BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 2, Problem 4T

(a)

To determine

To find: The value of R(2) and R(4) and describe represented values.

Expert Solution

Answer to Problem 4T

The value of R(x) at R(2) is 4000 and at R(4) is 2000  Value represent the total sale revenue with prices of $2 and $4 .

Explanation of Solution

Given:

The given function is,

R(x)=500x2+3000x

Calculations:

The given function is,

R(x)=500x2+3000x

Substitute 2 for x in the above function to find the value of R(2) ,

R(2)=500×(2)2+3000×2=500×4+3000×2=4000

The value of R(x) at R(2) is 4000 .

Substitute 4 for x in the above function to find the value of R(4) ,

R(4)=500×(4)2+3000×4=500×16+3000×2=2000

The value of R(x) at R(4) is 2000 .

Thus, the value of R(x) at R(2) is 4000 and at R(4) is 2000 . Value represent the total sale revenue with prices of $2 and $4 .

(b)

To determine

To sketch: The graph of function R and what happens to revenue as the price increases from 0 to 5 dollars.

Expert Solution

Explanation of Solution

The given function is,

R(x)=500x2+3000x

Substitute some value of x and make a table for x and R(x).

x R(x)
0 0
1 2500
2 4000
3 4500
4 4000
5 2500

The graph of function R(x) is shown in Figure (1),

Precalculus: Mathematics for Calculus - 6th Edition, Chapter 2, Problem 4T , additional homework tip  1

Figure (1)

Hence, the revenue increase up to $3 and decreases after $3 .

(c)

To determine

To find: The maximum revenue, and price.

Expert Solution

Answer to Problem 4T

The maximum revenue is $4500 and it can achieve at price $3 .

Explanation of Solution

The given function is,

R(x)=500x2+3000x

The graph of function R(x) is shown in Figure (1).

Precalculus: Mathematics for Calculus - 6th Edition, Chapter 2, Problem 4T , additional homework tip  2

Figure (1)

Form Figure (1), the $4500 is the peak value of the graph at R(3) .

Hence, the maximum revenue is $4500 and it can achieve at price $3 .

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