Chapter 5.2, Problem 56E

### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

Chapter
Section

### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Supply In Exercises 53 and 54, find the supply function x   =   f ( p ) that satisfies the initial condition. d x d p  =  400 ( 0.02 p   −   1 ) 3 x  = 10,000 when  p   =   $100 To determine To calculate: The demand function x=f(p) for which dxdp=400(0.02p1)3 that satisfies the initial condition x=10,000 when p=$100.

Explanation

Given Information:

The provided equation of derivative of the demand function is dxdp=400(0.02p1)3 and the initial

condition is x=10,000, when p=\$100.

Formula Used:

According to the general power rule for integration,

If u is a differentiable function of x, then

undu=un+1n+1+C

where n1

Calculation:

Consider the equation dxdp=400(0.02p1)3.

Integrate dxdp to obtain x.

x=400(0.02p1)3dp

x=20,0000.02(0.02p1)3dp …… (1)

Let 0.02p1=u.

Differentiate with respect to p,

ddp(0.02p1)=dudpddp(0.02p)ddp(1)=dudp0.02=dudp0.02dp=du

Substitute the value of 0.02dp=du and 0

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