Show that the height of the cylinder of max volume that can be inscribed in a sphere of radius a

Show that the height of the cylinder of max volume that can...

Key Concepts

Geometric Constraints in Inscribed Figures

When one solid is inscribed in another, the dimensions of the inner solid are determined by the boundary of the outer solid. This involves using geometric relationships to relate different dimensions, such as relating the height and base radius of a cylinder to the radius of a sphere. These constraints are essential for setting up equations that describe how the inscribed solid fits within the larger one.

Formulating the Volume Function

The process begins by expressing the volume of the inscribed solid (in this case, the cylinder) in terms of its dimensions. By using the geometric constraints that relate these dimensions, the volume can often be written as a function of a single variable. This single-variable representation is crucial for applying calculus techniques to the optimization problem.

Optimization Using Calculus

Once the volume function is formulated, calculus is used to optimize it. This involves taking the derivative of the volume function with respect to the chosen variable, setting the derivative equal to zero, and solving for the critical points. This procedure identifies the dimensions that maximize the volume of the inscribed solid, which is the core of many optimization problems in geometry.

Transcript

00:01 So in this case we need to show that height of the cylinder of a maximum volume that can be described in a sphere of radius a is 2a upon root 3.

00:11 So i have just created a rough sketch for this purpose.

00:15 This whole height is h.

00:18 This radius is a.

00:20 It is given and this base radius is r.

00:24 So if we talk about om, so it is nothing but h upon 2.

00:29 So basically first we apply pythagros theorem in this triangle that is amo and it is nothing but r square plus h square upon 4 is a square.

00:46 So from here we can conclude that h square upon 4 is a square minus r square and from here h is nothing but two times of root of a square minus r square.

01:00 R square.

01:03 Now if we talk about the volume, volume is nothing but volume of cylinder, volume of cylinder is nothing but pi square into h and that is something pi r square into two times of a square minus r square.

01:22 Now v is something that is 2 pi r square times root of a square minus r square now we can simplify it further consider z equals to b square and that is something four pi square r to the power four a square minus r square and that is something four pi square a square a square r to the power four minus r to the power six so as v is maximum minimum so jad will be also maximum minimum because b square will be maximum minimum so not differentiating with respect to r we get four pi square times four a square r q minus six r to the power five so second derivative would be d2 z upon d r2 that is four pi square bracket this would be 12 a square r square minus 30 r to the power 4 for the maximum minimum d z upon d r is 0 by that that region this bracket must be zero so we can say that 4a square r square r cube minus 6 r to the power 5 is equal to 0 so if we take 2 2 r cube common so we will get 2a square minus 3 r square equals to 0 r cannot be 0 so 2a square equals to 3 r square so from here we can conclude that r is nothing but root 2 upon 3 of a...

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