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I am from a family which has been running a business for years now. It's a three-generation old business. My grandfather started it when he was young, my dad took over and eventually my brother now. I am quite young to run a business, but wish to join hands soon.

We deal in showpieces of various shapes, sizes, materials. We have a big shop at the market and a factory at the outskirts. Our business never remained restricted. It has evolved from time to time. If one visits our shop, he can spot a variety of showpieces from antiques to the latest ones.

Nowadays many connoisseurs of home decor often visit our shops. It is very interesting to hear them explain their requirements and at times asking for customized products. Getting them their satisfactory showpieces is not an easy job. Afterall, showpieces define the look of an architecture. There are also antique fanciers. People of all types come and look for things to make a decor complete.

A few days back there was a sudden hype of a certain type of showpiece. It's a glass sphere with a certain shape inside. It has a light attached and a music plays every time the button is pressed.

This IA is based on guessing the shape of the object that when kept inside the sphere would utilize the maximum space, thereby reducing space wastage.

The main motive of this IA is to study the optimized volume of different shaped blocks when placed in a sphere of radius r. Another motive of this IA, is to develop a comparative relationship with percentages of volume consumed by the block in the sphere and also percentages of the volume that has been wasted or not used by the block in the sphere.

Which will be the most efficient block to keep inside a sphere to obtain maximum volume of the sphere and possessing least wastage of volume out of Cuboidal, and Cylindrical?

Showpieces enhance the beauty of any architectural work. Room decor is incomplete without proper showpieces. They play a vital role in delivering the look one chooses to. They even serve as good gift options for many people. They come at different rates, various shapes and sizes. They are both machines made as well as hand made. A variety of materials are used in making showpieces. They can be made of glass, plastic, bronze, silver, still and also can be made of clay.

I have targeted to design a showpiece which will be sphere in shape and will have an element inside it. In this IA, I will try to guess the shape of the inside element so that maximum space of the sphere can be used, hence wastage of space can be minimum.

This, if designed properly can not only serve the public demand of such showpieces but also prove to be a profitable business for my family.

Let us assume that the principle sphere is of radius r and it is placed with the centre at the origin of the Cartesian Plane. Therefore, let the distance of each of the three sides of the cuboid circumscribed by the sphere of radius r from the centre be x, y, and z. Therefore,

Length of Cuboid = 2x

Breadth of Cuboid = 2y

Height of Cuboid = 2z

Therefore, **Volume of the Cuboid = (2x . 2y. 2z) = 8xyz**

In order to do the optimisation, we have to find the critical points of the cuboidal block present inside the sphere and as optimisation using all three variables is difficult to obtain, that’s the reason why, I will substitute the variable z by an expression having x, y, and r. As r is the radius of sphere, the value is constant.

As, the Sphere is drawn with its centre at the origin of the cartesian plane, the equation of the sphere will be:

*x*^{2 }+ *y*^{2 }+* z*^{2 }=* r*^{2}

=> *z*^{2 }= *r*^{2 }- *x*^{2 }- *y*^{2}

=> z = \(\sqrt{r^2-x^2-y^2}\)

Volume of Cuboid (V) = 8xy\(\sqrt{r^2-x^2-y^2}\)

In order to find the Maximum Value of V, we will use Calculus (Maxima Concept). I will find the partial derivative of the equation of volume with respect to x and y separately and then find the values of x, y and z for maximum volume.

\(\frac{∂V}{∂x}=8y\sqrt{r^2-x^2-y^2}+\frac{8xy . (-2x)}{2\sqrt{r^2-x^2-y^2}}\)

=> \(\frac{∂V}{∂x}=8y\sqrt{r^2-x^2-y^2}-\frac{8x^2 y}{2\sqrt{r^2-x^2-y^2}}\)

=> \(\frac{∂V}{∂x}=8y(\frac{r^2-x^2-y^2-x^2}{\sqrt{r^2-x^2-y^2}})\)

=> \(\frac{∂V}{∂x}=8y(\frac{r^2-2x^2-y^2}{\sqrt{r^2-x^2-y^2}})\)

\(\frac{∂V}{∂y}=8y\sqrt{r^2-x^2-y^2}+\frac{8xy.(-2y)}{2\sqrt{r^2-x^2-y^2}}\)

=> \(\frac{∂V}{∂y}=8x\sqrt{r^2-x^2-y^2}+\frac{8xy^2}{\sqrt{r^2-x^2-y^2}}\)

=> \(\frac{∂V}{∂y}=8x(\frac{r^2-x^2-y^2-y^2}{\sqrt{r^2-x^2-y^2}})\)

=>\(\frac{∂V}{∂y}=8y(\frac{r^2-2y^2-x^2}{\sqrt{r^2-x^2-y^2}})\)

According to the principle of Maxima, if the partial derivatives are zero, then only we can have the maximum values of the respective variables.

