# Calculus- Economics?

A property management company manages an apartment block containing 150 units. All 150 units are rented at a monthly rate of \$460 per unit and each unit costs the property management company \$72.50/month for utilities and repairs. For every \$25 rent increase, four fewer apartments are occupied. What rent should be charged in order to realize the most profit?

Relevance

Profit is revenue minus cost.

So when at full capacity, the revenue and costs are:

R() = 150 * 460

R() = 69000

C() = 72.5 * 150

C() = 10875

For every \$25 rent increase, 4 fewer apartments are rented. Since 150 is the max, we can't have a \$25 decrease from this base.

So if we let "x" be the number of \$25 rent increases, then the rent becomes: (460 + 25x) and the number of units rented will be (150 - 4x)

With this we can come up with updated revenue and cost functions and come up with a profit function in terms of "x":

R(x) = (460 + 25x)(150 - 4x)

C(x) = 72.5(150 - 4x)

Now let's come up with the profit function and simplify:

P(x) = R(x) - C(x)

P(x) = (460 + 25x)(150 - 4x) - 72.5(150 - 4x)

P(x) = 69000 - 1840x + 3750x - 100x² - 10875 + 290x

P(x) = -100x² + 2200x + 58125

Now that we have a profit function where "x" is the input and it's a quadratic we can solve for the maximum by solving for the zero of the first derivative:

P'(x) = -200x + 2200

0 = -200x + 2200

200x = 2200

x = 11

So the maximum profit will be had when there are 11 increases of \$25. This amount becomes:

460 + 25x

460 + 25 * 11

460 + 275

\$735 per month

And they will rent out:

150 - 4x

150 - 4 * 11

150 - 44

106 units

• .

Monthly rental p for each unit if n units are occupied

( 150, 460 )

( 146, 485 )

( p, n )

( p - 485 )( 146 - 150 ) = ( 485 - 460 ) ( n - 146 )

p = ( 1397.50 - 6.25n )

Cost of occupying n units; C = 72.50n

Revenue from n units = np = ( 1372.50n - 6.25n² )

Profit, P:

P = ( 1397.50n - 6.25n² ) - 72.50n

P = 1325n - 6.25n²

P’(n) = 1325 - 12.5n

1325 - 12.5n = 0

∴ n = 106

p = ( 1397.50 - 6.25 * 106 )

p = \$735

————

• Points given are (460,150) and (485,146)

x = Rent, y = Units occupied

Slope = -4/25 = -0.16

Linear equation

y-150 = -0.16(x-460)

y = -0.16x + 223.6

R = Revenue = xy = -0.16x^2+223.6x

C = Costs = 72.5y = -11.6x+16211

P = Profit = R-C

P = -0.16x^2+223.6x+11.6x-16211

P = -0.16x^2+235.2x-16211

dP/dx = -0.32x+235.2 = 0 for maximum

x = 235.2/0.32 = 735