Finding Max & Min of Optimization Problem: How To Distinguish

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SUMMARY

Finding the maximum and minimum values of an optimization problem involves identifying critical values through the first and second derivative tests. The first derivative test determines critical points where the derivative equals zero, while the second derivative test assesses the concavity of the function at these points. Specifically, if the second derivative is greater than zero at a critical point, it indicates a local minimum, whereas a value less than zero indicates a local maximum. Understanding these tests is crucial for distinguishing between maxima and minima in optimization problems.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives
  • Familiarity with critical points in functions
  • Knowledge of optimization techniques in mathematical analysis
  • Experience with evaluating functions over intervals
NEXT STEPS
  • Study the application of the first derivative test in optimization problems
  • Explore the second derivative test for identifying concavity and extrema
  • Learn about the implications of critical points in real-world optimization scenarios
  • Investigate advanced optimization techniques, such as Lagrange multipliers
USEFUL FOR

Students and professionals in mathematics, particularly those focused on calculus, optimization, and mathematical analysis. This discussion is beneficial for anyone seeking to deepen their understanding of extrema in functions.

thinkbot
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How are Finding the max and min of a optimization problem different. and how do you distinguish them in an equation?
 
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Finding the critical values is the same, and then we can use either the first or second derivative tests to determine the nature of the extremum, or whether it is actually an extremum or not.

Are you familiar with these two tests?
 
Yes I am familiar
 
thinkbot said:
Yes I am familiar

Can you briefly explain how they work? :D
 
find f^1(x) = 0 x('s)= critical numbers plus some points (a,b) or [a,b]
f(x,a,b) = extrema Larget # max smallest # min
Using f^2(x) for the critical #'s f2(x) > 0 then min and ect.
 
thinkbot said:
find f^1(x) = 0 x('s)= critical numbers plus some points (a,b) or [a,b]
f(x,a,b) = extrema Larget # max smallest # min
Using f^2(x) for the critical #'s f2(x) > 0 then min and ect.

I meant can you explain how and why the first and second derivative tests for relative extrema work, i.e., the rationale behind them.
 

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