\(\frac{∂V}{∂y}=8y(\frac{r^2-2x^2-x^2}{\sqrt{r^2-x^2-y^2}})\)

=> 0 = 8y \((\frac{r^2-2x^2-y^2}{\sqrt{r^2-x^2-y^2}})\)

=> 0 = 8y (*r*^{2 }- 2*x*^{2 }- *y*^{2})

Either, 8y = 0

Or, (*r*^{2 }- 2*x*^{2 }- *y*^{2}) = 0

Since, it is absurd to get the value of y = 0 because, y is one of the dimensions of the cuboidal block and thus if y is equal to zero no cuboidal structure is possible.

∴ (r^{2 }- 2x^{2 }- y^{2}) = 0

=> 2x^{2 }+ y^{2 }= r^{2}……………………(i)

\(\frac{∂V}{∂y}=8y(\frac{r^2-2y^2-x^2}{\sqrt{r^2-x^2-y^2}})\)

=> 0 = 8x \((\frac{r^2-2y^2-x^2}{\sqrt{r^2-x^2-y^2}})\)

=>0 = 8x (*r*^{2 }- 2*y*^{2 }-* x*^{2})

Either, 8*x* = 0

Or, (*r*^{2 }- 2*y*^{2 }- *x*^{2}) = 0

Since, it is absurd to get the value of y = 0 because, y is one of the dimensions of the cuboidal block and thus if y is equal to zero no cuboidal structure is possible.

∴ (*r*^{2 }- 2*y*^{2 }- *x*^{2}) = 0

=> *x*^{2 }+ 2*y*^{2 }= *r*^{2}……………………(ii)

By (i) – 2(ii), we get,

-3*y*^{2 }= - *r*^{2}

=> *y*^{2} = \(\frac{r^2}{3}\)

=> *y* = \(\frac{r}{\sqrt3}\)

Putting the value of y in equation (i), we get,

2*x*^{2} + \(\frac{r^2}{3}\) = r^{2}

=> 2*x*^{2} = *r*^{2} - \(\frac{r^2}{3}\)

=> 2*x*^{2} = \(\frac{2r^2}{3}\)

=> *x *= \(\frac{r}{\sqrt3}\)

Since, we know that,

*z* = \(\sqrt{r^2-\frac{r^2}{3}-\frac{r^2}{3}}\)

=> *z* = \(\frac{r}{\sqrt3}\)

Therefore, the critical points of the cuboid placed inside a sphere of radius r are as follows:

*x* = \(\frac{r}{\sqrt3}\)

*y* = \(\frac{r}{\sqrt3}\)

z = \(\frac{r}{\sqrt3}\)

Therefore, the dimensions of the Cuboid for its volume be maximum are:

Length = \(\frac{2r}{\sqrt3}\)

Breadth = \(\frac{2r}{\sqrt3}\)

Height = \(\frac{2r}{\sqrt3}\)

Volume of Cuboid (V) = 8xyz

∴ Volume of Cuboid V = 8 \(\frac{r}{\sqrt{3}}.\frac{r}{\sqrt{3}}.\frac{r}{\sqrt{3}}=\frac{8r^3}{3\sqrt{3}}\) = 1.539r^{3}

Let us assume that the principle sphere is of radius r and it is placed with the centre at the origin of the Cartesian Plane. Therefore, let the height of the cylinder be h and the radius of base of cylinder be R. Therefore,

Therefore, Volume of the Cylinder = πR2h

In order to do the optimisation, we have to find the critical points of the cylindrical block present inside the sphere and as optimisation using all two variables is difficult to obtain, that’s the reason why, I will substitute the variable R by an expression having h and r. As r is the radius of sphere, the value is constant.

Therefore, we can write,

\((\frac{h}{2})^2+R^2=r^2\)

=> R^{2 }= r^{2 }- \(\frac{h^2}{4}\)

Volume of Cylinder (V) = π \(\biggl(r^2-\frac{h^2}{4}\biggl)h=\pi(r^2h-\frac{h^3}{4})\)

In order to find the Maximum Value of V, we will use Calculus (Maxima Concept). I will find the partial derivative of the equation of volume with respect to h separately and then find the values of h and R for maximum volume.

\(\frac{∂V}{∂h}=\pi(r^2-\frac{3h^2}{4})\)

According to the principle of Maxima, if the partial derivatives are zero, then only we can have the maximum values of the respective variables.

