3. Practice similar 4 attempts remaining. The manager of a large apartment complex knows from experience that 120 units will be occupied if the rent is $\$ 440$ per month. A market survey suggests that, on the average, one additional unit will remain vacant for each $\$ 10$ increase in rent. Similarly, one additional unit will be occupied for each $\$ 10$ decrease in rent. (Round your answers to the nearest $\mathrm{c}$ or cent, but do not round until your final computation.) a. If $x$ is the number of units rented, and $p$ is the rent per unit in dollars, what is the price-demand equation (assuming it is linear)? \[ p(x)= \] b. What is the monthly revenue function for the manager? \[ R(x)= \] c. How many apartment units should be rented to maximize the monthly revenue? Apartment units: d. What is the maximum monthly revenue for the manager? Maximum revenue: $\$$ e. What rent should the manager charge to maximize the monthly revenue? Rent: $\$$ \& per unit Submit answer

Step 1 ：Given that the price-demand equation is a linear equation that relates the price per unit (p) and the number of units rented (x). We know that 120 units will be rented if the rent is $440 per month. For each $10 increase in rent, one less unit will be rented, and for each $10 decrease in rent, one more unit will be rented. This implies that the slope of the price-demand equation is -10.

Step 2 ：We can write the equation in the form \(p = mx + b\), where m is the slope and b is the y-intercept. Substituting the given values into the equation to solve for b: \(440 = -10*120 + b\), we get \(b = 440 + 1200 = 1640\).

Step 3 ：So, the price-demand equation is: \(p(x) = -10x + 1640\).

Step 4 ：The monthly revenue function for the manager is the product of the number of units rented and the rent per unit. In terms of x and p, this is: \(R(x) = x*p(x) = x*(-10x + 1640) = -10x^2 + 1640x\).

Step 5 ：To find the number of apartment units that should be rented to maximize the monthly revenue, we need to find the maximum of the revenue function. This occurs at the vertex of the parabola represented by the function. The x-coordinate of the vertex of a parabola given by the equation \(y = ax^2 + bx + c\) is \(-b/2a\). In this case, \(a = -10\) and \(b = 1640\), so: \(x = -1640/(2*-10) = 82\).

Step 6 ：So, \(\boxed{82}\) apartment units should be rented to maximize the monthly revenue.

Step 7 ：To find the maximum monthly revenue, we substitute \(x = 82\) into the revenue function: \(R(82) = -10*(82)^2 + 1640*82 = $67,280\).

Step 8 ：So, the maximum monthly revenue for the manager is \(\boxed{67280}\) dollars.

Step 9 ：To find the rent that the manager should charge to maximize the monthly revenue, we substitute \(x = 82\) into the price-demand equation: \(p(82) = -10*82 + 1640 = $500\).

Step 10 ：So, the manager should charge \(\boxed{500}\) dollars per unit to maximize the monthly revenue.