   Chapter 9.1, Problem 60E ### 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

# Revenue If the revenue for a product is R ( x ) = 100 x − 0.1 x 2 and the average revenue per unit is R ¯ ( x ) = R ( x ) x ,   x > 0 find (a) lim x → 100 = R ( x ) x and (b) lim x → 0 + = R ( x ) x

(a)

To determine

To calculate: The value of the limit limx100R(x)x where R(x)=100x0.1x2.

Explanation

Given Information:

The limit is limx100R(x)x. If the revenue for a product is R(x)=100x0.1x2 then the average revenue per unit is R¯(x)=R(x)x, where x>0.

Formula used:

For a polynomial function f(x)=anxn+an1xn1++a1x+a0 where an0 and n is a positive integer,

limxcf(x)=f(c)

This is defined for all values of c.

Calculation:

Consider the provided limit,

limx100R(x)x

Substitute 100x0

(b)

To determine

To calculate: The value of the limit, limx0+R(x)x where R(x)=100x0.1x2.

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