Results 1 to 6 of 6

Math Help - Business Calculus

  1. #1
    Member
    Joined
    May 2009
    Posts
    93

    Lightbulb Business Calculus

    For speeds between 40 and 65 miles per hour, a truck gets 480/x miles per gallon when driven at a constant speed of x miles per hour. Diesel gasoline costs $2.23 per gallon, and the driver is paid $15.10 per hour. What is the most economical constant speed between 40 and 65 miles per hour at which to drive the truck?


    I would just like to know if I calculated this right...thanks!



    D(S)=P/S
    C(S) ?= G/(480/S)
    C(48) ?= G/(480/48)
    C(48) ?= G/10
    At 10 MPG, each mile costs 1/10 of a gallon of gas
    C(S)=SG/480
    F(S)=P/S+SG/480
    F'(S)= -P/S^2+G/480
    P/S = PS^(-1), DS/DY(PS^(-1))=-1PS^(-2)
    DS/DY(SG/480)=G/480
    Min[F'(S)] @ 0=-P/S^2+G/480, solved for S

    P/S^2 = G/480
    S^2/P = 480/G
    S^2 = 480P/G

    S=|SQRT(480P/G)
    For P=15.10 and G=2.23, we get S=SQRT(480*15.1/2.23)= ~57.011 MPG.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,928
    Thanks
    757
    Quote Originally Posted by lisa1984wilson View Post
    For speeds between 40 and 65 miles per hour, a truck gets 480/x miles per gallon when driven at a constant speed of x miles per hour. Diesel gasoline costs $2.23 per gallon, and the driver is paid $15.10 per hour. What is the most economical constant speed between 40 and 65 miles per hour at which to drive the truck?


    I would just like to know if I calculated this right...thanks!



    D(S)=P/S
    C(S) ?= G/(480/S)
    C(48) ?= G/(480/48)
    C(48) ?= G/10
    At 10 MPG, each mile costs 1/10 of a gallon of gas
    C(S)=SG/480
    F(S)=P/S+SG/480
    F'(S)= -P/S^2+G/480
    P/S = PS^(-1), DS/DY(PS^(-1))=-1PS^(-2)
    DS/DY(SG/480)=G/480
    Min[F'(S)] @ 0=-P/S^2+G/480, solved for S

    P/S^2 = G/480
    S^2/P = 480/G
    S^2 = 480P/G

    S=|SQRT(480P/G)
    For P=15.10 and G=2.23, we get S=SQRT(480*15.1/2.23)= ~57.011 MPG.
    I get 57.01 mph
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2009
    Posts
    93

    Unhappy

    The teacher said this was wrong and this is what she said....please help!


    is still way off....Start with
    Let D be the distance traveled in miles and x the speed in miles per hour. The time of travel is
    t = D/x and the driver is paid $15.10D / x. The number of gallons of gasoline is
    D miles / (480/x miles/gallon) = Dx / 480 gallons.
    The cost of travel is C = 15.1D / x 2.23Dx/480
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by lisa1984wilson View Post
    The teacher said this was wrong and this is what she said....please help!


    is still way off....Start with
    Let D be the distance traveled in miles and x the speed in miles per hour. The time of travel is
    t = D/x and the driver is paid $15.10D / x. The number of gallons of gasoline is
    D miles / (480/x miles/gallon) = Dx / 480 gallons.
    The cost of travel is C = 15.1D / x 2.23Dx/480
    Which is wrong the cost of travel is:

    C(x,D)=\frac{15.10 D}{x} + \frac{2.23 D x}{480}

    You can set D=1 as everything on the right is proportional to D and the optimum speed is the same whatever the distance.

    Then you need to solve:

    \frac{d}{dx} C(x,1)=-\frac{15.10}{x^2}+\frac{2.23}{480}=0

    which appears to give x=57.0107 mph

    CB
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by lisa1984wilson View Post
    For speeds between 40 and 65 miles per hour, a truck gets 480/x miles per gallon when driven at a constant speed of x miles per hour. Diesel gasoline costs $2.23 per gallon, and the driver is paid $15.10 per hour. What is the most economical constant speed between 40 and 65 miles per hour at which to drive the truck?


    I would just like to know if I calculated this right...thanks!



    D(S)=P/S
    C(S) ?= G/(480/S)
    C(48) ?= G/(480/48)
    C(48) ?= G/10
    At 10 MPG, each mile costs 1/10 of a gallon of gas
    C(S)=SG/480
    F(S)=P/S+SG/480
    F'(S)= -P/S^2+G/480
    P/S = PS^(-1), DS/DY(PS^(-1))=-1PS^(-2)
    DS/DY(SG/480)=G/480
    Min[F'(S)] @ 0=-P/S^2+G/480, solved for S

    P/S^2 = G/480
    S^2/P = 480/G
    S^2 = 480P/G

    S=|SQRT(480P/G)
    For P=15.10 and G=2.23, we get S=SQRT(480*15.1/2.23)= ~57.011 MPG.
    You won't get full marks for this becaise:

    1. You don't explain what each variable/symbol means before or when you use it.

    2. You don't explain what you think you are doing at each major step.

    What you are presenting is just a jumble of calculation and leaving it to the reader to sort out what they think you are trying to do.

    CB
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,928
    Thanks
    757
    look at the problem in terms of units ...

    cost for the truck ...

    \frac{2.23 \, dollars/gal}{(480/x) \, miles/gal} = \frac{2.23x}{480} \, dollars/mile

    cost for the driver ...

    \frac{15.10 \, dollars/hr}{x \, miles/hr} = \frac{15.10}{x} \, dollars/mile

    total cost as a function of speed in units of dollars per mile ...

    C(x) = \frac{2.23x}{480} + \frac{15.10}{x}

    C'(x) = \frac{2.23}{480} - \frac{15.10}{x^2}

    set C'(x) = 0 ...

    \frac{2.23}{480} = \frac{15.10}{x^2}

    x = \sqrt{\frac{480 \cdot 15.10}{2.23}} = 57.01 mph

    since C''(x) > 0 , this value yields a minimum
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Business application calculus thread 1?
    Posted in the Calculus Forum
    Replies: 16
    Last Post: August 7th 2011, 08:24 AM
  2. Business Calculus Problem
    Posted in the Business Math Forum
    Replies: 2
    Last Post: December 2nd 2010, 08:07 PM
  3. Business Calculus (homework question)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 6th 2008, 11:45 AM
  4. Differentation (Business Calculus)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 3rd 2008, 12:28 AM
  5. Replies: 0
    Last Post: September 10th 2006, 06:33 PM

Search Tags


/mathhelpforum @mathhelpforum