Results 1 to 3 of 3

Math Help - Calculus question (app. of differentiation)

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    23

    Calculus question (app. of differentiation)

    Hi Guys:

    1. A company produces two products whose production levels are represented by
    x

    and
    y. As a function of these two variables profit is given by:

    P(x,y)
    = 900x 30x^2 + 30xy 100y 10y^2 + 2,000
    a) What values of
    x and y maximize the profit?
    b) What is the maximum value of the profit?

    c) Verify that the solution is indeed a local maximum.

    Just want to make sure I am correct...

    a) x = 5 and y = -25
    b) -$1,750
    c) I am not sure how to go about getting this

    2. A local inventor has developed a new camera technology that uses special tubes to
    improve the quality of the pictures it takes. To produce, each camera will cost $1,000
    of fixed costs, $25 for every tube used, and $400 divided by the number of tubes used
    in each camera. The camera sells for $2,025.
    a) Write the function
    P(x) for the profit on a camera, where x represents the number
    of tubes used in each camera.
    b) What obvious restriction must be placed on
    x?
    c) Use the function in a) to determine the number of tubes that will maximize profit.

    d) Show that
    P(x) is maximized at the stationery point.

    Answers (again, tell me if and where I went wrong)

    a) R(x) = 2,025x
    C(x) = 25x + 400/x
    Revenue - Cost = Profit

    2,000 x + 400 / x

    b) x > 0

    I can't figure out C and D to see as I am pretty sure my equation is wrong haha.


    Thanks

    Ibrox
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,689
    Thanks
    617
    Hello, ibrox!

    Sorry ... your answers are way WAY off.


    1. A company produces two products whose production levels are x and y.
    The profit is given by: . P(x,y) \:=\: 900x - 30x^2 + 30xy - 100y -10y^2 + 2000

    (a) What values of x and y maximize the profit?
    (b) What is the maximum value of the profit?
    (c) Verify that the solution is indeed a local maximum.

    (a) Set the partial derivatives equal to zero, and solve.

    . . \begin{array}{ccccccc}P_x \:=\:900 - 60x + 30y \:=\:0 & \Longrightarrow & 4x - 2y \:=\:60 & {\color{blue}[1]} \\<br />
P_y \:=\:30x - 100 - 20y \:=\:0 & \Longrightarrow & 3x-2y \:=\:10 & {\color{blue}[2]} \end{array}

    Subtract [2] from [1]: . \boxed{x \:=\:50}

    Substitute into [1]: . 4(50) - 2y \:=\:60 \quad\Rightarrow\quad \boxed{y \:=\:70}

    To maximize profit, the company should produce: . \begin{array}{c}\text{50 units of product X}\\ \text{70 units of product Y}\end{array}


    (b) The maximum profit will be:
    . . P(50,70) \:=\:900(50) - 30(50^2) + 30(50)(70) - 100(70) - 10(70^2) + 2000 \;=\;\boxed{21,\!000}


    (c) To verify an extreme value, we use the Second Partials Test.

    Form the expression: . D \:=\:(P_{xx})(P_{yy}) - (P_{xy})^2
    . . and evaluate it at the critical point.

    If D < 0, saddle point.

    If D > 0, there is an extreme point.
    . . If f_{x\!x},f_{yy} < 0, maximum point.
    . . If f_{x\!x},f_{yy} > 0, minimum point.


    We have:. . P_{xx}=\text{-}60,\;P_{yy} = \text{-}20,\;P_{xy} = 30

    Hence: . D \:=\:(\text{-}60)(\text{-}20) - 30^2 \:=\:+300 . . . There is an extreme point.

    Since P_{xx} = \text{-}60 (negative), it is a maximum.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2009
    Posts
    23
    Soroboth,

    Thanks, I see what I did wrong, I screwed up getting the derivative, forgot to multiple by 2 lol.

    In regards to question 2, I took a stab at it and got:

    a) -25x + 400/x + 1025
    b) x > 0
    c) x = 4
    d) Not sure what is meant by show it is maximized at stationary point.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Calculus: Differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 27th 2010, 04:13 AM
  2. Calculus Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 28th 2009, 06:30 PM
  3. Differentiation - Calculus help!
    Posted in the Business Math Forum
    Replies: 1
    Last Post: October 13th 2008, 04:31 AM
  4. Calculus - Implicit differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 27th 2008, 04:17 AM
  5. Calculus differentiation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 23rd 2008, 06:37 PM

Search Tags


/mathhelpforum @mathhelpforum