Sunday, October 11, 2009

8th post

this week was about reviewing and doing a billion packets.......

review:

The derivative:
of a function at a chosen input value describes the best linear approximation of the function near that input value. For a real-valued function of a single real variable, the derivative at a point equals the slope of the tangent line to the graph of the function at that point.

optimazation:
In the simplest case, this means solving problems in which one seeks to minimize or maximize a real function by systematically choosing the values of real or integer variables from within an allowed set. This formulation, using a scalar, real-valued objective function, is probably the simplest example; the generalization of optimization theory and techniques to other formulations comprises a large area of applied mathematics. More generally, it means finding "best available" values of some objective function given a defined domain, including a variety of different types of objective functions and different types of domains.

Rolle's theorem:
essentially states that a differentiable and continuous function, which attains equal values at two points, must have a point somewhere between them where the slope of the tangent line to the graph of the function is zero.

the mean value theorem:
states, roughly, that given a section of a smooth (differentiable) curve, there is at least one point on that section at which the derivative (slope) of the curve is equal (parallel) to the "average" derivative of the section.[1] It is used to prove theorems that make global conclusions about a function on an interval starting from local hypotheses about derivatives at points of the interval.


okay so im still having problems with optimazation so if anyone can put examples up that would be helpful.

1 comment:

  1. ok for the optimization question you have...lets look at number four on the test we just got back.....[3.7 Homework Quiz .. out of 20 points]

    so we know that we have a perimeter of 24 right...well to make your secondary and primary equation you make two squares...we know that the area of a square is lenth times width! and that there is two lengths and two widths for each square so your seconday equation will be 2(2L+2w)...that's your perimater because you're adding them...and since we know that the sum of the two perimeters is 24 you set 2(2L+2w) equal to 0!
    so now...dealing with optimization:
    take your secondary equation and solve for...lets go with length
    2[2(L)+2(w)] = 0
    24=4(L)+4(w)
    24-4(w)=4(L)
    (L)=6-w
    so now plug in L to the primary equation which is your area...so that's
    A=(L)(w)
    A=(6-w)(w)
    A=6w-(w)^2
    TAKE THE DERIVATIVE OF THAT:
    A'=6-2w
    -2w=-6
    w=3
    SO NOW TAKE WHAT YOU KNOW....W=3....and PLUG IT INTO THE SECONDARY FORMULA:
    24=2(2(L)+2(w)
    24=2(2(L)+2(3)
    24=4L+12
    12=4L
    L=3

    so knowing your W is 3 and your L is 3
    [it's a square so they had to be the same]
    you know that the width is three and the length is three!!

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