P$ and $C$ as:

$$P-120=-\frac{20}{3}(C-30)$$

which we may arrange in slope-intercept form as:

$$P(C)=-\frac{20}{3}C+320$$

Now, the revenue $R$ for the hall is the product of the charge per person times the number of people attending, hence:

$$R(C)=C\cdot P(C)=C\left(-\frac{20}{3}C+320 \right)=-\frac{20}{3}C(C-48)$$

We know the vertex of this parabolic revenue function will be on the axis of symmetry, which will be midway between the two roots, at $C=0,\,48$, which means the axis of symmetry is the line $C=\dfrac{0+48}{2}=24$.

Thus, revenue is maximized when the number of people attending is given by:

$$P(24)=-\frac{20}{3}\cdot24+320=-160+320=160$$

Thus, when 160 people attend, revenue is maximized.