If the circles of the maximum area inscribed in the region...

Key Concepts

Inscribed Circle Optimization

This concept involves finding a circle with the largest possible area that can fit entirely inside a given region. It typically requires an understanding of how boundaries constrain the circle’s placement and size, and may involve techniques from calculus, geometry, or optimization to determine the optimal position and radius.

Regions Defined by Inequalities

Regions in the plane can be described by systems of inequalities that specify which points belong to the area of interest. Understanding these regions involves determining the boundaries set by the equalities and analyzing how multiple such conditions interact to shape the overall area.

Quadratic Functions and Parabolas

Quadratic functions form parabolic curves whose properties—such as vertices, axes of symmetry, and intersections with other curves—are crucial for defining the boundaries of regions. Analyzing these curves helps in determining the limits of the region in which a shape, like an inscribed circle, must fit.

Area Calculation

Calculating the area of a bounded region is a fundamental task in geometry and calculus. Whether through integration or geometric formulas, finding the area of complex regions defined by multiple curves is essential, particularly when optimizing shapes within those regions.

Intersection of Curves

The points where curves intersect are critical in defining the boundaries of the regions under study. By determining the intersections, one can accurately map the limits of the region, which is necessary both for understanding the area available and for solving optimization problems involving inscribed figures.

Transcript

00:01 In the first part, we find out all the value of the value of a, we have a equation and zero, but the area of the bounded region flows between the parabola, y -x -negative, a -x -square, and a -y -quotes -x -quil is maximum.

00:14 So we'll work on that now.

00:16 So let's work on this, just for the graph of this and this here.

00:20 So we start with y -quels -x -negative a x -square.

00:23 So we'll get a -x -quire -negative x equals negative y.

00:28 And if we just go for committing square we just we get here x -square -negative x over a equals negative 1 over a y let's go out here so committing square now x -negative x over a then we have positive 1 over 2a square negative 1 over 4a square equals negative y over a so we have to be x negative 1 over 2 a whole square equals we have negative 1 over a y -negative 1 over 4 so the graph is here we have the vertex shifted vertex will shift so let me go to the graph here this is y this is x and y plus negative 1 over 2a y negative 1 over 4a so here the vortex like this with a vertex here and the graph is passing through origin we have x equal to 0 y equal to 0 it will become 1 over 4a square so it'll find through origin the graph here we have opening downward negative here for opening downwards like this the required graph here we have this is for white 12s negative a xx2.

01:45 Next graph is you know we have y12 x square over a why equals x square over a the graph we have given like this why equals xpure over a find the point of intersection here and point of detection is 0 .0 second point of intersection is equal here x square over a equals x negative a xpere so when just simplify this we get x equal to 0 and a over 1 plus a square because the point here so just coming out to this point here this point is x a over 1 plus a square from a y1 so if it is being y1 for you know why this require x here we have a over 1 plus a square the maximum area we have find the area here between this parabola say we have given area is 02 to source to you 02 a over 1 plus a sq we have so this parabola this one that is x negative a x square negative x square over a dx so we just integrate it now and solve this so let me give you an inch here so want to be x square over 2 negative a x cube over 3 negative 1 over a x cube over 3 negative 1 over a x cube over three and four from 0 2 a over 1 plus a squared in this and you will get result test equal to a square over 6 1 plus a square it's all here now the area here we have given as a is equal to this so we'll define a maximum area the value of a for which next a million will define up d a over a we just apply here uh there is maximum to put it is equal to so it's coming out to be we have applied here portion 2.

03:49 There will be 1 plus a square whole square times 2a negative a square times 2.

03:56 1 plus a square times 2a which will go over 1 plus a square whole square.

04:05 This equals 0.

04:07 So, by solving this we will get a and afterizing it and solving this you will get the value of a that is coming out to be.

04:18 Be equal to positive negative one we have so a greater than zero a is greater than zero we have so a is so should be one it is given the portion a greater than zero so we take a as equal to one if you go for d2a by d sq so i just not fall completely d2 a over d square i've just told me to show that the a equal to one show the maximum so what i can do here let's take here uh so might be 1 over 6 we have and then 1 plus a square whole to the power 4 times derivative of numerator so that's going to be 2a times 2 1 plus a square times 2a positive 1 plus a square whole square times 2 4 this being have got here the derivative of this and so this is bracket here next we'll have a negative is coming up to be 4 a cube times we have 2a negative 1 plus a square times we have 12 a square this we have got here and the next part is coming up to be we'll get negative 2a 1 plus a square whole square negative a square 2a dot 2 1 plus a square times 4 1 plus a square to the part 3 and next we have 2 to 8 so this we have got here all divide by whatever denominator that's all always positive i put now a equal to 1 this is positive here we have this is coming out to be a positive number then it's going to be negative number here we have negative 8 and then it's going to be negative this and then this is negative this will be positive overall you will get a negative number it's going to be less than 0 take it out it should be d2a over d s squared, it's going to be less than 0 negative.

06:36 So you will prove that a equal to 1 is the maximum.

06:41 It's the maximum.

06:43 And then define the value.

06:46 Maxim area is coming out to be.

06:47 So you go to the area that is a, a square over 6, 1 plus a square also.

06:54 This we've already got earlier.

06:56 So we just report here as1.

06:59 So a is coming out to be 1 over 6, then 2 square that is giving 1 over 24...