Finding Revenue Function

Question

An apartment complex is currently renting out 200 units at $1500/month each. For every $50 increase in monthly

rent, the number of units rented decreases by 5. Find the revenue function.

Answer 1

Typically revenue functions are revenue as a function of the number of units sold (revenue is the "y" value and number of units is the "x" value). If we use that here, we can create a table like this with the information we are told:

Meaning the slope is (1550 - 1500)/(195 - 200) = -10 making the equation y = mx + b into y = -10x + b where y is the rent charged and x is the number of units rented. To find "b" we can plug in a point (I'll pick (200, 1500)) so:

1500 = -10(200) + b or b = 3500

So y = -10x + 3500

But revenue is the number of units rented times the rent charged or x•y. Since "y" is the same as "-10x + 3500" we can say R(x) = x(-10x + 3500) or R(x) = -10x² + 3500x

However, in this case, it seems better to flip the revenue equation around and base it off the rent you charge since that is your independent variable. So, let's flip the table above around and make the slope be the change in units rented divided by the rent charged or slope = (195 - 200)/(1550 - 1500) = -1/10

In this case y is the number of units rented and x is the price charged making the linear equation y = -1/10x + b and plugging in the point (1500, 200) gives us 200 = -1/10(1500) + b or b = 350 and the linear equation is then y = -1/10x + 350 or y = -0.1x + 350. The revenue equation is then (-0.1x + 350)(x) or:

R(x) = -0.1x² + 350x where R(x) the revenue and "x" is the rent you charge.

Since you don't specify, either could be right.