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Interactive video lesson plan for: optimization and modelling ap calculus pt V

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For more cool math videos visit our site at http://mathgotserved.com or http://youtube.com/mathsgotserved
Good day students welcome to mathgotserved.com in this clip were going to be going over an example on how to carry out to my station on Madeleine on using derivatives this is going to be part five of our multi part series on optimization and motherly feel free to visit our website@mathgotserved.com for access to a wide the rights you of tutorials ranging from algebra all the way to calculus listed below are the steps that we going to use to carry out our optimization and motherly process so step number one you create the model by using a diagram and on the optimization function in the constraint functions altogether step two you are going to narrow value function is to just one function by substituting values from your constraint function into your on optimization of primary function in step three you find the potential extrema the endpoints and the critical points in part four
classify the extrema using the second derivative By Evaluating the Value of the Function of the Critical Points in the Endpoints and the Lastly You Interpret Your Result by Concluding with the Indication of What Your Result Means within the Context of the Problem.
Our right so for problem number five the problem is as follows Michael once you the rectangle can for his chickens is ten rugby fields to equals the region were one side is a pre-existing will so we going to be building a three sided fence that's connected to a wall right is Michael has 500 feet fencing material what are the dimensions of the can that we yield the largest area and what is the largest area right okay so on this groundbreaking up using the steps like you do earlier you have to do this we by any means this is how I like to organize the problem price step number one we going to create a lot of other cool model of the situation so the first you need to look at is the optimization function right so optimization function okay so what are we optimizing we looking for the largest area okay what is the shape we do with the rectangle okay so we're looking for the largest area of the rectangle so the area of our rectangle is length times with our right to this is our optimization function and then we Constraint on the dimensions of our parents right so what are the constraints while in order to generate this we need to look at the all shape of the situation right so let's go ahead and sketch it we have this is the wall of the that's you alright there be pre-existing wall and anyhow the rectangular pan like this on the wall okay now how can we label on our rectangular and Scotus the wall both can call this X in X-Men that the the length and then what the with going to be is the rectangle won't is made up of fencing that is 500 feet long be take out X and X from X what will you have left you will have five hundred minus the two X in the case of the entire length of the fence which is five hundred minus the two of opposite sides two X okay so W with I so this is our length right here in this is our with so we can use these are constraints on the material for building our fence to formulate our constraint functions okay so we have the length is equal to X and then the with is equal to five hundred minus two X I never optimization function in our constraints we done with part one of the best part two O part two we need to carry out the substitution that will allow us to express our optimization function in terms of exactly one variable okay we just need one independent variable here this area is dependent into variables here we need to narrow it down this to one okay so our area of the function of X is going to be length which is X times the with which is five hundred minus two X we simplify that will have area equals five hundred X minus two X square right we also have to state the domain our independent variable which is X so let's go ahead and stated domain not look at our fence also can X the X can be zero right this is such thing as the negative length for fencing so X can be zero in all you have is one long side this I just considered all the material in this portion is gone you be left with five hundred eight fencing material just of cosine zero area here okay so the smallest possible value for X is zero right so zero is less than or equal to X what is the biggest possible value X the X and X consumer all the material one how long is it going to be to see have five hundred and these to consume all the material when a minute we know that limit take the entire material 500 feet of fencing and split it into two sides which is going to be two hundred and fifty so the upper bound four X these two h

Tagged under: Calculus Optimization Maximum,increasing,Minimum,global,volume,point inflection,area,patrickJMT fencing,Khan Academy box,related rates domain,feasible,open,primary,secondary,endpoints,critical points,applicaton,distance,point,inscribed,cone,sphere,surface,absolute,antiderivative,concave ,concace

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