anonymous
  • anonymous
The manager of a large apartment complex knows from experience that 80 units will be occupied if the rent is 360 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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[(80-x)(360+10x)\] will be the money earned if \(x\) represents each $10 increase maximize that expand, max is at \(-\frac{b}{2a}\)
anonymous
  • anonymous
I got x=22. what do i do with that answer?
anonymous
  • anonymous
@satellite73 i got x=22. what do i do with that answer?

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anonymous
  • anonymous
Ok let's see now. We can model the situation with a function. Let's call the number of occupied units 'u' and the rent 'r'. Revenue is given by: R = ur, where u is the occupied units and r is the rent. Let's make u a function of r like this: u = mr + b We know that when r increases by 10, u decreases by 1. We also know one point that the function goes through and that is (360, 80). Another point would be (370, 79). We can now find m, the slope:\[\bf slope = \frac{ 79-80 }{370-360 }=-\frac{1}{10}\]Now let's slope for 'b' by plugging in one of the points we already have:\[\bf 80=-\frac{1}{10}(360)+b \implies b = 116\]So our function 'u' is given by:\[\bf u=-\frac{1}{10}r+116\]Plugging this in for 'u' in our Revenue relation gives us:\[\bf R(u,r)=ur \rightarrow R(r)=r \left( -\frac{1}{10}r+116 \right)=-\frac{ 1 }{ 10 }r^2+116r\]Now we have expressed the Revenue as a quadratic function in terms of a single variable. We must now find the maximum of this function in order to maximise the Revenue. We take the derivative and find where it will be 0; since this a quadratic function, there will only be a single such x at which the derivative is 0 and that will give us our maximum Revenue:\[\bf R'(r)=-\frac{ 1 }{ 5 }r+116 \rightarrow -\frac{ 1 }{ 5 }r+116 = 0 \implies r = 580\] Now plug this r value back into our Revenue function R(r) and get the maximum Revenue. That's it.
anonymous
  • anonymous
if \(x=22\) that means you should have 22 $10 increases
anonymous
  • anonymous
@satellite73 oh okay! thanks! makes sense!
anonymous
  • anonymous
in other words, instead of charging $360 you should charge \(360+220=580\)
anonymous
  • anonymous
you will make $33,640 a month
anonymous
  • anonymous
looks like @genius12 got the same answer using a different method, which i don't know
anonymous
  • anonymous
@genius12 @satellite73 you both were a bunch of help! thanks to both of you!
anonymous
  • anonymous
no problem

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