uadratic Functions Project - Profit Parabolas One of the many applications of quadratic functions in called the Profit Parabola. The Profit Parabola can be seen if we investigate the following scenario: 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 and attach it to the back of this project. 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. SHOW ALL YOUR WORK! Rent hikes NONE 1 2 3 X Rent per apartment 400 400+1(20) 400+2(20) 400+3(20) 400+x(20) Number of rentals 90 90-1(1) 90-2(1) 90-3(1) 90-x(1) Total revenue $400x90= $36,000 $420x89=$37,380 $440x88=$38,720 $460x87=$40.020 (400+20x)(90-1x) What is the formula for revenues R after x $20 price hikes in the apartment building? R(X)= (400+20X)(9O-X) -20x^2+1400x+36000 Graph the function and attach the graph. Find the maximum revenue (or income) of the apartment building. R(x)=(400+20x)(90-1x)=36000 3600+-400x+1800x+-20x^2 -20x^2+1400x+36000 H=-b/2a=-1400/2(-20)=-1400/-40=35 K+r(h)=-20(35)^2+1400(35)+36000 -24500-49000+36000=60500 MTH109 What is the rent that coincides with this maximum revenue? $1100.00 What is the outcome if the rent hike of $20 results in 2 additional vacancies instead of 1? (400+20x)(90-2x) 3600-800x+1800x-40x^2 -40x^2+1000x+36000 -1000/2(-40) -1000/-80=12.5 -40(12.5)^2+1000(12.5)+36000 -6250+12500+36000=42250 400+20(12.5) 400+250=$650 90-12.5(2)=65 Set up a similar scenario, of your own invention, using a business that you are interested in. Write up the scenario (problem) and the solution process involved. Find the solution to the problem you invented. Graph the function and attach it. MTH109