Application Problem 2 - Solutions (1)

This application problem is due on Friday 9/15 at 11:59pm. Please see the Specifications and Sample Solution on Canvas for a guide to what your submission should look like. As a reminder, it can be hand-written or typed, and you should upload your submission to Canvas. Application Problem 2 An apartment manager keeps careful records of how the rent charged per unit corresponds to the total number of occupied units in a large apartment complex. This collected data is shown in the table below: Monthly Rent $650 $700 $750 $800 $850 $900 Occupied Apartmen ts 203 196 189 182 175 168 1. Why is it reasonable to say that the number of occupied apartments is a linear function of rent? a. Yes, it is reasonable to say that the number of occupied apartments is a linear function of rent. They are directly interconnected. The amount of occupied apartments (y-axis) is dependent on the cost of the monthly rent (x axis). The higher the rent gets, the less tenants there are in the apartment. The lower the rent is, the more tenants there are in the facility. This function, however, is linear as well as decreasing with a rate of change of -7/50. As the rent increases by $50, a group of 7 tenants leaves. This can be calculated through the rate of change formula which is f(b)-f(a)/b-a. Plugging the values in, I came up with 196-203/700-650 which then simplifies into -7/50. With this answer and reasoning, we can definitely say that the number of occupied apartments is a linear function of rent. 2. Let A be the number of occupied apartments and R the monthly rent charged (in dollars). What is the slope of the linear function A ( R )? What is the meaning of the slope in the context of this question? The slope is negative – why does that make sense in this context? a. The slope of the linear function A(R) is equal to -7/50. Once again, this is calculated through the rate of change formula which is f(b)-f(a)/b-a. Plugging the values in, I came up with 196-203/700-650 which then simplifies into -7/50.This is the slope as well as the rate of change for the function. This slope represents the relationship between the people leaving the apartment and the increase of the rent. This slope being negative makes sense in this context because the amount of tenants living in the apartment complex (y-axis) is decreasing as the rent (x-axis) increases. 3. Determine a formula for the function A ( R ). At what monthly rent will there be no interested renters?

a. The formula for the function A(R) is A(R)= -7/50R + 294. We have already established the slope of the function which is -7/50, now we need to determine the y-intercept to position the line correctly. In order to do this, we can use the formula A-A o =m(R-R o ). The m stands for the slope, so we can plug the found slope into the equation to create A-A o =-50/7(R-R o ). For the A o and R o , we only need to plug in values from the table. Considering the number of residents is dependent on the cost of rent, we can assume that the cost (R) represents the x axis and the number of residents (A) represents the y-axis. Therefore, we can further this equation by plugging in points. The points I used were (650, 203). This gives us the equation A-203=-7/50(R-650). Using the distributive property to multiply -7/50 with R and -650, we are left with A-203=-7/50R+91. We now need to isolate A, we do this by transferring the number -203 to the other side of the equation (note: this also makes the number positive instead of negative). Adding the 230 to the only like number (number without a variable: 91) we are left with the sum 294. This is what the equation looks like now A=-7/50R+294. This is the completed function for A(R). I also confirmed this in desmos and all of the points in the table line up correctly when graphed. The monthly rent cost in which no buyers would be interested is $2100. We have already found the function of A(R) and we can use that to find our point where there will be no buyers. Setting our function equation equal to zero (0=-7/50R+294) we have replaced the number of tenants (A) with 0 tenants. This will tell us the cost of apartments that no buyer would be interested in. We now need to subtract 294 from both sides to isolate the slope (-294=-7/50R). Now we need to divide both sides of the equation by -7/50 to isolate the variable R (-294/-7/50=-7/50R/-7/50). This leaves us with 2100=R. R stands for the cost of rent, which means $2100 is the cost of rent that no buyer is interested in. 4. What do you think is a reasonable domain for the function? Why? a. A reasonable domain for this function would be [0, 2100]. Considering it is not realistic for our domain to contain negative numbers in this instance (as it is impossible to charge negative amounts of rent for each apartment), the lowest number in our domain must be 0 (lowest price must be 0). If we plug 0 into the R (meaning the rent is zero, the minimum), we then get the amount of renters that would be interested in moving in, 294 people. I did this by, again plugging 0 into R which gave me -7/50(0)+294. 0x-7/50 is equal to zero, and 0+294 is 294. This means that A(R) (the number of interested renters) is 294, which is the maximum number of people that can live in the apartment because the price is at the lowest it can be. I can then find the second number of the domain by finding the maximum price that the landlord can charge where no buyers are interested by setting the function A(R)=0 (meaning that the number of renters is 0), this will give us our second (maximum) price. Plugging in zero to the y-value, we get 0=-7/50R+294. First, I subtracted 294 from both sides to isolate the slope (-294=-7/50R). Now we need to divide both sides of the equation by -7/50 to isolate the variable R (-294/-7/50=-7/50R/-7/50). This leaves us with 2100=R. With this method, I was able to find the amount of money that could be charged per apartment where there would be no interested buyers. This then solidifies my domain of the minimum and maximum amount of money that could be charged: [0, 2100]. 5. If the rent were increased to $1000, how many occupied apartments should the