   Chapter 5, Problem 33RE ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

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

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
20 views

# Production The rate of change of the output of a small sawmill is modeled by d P d t = 2 t ( 0.001 t 2 + 0.5 ) 1 / 4 ,         0 ≤ t ≤ 40 where t is the time (in hours) and P is the output (in board-feet). Find the output of the sawmill after (a) 6 hours and (b) 12 hours.

(a)

To determine

To calculate: The output of sawmill after 6 hours that rate of change of output given by,

dPdt=2t(0.001t2+0.5)1/4, 0t40.

Explanation

Given Information:

The rate of change of the output of sawmill;

dPdt=2t(0.001t2+0.5)1/4, 0t40.

The initial condition P(0)=0

Here time (in hours) is t and output (in hours) is P.

Formula used:

The power rule of integrals:

undu=un+1n+1+C (for n1)

Here, u is function of x.

The property of Intro-differential:

df(x)dxdx=f(x)

Calculation:

Consider the derivative:

dPdt=2t(0.001t2+0.5)1/4

Apply, integration on both sides:

dPdtdt=2t(0.001t2+0.5)1/4dt+C

Now apply, the power rule of integrals and the property of Intro-differential:

P(t)=1000(0.001t2+0.5)1/4d(0

(b)

To determine

To calculate: The output of sawmill after 12 hours that rate of change of output given by,

dPdt=2t(0.001t2+0.5)1/4, 0t40.

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Find more solutions based on key concepts 