optimization-inscribed-rectangle-unit-4-test-application-of-derivatives-ap-calc-caluclus-6

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Good day students welcome to mathgotserved.com in this clip were going to be going over question number six on the need for test on the home the application of derivatives in this example we going to be going over an optimization problem that involves the rectangle inscribed you in the parabola okay right let's take a look at the question is that the rectangle has to obits vertices on the X axes are the other two about the X axes on the parabola why equals twelve minus X square so we are to find the dimensions this out the rectangle with the maximum area so in this problem we are optimizing area okay before we get started let's go over the steps for solving optimization problems okay so step number one you want to model the situation by using the well label diagram we can then you get diagram to create the primary function and the constraint function or functions and then step number two we want to substitute units the constraint function into the optimization function in the state your domain on the constraint on the independent variable in part three one of find your extrema okay your potential extrema the combination of your critical point and your endpoints is the intervals is close step for you classify the extrema and then lastly state the meaning of your result right somebody applying the procedures here to this problem you can see little on guide I have on the right so first thing that going to do is create a model or diagram we have the rectangle inscribed on the problem so what does that we so first thing to do let's go ahead and sketch our problem on this Iraq sketch so we are modeling right now in step one which is to model you look at this equation right here is a problem that having reflected downwards and shifted up twelve units okay so can put this is twelve and any goes down this and then together we right okay this is just a rough sketch and then we have a rectangle that inscribed in this problem right here so let's go ahead and sketchy rectangle inside the problem Right so we trying to maximize the area of that rectangle that's the goal of this problem right so let's see what we have so we need to find the dimensions of this rectangle okay so let's see we can call this X right X sine X so the entire the with of our rectangle is going to be two X units long okay so how white is our rectangle is going to be two X units long so with this to X and then combine that with the height will look at the coordinates of this point right here important of any point of the problem is given by X, Y which is twelve minus X square so the y-coordinate represents the height of our is right rectangle okay so the go like this here here that's our height and is given by twelve minus X square okay right so let's go ahead and write down our primary function the primary function is on the area which is length times with okay and then our constraint for secondary functions are the length like this constraint by will give you doing with times height so we can also called is the length of this matter so the length is swell minus X square and then always how white is it to X okay now what the next part part two we going to make a substitution want to create the function in this one variable that we can use to find the extrema okay so step two is any substituting so let's see using substitution we can make here we just substituted X were to X for the with in the nineties is an area of the function of X okay so the area of the function of X two X times twelve minus X square where the constraints on my domain also okay X get X can get was zero nine and then how Beacon X get the maximum value X can attain is the X intercept of the parabola twelve minus X square right so to find the X intercept will simply set twelve minus X square equal to zero and then we'll have X square equals twelve in then you get X by itself you have plus a minus squared of twelve which is to roots three right so the maximum value that X can attain is to roots three okay I so there goes the domain of the independent variable right now going to proceed to step three we're going to find the potential extrema okay where you have to potential extrema of the endpoints your answer is three now we need to find the critical points price critical points critical points are where all the derivative is equal to zero or does not exist okay our undefined so let's see here we have the area function two X times twelve minus X square I wanted to the product rule so is distribute so we have the area have twenty-four X minus two X to the third so the differentiate this will have a prime of X equals twenty-four minus six X square okay apply the power rule to differentiate these two terms so now is

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