The business manager of a 90 unit apartment building is trying to determine the rent to be charged.

The business manager of a 90 unit apartment building is trying to determine the rent to be charged. From past experience with similar buildings, when rent is set at $400, all the units are full. For every $20 increase in rent, one additional unit remains vacant. What rent should be charged for maximum total revenue? What is that maximum total revenue? To help solve the above scenario, perform an internet search for Profit Parabola or Applications of Quadratic Functions. List the URL of one of the applications that you find. URL ___________________________________________________________________

Go to http://www.purplemath.com/modules/quadprob3.htm to see the process used for determining the quadratic function for revenues R(x) as a function of price hikes x on page 3 with the canoe-rental business problem. Use this process to determine the quadratic function that models the revenues R(x) as a function of price hikes x in the apartment building scenario above.

Expert Solution

When rent is set at $400, all the units are full. For every $20 increase in rent, one additional unit remains vacant. Let the increase in rent per apartment be $ x, where x is a multiple of 20. Then the no. of apartments rented is 90-(x/20). The total revenue R(x)= no. of apartments rented* rent per apartment = [90-(x/20)](400+x) i.e. R(x) = (1/20)(-x²+1400x+720000) = (-1/20)( x² -1400x) +720000/20 = (-1/20)( x² -2*700x +700²) +36000+ (700)²/20 = (-1/20)( x-700)²+36000+24500 =(-1/20)( x-700)²+60500.

This the equation of a downwards opening parabola with vertex at (700, 60500). Since the vertex is the highest point of a downwards opening parabola, hence a rent of $ 400+700 = $ 1100 per apartment should be charged for maximum total revenue. The maximum total revenue is $ 60500.

Alternatively, R(x) is maximum, when dR/dx = 0 and d²R/dx² is negative. Here, dR/dx = (1/20)( -2x+1400) which is equal to 0 if x = 700. Also, d²R/dx² = (-1/20)*2 = -1/10 which is always negative regardless of the value of x. Thus, the total revenue R(x) is maximum when the increase in rent per apartment is $ 700 i.e. the rent per apartment is $ 1100. The maximum total revenue is $ 60500 as above.

milcah answered 2 years ago

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