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Math Help - Optimization and calculus

  1. #1
    GoGo
    Guest

    Help??? - Optimisation and calculus

    please if anyone could help that would be great!
    The question is about optimisation/calculus.
    Attached Thumbnails Attached Thumbnails Optimization and calculus-maths-problem.jpg  
    Last edited by GoGo; August 13th 2005 at 02:11 PM. Reason: Forgot to add attachment
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    By your "...about optimisation/calculus." here, you mean using Calculus to find the maximum F based on v. The Calculus part is getting the 1st derivative of F with respect to v. This is dF/dv, or just F'.
    This F' is also the slope of the tangent line. When this slope is zero, the tangent line is horizontal and the F is either maximum or minimum. So, we set F' to zero to get the optimum F.
    I assume you know all that, and you know how to get derivatives.

    F = (0.27v) / (4.2 +0.1905v +0.006v^2)

    Differentiate both sides with respect to v,

    dF/dv = F' = [(4.2 +0.1905v +0.006v^2)*{0.27} -(0.27v){0.1905
    +0.006(2v)] / [(4.2 +0.1905v +0.006v^2)^2]

    F' = [1.134 +0.051435v +0.00162v^2 -0.051435v -0.00324v^2] / [(4.2 +0.1905v +0.006v^2)^2]

    F' = [1.134 -0.00162v^2] / [(4.2 +0.1905v +0.006v^2)^2]

    Set that to zero,
    0 = [1.134 -0.00162v^2] / [(4.2 +0.1905v +0.006v^2)^2]
    Multiply both sides by [(4.2 +0.1905v +0.006v^2)^2],
    0 = 1.134 -0.00162v^2
    0.00162v^2 = 1.134
    v^2 = 1.134 / 0.00162 = 700
    v = sqrt(700) = 26.46 km/hr. ----the v for max or min F

    To see if F is maximum or minimum, we check the slope of the tangent line before and after v = 26.46 km/hr.

    At v = 26 km/hr,
    F'(26) = [1.134 -0.00162(26^2)] / [(4.2 +0.1905(26) +0.006(26^2))^2]
    F'(26) = [0.0389] / [174.478] = 0.000223 ----------(positive)***

    At v = 27 km/hr,
    F'(27) = [1.134 -0.00162(27^2)] / [(4.2 +0.1905(27) +0.006(27^2))^2]
    F'(27) = [-0.0470] / [188.170] = -0.00025 --------(negative)***

    That means as we go past the point where v = 26.46, the slope of the tangent line changes from positive to negative.
    So at v = 26.46, the F is "top of the hill" or maximum.

    Therefore, for maximum F, the speed v must be 26.46 km/hr. But since speed limits are in whole numbers, then the speed limit to be posted should be 26 km/hr. -----answer.

    **Correction:
    I think speed limits are in intervals of 5, so the speed limit should be 25 kph.

    ----------------------
    Task 10.

    Show how the formula
    F = (0.27v) / (4.2 +0.1905v +0.006v^2) -----(i)
    was developed from
    (No. of cars)/sec = speed / (total stopping distance) ---(ii)

    What is the "total stopping distance"?
    I'd say it is the Reaction Distance plus the Braking Distance.

    Anayzing the table as shown, we can see that:
    >>>Reaction Distance = 0.1905 times v = 0.1905v.
    >>>Braking Distance = 0.006 times v^2 = 0.006v^2.
    So, for me, the total stopping distance is
    = 0.1905v +0.006v^2
    Hence, Eq.(ii) would be:
    The theoretical formula, based on Eq.(ii) is
    F = v / [0.1905v +0.006v^2] ---***

    My questions now are, in the final formula or in Eq.(i),
    >>>why is there a coefficient of 0.27 slapped on the speed in the numerator?
    >>>and why is 4.2 m was added in the total stopping distance in the denominator?

    Apparently, these 0.27 and 4.2 were introduced in the final formula as "regulating factors" on the theoretical formula. Could be that a design Code was applied, and in so doing, the 0.27 and 4.2 were incorporated.

    The addition of the 4.2 m in the final denominator could be explained by "insuring" that the 4.2 m distance between any two cars is maintained even if the car infront failed to react and brake.

    While that of the 0.27 could be for density per the passing/diversion lane. It could have something to do with how far apart or near apart are the cars sideways in the passing/diversion lane. Or, whatever.
    Last edited by ticbol; August 13th 2005 at 11:47 PM.
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