Results 1 to 2 of 2

Math Help - Indefinite Integration

  1. #1
    Member
    Joined
    Nov 2007
    Posts
    232

    Indefinite Integration

    Let f(x) represent the total number of items a subject has memorized x minutes after being presented with a long list of items to learn. Psychologists refer to the graph of y = f(x) as a learning curve and to f'(x) as the learning rate. The time of peak efficiency is the time when the learning rate is maximized. Suppose the learning rate is

    f'(x) = 0.1 (10 + 12x - 0.6x^2) for 0 \leq x \leq 25

    a) When does peak efficiency occur? What is the learning rate of peak efficiency?

    b) What is f(x)?

    a) Find f(0), f(25)

    0 = 0.1(10 + 12x - 0.6x^2)

    10 + 12x - 0.6x^2 = 0

    12x - 0.6x^2 = -10

    x = 0

    1 - 200x = 0

    1 = 200x

    \frac{1}{200} = x

    ...but the answer is x = 10 and I don't even know where this number comes from! Plus, the critical points I found are outside the restrictions...

    b) \int 1 \, dx + \int 1.2x \, dx - \int 0.06x^2 \, dx

    = x + 0.6x^2 - 0.02x^3 + c

    ...my question is why is there no "c" (arbitrary constant)??



    =====================


    A manufacturer estimates marginal revenue to be 200q^{-\frac{1}{2}} dollars per unit when the level of production is q units. The corresponding marginal cost has been found to be 0.4q dollars per unit. If the manufacturer's profit is $2,000 when the level of population is 25 units, what is the profit when the level of production is 36 units?

    \int  200q^{-\frac{1}{2}} + 0.4q\, dq

    = 400q^{\frac{1}{2}} + 0.2q^2 + c

    2000 = 400(25)^{\frac{1}{2}} + 0.2(25)^2 + c

    2000 = 2125 + c

    -125 = c

    R(q) = 400q^{\frac{1}{2}} + 0.2q^2 - 125

    R(36) = 2534.20

    ...the right answer is $2265.80, what am I doing wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jan 2009
    From
    Australia
    Posts
    59
    Quote Originally Posted by Macleef View Post
    Let f(x) represent the total number of items a subject has memorized x minutes after being presented with a long list of items to learn. Psychologists refer to the graph of y = f(x) as a learning curve and to f'(x) as the learning rate. The time of peak efficiency is the time when the learning rate is maximized. Suppose the learning rate is

    f'(x) = 0.1 (10 + 12x - 0.6x^2) for 0 \leq x \leq 25

    a) When does peak efficiency occur? What is the learning rate of peak efficiency?

    b) What is f(x)?

    a) Find f(0), f(25)

    0 = 0.1(10 + 12x - 0.6x^2)

    10 + 12x - 0.6x^2 = 0

    12x - 0.6x^2 = -10

    x = 0

    1 - 200x = 0

    1 = 200x

    \frac{1}{200} = x

    ...but the answer is x = 10 and I don't even know where this number comes from! Plus, the critical points I found are outside the restrictions...
    You need to find the maximum in f'(x) so what you need to calculate is the value of x for which f''(x) = 0

    f'(x) = 0.1 (10 + 12x - 0.6x^2)

    f''(x) = 0.1 (12 - 1.2x)

    So solving for f''(x) = 0 you get x=10 so the only critical point occurs for this value. Now determine if it is a maximum or minimum you need to look at the next derivative, i.e. f'''(x) you will find the sign is negative and therefore x=10 is a global maximum.

    The rate of peak efficiency is just

    f'(10) =  0.1 (10 + 12\times 10 - 0.6\times 10^2)

    Quote Originally Posted by Macleef View Post
    b) \int 1 \, dx + \int 1.2x \, dx - \int 0.06x^2 \, dx

    = x + 0.6x^2 - 0.02x^3 + c

    ...my question is why is there no "c" (arbitrary constant)??
    One may assume that at time x=0 the subject has not memorised anything so

    f(0) = 0

    This can be used to eliminate the constant. ( c=0).

    Quote Originally Posted by Macleef View Post
    A manufacturer estimates marginal revenue to be 200q^{-\frac{1}{2}} dollars per unit when the level of production is q units. The corresponding marginal cost has been found to be 0.4q dollars per unit. If the manufacturer's profit is $2,000 when the level of population is 25 units, what is the profit when the level of production is 36 units?

    \int  200q^{-\frac{1}{2}} + 0.4q\, dq

    = 400q^{\frac{1}{2}} + 0.2q^2 + c

    2000 = 400(25)^{\frac{1}{2}} + 0.2(25)^2 + c

    2000 = 2125 + c

    -125 = c

    R(q) = 400q^{\frac{1}{2}} + 0.2q^2 - 125

    R(36) = 2534.20

    ...the right answer is $2265.80, what am I doing wrong?
    The only thing you have done wrong is to have the wrong sign for the marginal cost. As this is a cost term it should be negative. So start with

    \int  200q^{-\frac{1}{2}} - 0.4q\, dq

    and you should get the correct answer.

    Hope this helps.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Indefinite Integration
    Posted in the Calculus Forum
    Replies: 7
    Last Post: October 28th 2010, 04:21 PM
  2. Indefinite integration with u-substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 24th 2010, 07:52 AM
  3. Indefinite Integration: Sub
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 8th 2009, 03:01 PM
  4. indefinite integration
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 19th 2008, 03:05 AM
  5. Help on indefinite integration needed!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 2nd 2008, 04:06 AM

Search Tags


/mathhelpforum @mathhelpforum