Maximum revenue

Question

The manager of a large apartment complex knows from experience that 110 units will be occupied if the rent is 420 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 10 dollar increase in rent. Similarly, one additional unit will be occupied for each 10 dollar decrease in rent. What rent should the manager charge to maximize revenue?

Answer 1

Hi Sidney,

 

Let's begin by writing the Revenue Function, R(x), we have to maximize.  It will be based upon the number of units (x) less than the initial 110 units with a corresponding increase in revenue dollars of 10x above the initial revenue of $46,200 (110 x 420):

    R(x) = (110 - x)(420 + 10x)

 

Next, let's put this function into standard quadratic form (ax² + bx + c):

    R(x) = (110 - x)(420 + 10x)

    R(x) = (110)(420) + (110)(10x) - x(420) - x(10x)

    R(x) = 46,200 + 1100x - 420x - 10x²

    R(x) = -10x² + 680x + 46,200

 

Next, let's use the axis of symmetry formula

(x = -b/2a) for a parabola to determine the x coordinate of the function's vertex (maximum point):

    R(x) = -10x² + 680x + 46,200 

        a = -10

        b = 680

        x = -b/2a

        x = -680/2(-10)

        x = -680/-20

        x = 34 (# of units less than initial 110 units)

 

Next, let's calculate the y coordinate of the vertex by plugging in our value of 34 for x into our Revenue Function, R(x):

    R(x) = -10x² + 680x + 46,200

    R(x) = -10(34)² + 680(34) + 46,200

    R(x) = -10(1156) + 23,120 + 46,200

    R(x) = -11,560 + 23,120 + 46,200

    R(x) = 69,320 - 11,560

    R(x) = $57,760 (maximum revenue dollars)

 

Finally, let's calculate the optimum monthly rent based upon the maximum revenue dollars divided by by the previously determined number of units (34) less than the initial 110 units:

    Monthly Rent = 57,760/(110 - 34)

    Monthly Rent = 57,760/76

    Monthly Rent = $760

 

We can check our answer by seeing if the immediate higher number of units (77) and the immediate lower number of units (75) both yield a lower revenue than our optimum number of units (76): 

    77[420 + 10(110 - 77)] = 77(750) = $57,750

    76[420 + 10(110 - 76)] = 76(760) = $57,760

    75[420 + 10(110 - 75)] = 75(770) = $57,750

    

As we can see from these immediate higher and lower unit calculations, our answer of 76 units being rented at $760/month, is confirmed as yielding the maximum revenue.

 

Thanks for submitting this problem and glad to help.

 

God bless, Jordan (Romans 5:8)