## anonymous 5 years ago The manager of a large apartment complex knows from experience that 90 units will be occupied if the rent is 496 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 8 dollar increase in rent. Similarly, one additional unit will be occupied for each 8 dollar decrease in rent. What rent should the manager charge to maximize revenue?

1. anonymous

Let x be the NUMBER of dollars (not actual dollars, but the NUMBER of dollars/units) we increase the rent by. The total revenue will be given by$R(x)=(no. units) \times (rent/unit)$We're told that the number of units decreases by 1 for every $8 increase; i.e. for 8 x$1, or$(no. units) = (90-\frac{x}{8})$(note when x=8 (the NUMBER of dollars increased), the number of units falls by 1), Now, the rent per unit will be $(rent/unit)=(496+x)$(again, x is the NUMBER of dollars). Hence,$R(x)=(90-\frac{x}{8})(496+x)$I have to post before completing since the site is awkward to use...

2. anonymous

Expanding,$R(x)=-\frac{x^2}{8}+28x+44640$

3. anonymous

This is a quadratic with negative coefficient for x^2, so the extremum found will be a maximum. To find it, we differentiate R with respect to x, set the result to zero and solve for x. Doing this,$R'(x)=28-\frac{x}{4}:=0 \rightarrow x=112$

4. anonymous

The rent should be $496+112=608$per month. Lucky tenants!

5. anonymous

Feel free to 'fan' me - would appreciate the point ;)

6. anonymous

Thank you I need help with another if you don't mind A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. What is the area of the largest possible Norman window with a perimeter of 40 feet? I know we are dealing with the Perimeter and the area=area of rectangle+area of 1/2cricle

7. anonymous

Okay. Just give me a moment - I need to go make something to eat...

8. anonymous

For this one, I'll write it out and scan it. It would take a lot longer to type it out. Okay?

9. anonymous

k

10. anonymous

It's coming...

11. anonymous

It took a little longer since I misread the question and found the dimensions instead of maximal area.

12. anonymous

13. anonymous

Thank you again...

14. anonymous

No worries :)