All units in a 100-unit apartment building are rented out when the monthly rent is set at r = $520 per month. Suppose that one unit becomes vacant with each $10 increase in rent and that each occupied

Let x be the number of units rented out at a rent of r, then the number of vacant units is 100 - x. The revenue earned from rentals is therefore r*x. As the rent increases by $10, one unit becomes vacant. The revenue from this vacant unit is 0, but the maintenance cost remains at $80. Thus, the profit from each $10 increase in rent is (r+10)*x - r*(x-1) - 80. The total monthly profit P is therefore: P = r*x - 80*(100-x) + (r+10)*(x-1) - r*x Simplifying, we get: P = -80x + 10x - 800 P = -70x - 800 To maximize profit, we need to find the value of x that gives the maximum value of P. Taking the derivative of P with respect to x and setting it to 0, we get: dP/dx = -70 = 0 Therefore, x = 100/7. Since we can only rent out whole units, we round down to x = 14. The corresponding rent is r = $680. Thus, the monthly profit is: P = 680*14 - 80*86 + 690*13 - 680*14 P = $6,280 Therefore, the rent that maximizes monthly profit is $680, and the maximum monthly profit is $6,280.