   Chapter 10.3, Problem 10E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# A company handles an apartment building with 70 units. Experience has shown that if the rent for each of the units is $1080 per month, all the units will be filled, but 1 unit will become vacant for each$20 increase in the monthly rate. What rent should be charged to maximize the total revenue from the building if the upper limit on the rent is $1300 per month? To determine To calculate: The rent for maximum revenue from the building if there are 70 units in the building, all the units will be filled up if the rent for each unit is$1080.

Explanation

Given Information:

There are 70 units in the building, all the units will be filled up if the rent for each unit is $1080 but one unit will be vacant for each$20 increase in the monthly rate and the upper limit on the rent is \$1300.

Formula used:

If f(x) and g(x) are two differentiable functions then by the property of derivative:

ddx(f(x)+g(x))=ddxf(x)+ddxg(x)

And

ddxxn=nxn1

Where n is a constant and x is the variable.

Calculation:

Assume that there are x vacant units in the building then the total revenue from the building can be written as:

Revenue=(rent for one unit)(total number of units which are rented)=(1080+20x)(70x)R(x)=75,600+320x20x2

The absolute maxima and absolute minima will occur only at the critical points. To calculate the critical points of the revenue function find the first derivative of the function:

R(x)=75,600+320x20x2ddx(R)=ddx(75,600+320x20x2)

Use ddx(f(x)+g(x))=ddxf(x)+ddxg(x),

R(x)=ddx(75,600)+ddx(320x)&

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