My apologies for not getting around to this yesterday; I was unable to find the correct date. Anyway, yesterday, we continued working on optimization, looking at a problem with a wire. In order to maximize the area within the wire, we set the equation of the first derivative of the area equation equal to zero and solved. We also had to make sure to set the interval correctly in order to eliminate extraneous results. In the second part of the problem this was 10, because the wire left over from making the triangle to make the square was 10-x. Therefore, x could not be larger than 10 or smaller than zero in order to make sense. We also proved that the optimal points we found were extrema through using the two methods we learned earlier in class.
Finding absolute maximums and minimums on a closed interval.
A closed interval can be defined as a set of numbers [a,b] in which the points a and b are included in the set. An absolute maximum is defined as the highest given value in the function. An absolute minimum is the lowest given value in the function. A critical number is an x value of a function when the derivative 1. does not exist or 2. equals 0. More times than not, a critical number is a maximum or minimum.
Once critical points are found, we can then proceed to find absolute maximums or minimums. To find absolute extrema, look three places: 1. F'(x)= Does Not Exist 2. F'(x)= 0 3. End points (which would be "a" and "b" from the example above). Then at each of these points we can evaluate F(x) and determine whether the points are local max/min, absolute max/min, or boundaries. The highest value is the absolute maximum, the lowest value is the absolute minimum. All other non-boundary points are local min/max. If the boundaries are not the lowest or highest values then they are just simply boundaries.
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My apologies for not getting around to this yesterday; I was unable to find the correct date. Anyway, yesterday, we continued working on optimization, looking at a problem with a wire. In order to maximize the area within the wire, we set the equation of the first derivative of the area equation equal to zero and solved. We also had to make sure to set the interval correctly in order to eliminate extraneous results. In the second part of the problem this was 10, because the wire left over from making the triangle to make the square was 10-x. Therefore, x could not be larger than 10 or smaller than zero in order to make sense. We also proved that the optimal points we found were extrema through using the two methods we learned earlier in class.
Here are some good review concepts for the exam:
Finding absolute maximums and minimums on a closed interval.
A closed interval can be defined as a set of numbers [a,b] in which the points a and b are included in the set.
An absolute maximum is defined as the highest given value in the function. An absolute minimum is the lowest given value in the function.
A critical number is an x value of a function when the derivative 1. does not exist or 2. equals 0. More times than not, a critical number is a maximum or minimum.
Once critical points are found, we can then proceed to find absolute maximums or minimums.
To find absolute extrema, look three places:
1. F'(x)= Does Not Exist
2. F'(x)= 0
3. End points (which would be "a" and "b" from the example above).
Then at each of these points we can evaluate F(x) and determine whether the points are local max/min, absolute max/min, or boundaries. The highest value is the absolute maximum, the lowest value is the absolute minimum. All other non-boundary points are local min/max. If the boundaries are not the lowest or highest values then they are just simply boundaries.
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