Chapter 4.6, Problem 48E

To determine

To calculate:The manager charge to maximize revenue if the manager of a $100-\text{unit}$ apartment complex known from experience that all units will be occupied if the rent is $\$800$ per month. A market survey suggests that, on average, one additional unit will remain vacant for each $\$10$ increase in rent.

Answer to Problem 48E

The monthly rent should be set at $P\left(90\right)=\$900$ in order to maximize revenue

Explanation of Solution

Given information:

The manager of a $100-\text{unit}$ apartment complex known from experience that all units will be occupied if the rent is $\$800$ per month. A market survey suggests that, on average, one additional unit will remain vacant for each $\$10$ increase in rent.

Formula used:

Revenue function = The product of per unit times the number of unit sold

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

• If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
• If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate on both sides

Calculation:

As per the given problem

Let price $P$ as a function of occupied apartment $x$

Then the demand function which represents the cost of rent is given by

  $\begin{array}{l}P\left(x\right)=800+10\left(100-x\right)\\ =1800-10x\mathrm{......}\left(1\right)\end{array}$

Recall that,

Revenue function = The product of per unit times the number of unit sold

  $\begin{array}{l}R\left(x\right)=x\times P\left(x\right)\\ =x\times \left(1800-10x\right)\\ =1800x-10x^2\end{array}$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

• If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
• If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate on both sides,

  $R^{\prime }\left(x\right)=1800-20x\mathrm{......}\left(2\right)$

Solve for $R^{\prime }\left(x\right)=0$ , and simplified

  $\begin{array}{l}1800-20x=0\\ \Rightarrow 20x=1800\\ \Rightarrow x=90\end{array}$

Differentiate equation $\left(2\right)$ with respect to $x$

  $R^{″}\left(x\right)=-20$

Therefore,

  $R^{″}\left(x\right)=-20<0$

For $x=90$ is maximum

The price at $x=90$ in equation $\left(1\right)$

  $\begin{array}{l}P\left(x\right)=1800-10\times 90\\ =1800-900\\ =\$900\end{array}$

Conclusion:

Thus the monthly rent should be set at $P\left(90\right)=\$900$ in order to maximize revenue