\(\frac{∂V}{∂h}=\pi(r^2-\frac{3h^2}{4})\)

=> 0 = π\((r^2-\frac{3h^2}{4})\)

=> 0 = r^{2} - \(\frac{3h^2}{4}\)

=> \(\frac{3h^2}{4}\) = r^{2}

=> h = \(\frac{2r}{\sqrt{3}}\)

Since, we know that,

R^{2} = r^{2} - \(\frac{h^2}{4}\)

=> R^{2} = r^{2} - \(\frac{(\frac{2r}{\sqrt3})^2}{4}\)

=> R^{2} = r^{2} - \(\frac{r^2}{3}\)

=> R^{2} = \(\frac{2r^2}{3}\)

Volume of Cylinder (V) = πR^{2}h = π \(\frac{2r^2}{3}\frac{2r}{\sqrt{3}}=π\frac{4r^3}{3\sqrt{3}}\) = 2. 418r^{3}

We know that,

Volume of Sphere = \(\frac{4}{3}\pi\,r^3\) = 4.188r^{3}

**In case of a Cuboidal Block** which is inscribed by a sphere of Radius r:

∴ Volume of Cuboid (V) = 1.539r^{3} ~ 37%

Therefore, space or volume of the sphere which is not utilised:

Un - used Volume = (4.188 - 1.539) = 2.649r^{3} ~ 63%

**In case of a Cylindrical Block** which is inscribed by a sphere of Radius r:

Volume of Cylinder (V) = 2.418r^{3 }~ 58%

Therefore, space or volume of the sphere which is not utilised:

Unused Volume = (4.188 - 2.418) = 1.77r^{3} ~ 42%

**Cuboidal Block inside Sphere**

Volume of Sphere = \(\frac{4}{3}\)π*r*^{3} = 4.188*r*^{3}

Volume of Cuboid (V) = 1.539*r*^{3}

Volume of Sphere un - utilised (V_{loss}) = 4.1888*r*^{3 }- 1.5396*r*^{3 }= 2.6492*r*^{3}

For the derived expression, it is evident that with increase in radius of sphere, the loss in space or volume will increase.

Thus, we will find the rate of loss with respect to the increase in radius of sphere using differential calculus.

\(\frac{dV_{loss}}{dr}\) = 2.6492 *X* 3 *X r*^{2 }= 7.9476 *r*^{2}

**Cylindrical Block inside Sphere**

Volume of Sphere = \(\frac{4}{3}πr^3\) = 4.188*r*^{3}

Volume of Cylinder (V) = 2.418*r*^{3}

Volume of Sphere un - utilised (V_{loss}) = 4.1888*r*^{3 }- 2.418*r*^{3 }= 1.7708*r*^{3}

For the derived expression, it is evident that with increase in radius of sphere, the loss in space or volume will increase.

Thus, we will find the rate of loss with respect to the increase in radius of sphere using differential calculus.

\(\frac{dV_{loss}}{dr}\) = 1.7708 *X* 3 *X* *r*^{2 }= 5.3124* r*^{2}

In this IA, we have derived that cylinder is the shape of the element which when placed inside the sphere uses the maximum space resulting in minimum wastage. From the above study, I have derived that, percentage volume of sphere utilised by the largest right circular cylinder is 58%. The percentage volume of sphere utilised by cuboidal block or conical block is much less than that of cylindrical block. For cuboidal block, the percentage is 37% and that of conical block is 30%. From the above comparative study based on rate of increase in loss, we can conclude that, cuboidal block inscribed inside a sphere would incur more loss as compared to cylindrical block inscribed inside a sphere of same radius with increase in sphere.

This knowledge when applied to manufacturing of a certain type of showpiece will be quite profitable. Since we deal in showpieces, it is mandatory for us to keep ourselves updated with the demand and find ways for higher profits.

- The two most widely available shape of cartoons used to cover any product before sending is cylindrical and cuboidal shape. In this exploration, both the shapes were taken into consideration. This has improved the coherence of the exploration.
- The error calculation has been obtained using calculus to increase precision of the calculation. This has made the exploration more reliable and error free.

- In this exploration, the error incurred in terms of unused space in a spherical container when a cylindrical and a cuboidal cartoon is placed on it. However, there are different shapes of container. In this exploration, the outer container is kept constant as spherical. This has made the exploration less coherent.
- The spherical, cylindrical or cuboidal containers or cartoons that are considered in this exploration are perfectly oriented. However, in real-life scenario, no container is perfectly oriented as a regular structure. This assumption has made the exploration less coherent.

As a further extension of this exploration, the shape of the outer box could be changed to identify the unused space when any regular figure other than the shape of the outer box is placed inside it. Hence, in this case, the research question of the extension of this exploration could be framed as follows: How does the volume of unused space inside a cubical box depends on the shape of cartoon (spherical and cylindrical) used to cover any product with optimum dimension with respect to the outer cubical block?

- https://mathworld.wolfram.com/Sphere.html
- Tierney, John A. "Elementary techniques in maxima and minima." The Mathematics Teacher 46.7 (1953): 484-486.
- http://ncert.nic.in/textbook/textbook.htm?iemh1=0-15