# Optimization and Rent

The manager of a large apartment complex knows from experience that 110 units will be occupied if the rent is 342 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 9 dollar increase in rent. Similarly, one additional unit will be occupied for each 9 dollar decrease in rent. What rent should the manager charge to maximize revenue? I know the rate of change for people relative to rent, but how do I express it mathematically? How do I solve the equations with symbols.I would multiply by the number of increments.

If I was programming, this would be my pseudocode: Prompt rent, people Revenue_0 = rent * people; dp/dr = -1/9; for n = 0, n = (rent+9), n++ dR/dr = (rent*(dp/dr*n))+Revenue_0= 0 revmax = Revenue @ DR/dr = 0 display ("Revmax =",revmax) I am trying to find the maximum value for revenue, but don't know how to include the number of times the rent is increased by $9 into the formula. Is my reasoning correct, or am I mistaken? ## 2 Answers Let U(r) be the units occupied and r be the rent. Translate your conditions into symbols: you know for every 9 dollars r is increased, 1 more unit will be vacant, and for every 9 dollars it is decreased, one more will be occupied. So if you increase rent 9k dollars from a baseline of$r_0$, k more units will be vacant. So in symbols:$U(r_0+9k)=U(r_0)-k$and$U(342)=110$. Let$r_0=342$and$r_0+9k=r$(rent). Solve this second equation for$k$:$k=\frac{r-342}{9}$. You then have$U(r)=110-\frac{r-342}{9}$. Revenue is Price$\times$Quantity, or$rU(r)$. Let$R(r)=revenue =r\cdot U(r)\$. Now use the tools of calculus to optimize.

Let x be 0,1,2,3,... Then 9 dollars increase of rent leaving one extra apartment empty, means that profit = (342+9x)(110-x) Working out gives a parabola that opens down. The vertex is the maximum and that is what you are looking for. Use -b/2a from High school. No calculus needed!