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# Question...

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A company is deciding whether to package a ball in a cubic in a cubic box or a cylinder box. In either case, the ball will touch the bottom, top, and sides. What portion of the space inside the cylindrical box is empty?

Guest Apr 5, 2015

#2
+1037
+10

This site calculator will solve symbolic math. You can use it to test your solutions.

This returns the solution for the question above.

$$\underset{\,\,\,\,{{\rightarrow {\mathtt{c1, d, c2, c3}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{c1}}={\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{d}}}^{{\mathtt{3}}}}{{\mathtt{6}}}}\\ {\mathtt{c2}}={\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{d}}}^{{\mathtt{3}}}}{{\mathtt{4}}}}\\ {\mathtt{c3}}=\left({\mathtt{1}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{c1}}}{{\mathtt{c2}}}}\right)\right)\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{c1}} = {\mathtt{0}}\\ {\mathtt{d}} = {\mathtt{0}}\\ {\mathtt{c2}} = {\mathtt{0}}\\ {\mathtt{c3}} = {\mathtt{r12}}\\ {\mathtt{c1}} = {\frac{{{\mathtt{r13}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{6}}}}\\ {\mathtt{d}} = {\mathtt{r13}}\\ {\mathtt{c2}} = {\frac{{{\mathtt{r13}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{4}}}}\\ {\mathtt{c3}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}}\\ \end{array} \right\}$$

The result is the same as CPhill’s solution.

Note that all the Cs & Ds are not affecting the result. This calculator has a high resistance to CDD

Here’s another one

$$\underset{\,\,\,\,{{\rightarrow {\mathtt{c1, d, c2, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{c1}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{0.5}}{d}\\ {\mathtt{c2}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}\left({\mathtt{0.5}}{d}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)\\ {\mathtt{c2}}={\mathtt{c1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{c1}} = {\mathtt{r9}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\\ {\mathtt{d}} = {\mathtt{r9}}\\ {\mathtt{c2}} = {\mathtt{r9}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\\ {\mathtt{x}} = {\frac{{\mathtt{1}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}}\\ \end{array} \right\}$$

^ - - - Does anyone know what it is for? It’s easy.

This one is totally symbolic.

$$\underset{\,\,\,\,{{\rightarrow {\mathtt{c2}}}}}{{solve}}{\left(\begin{array}{l}{\frac{{\mathtt{c2}}{\mathtt{\,\times\,}}{\mathtt{d2}}}{{\mathtt{t2}}}}={\frac{{\mathtt{c1}}{\mathtt{\,\times\,}}{\mathtt{d1}}}{{\mathtt{t1}}}}\end{array}\right)} \Rightarrow {\mathtt{c2}} = {\frac{{\mathtt{c1}}{\mathtt{\,\times\,}}{\mathtt{d1}}{\mathtt{\,\times\,}}{\mathtt{t2}}}{\left({\mathtt{d2}}{\mathtt{\,\times\,}}{\mathtt{t1}}\right)}}$$

Nauseated  Apr 5, 2015
Sort:

#1
+83935
+10

The volume of the cylinder is

V = pi * r^2 * h   where r is the radius of the cylinder and h is the height.

And the radius of the ball = the radius of the cylinder.

But since the ball touches the sides, top and bottom, the height  of the cylinder must be 2r.

So the volume of the cylinder is

V = pi * r^2 * 2r  =  2pi *r^3

And the volume of the ball is

V = (4/3)pi* r^3

So..... the portion of the cylinder that is filled  by the ball (in terms of a ratio) is

[(4/3)pi * r^3] / [ 2 pi * r^3]  =  (4/3)/2 = 4/6 = 2/3 of the cylinder

So....the empty portion, in terms of ratio  = 1/3 of the cylinder

CPhill  Apr 5, 2015
#2
+1037
+10

This site calculator will solve symbolic math. You can use it to test your solutions.

This returns the solution for the question above.

$$\underset{\,\,\,\,{{\rightarrow {\mathtt{c1, d, c2, c3}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{c1}}={\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{d}}}^{{\mathtt{3}}}}{{\mathtt{6}}}}\\ {\mathtt{c2}}={\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{d}}}^{{\mathtt{3}}}}{{\mathtt{4}}}}\\ {\mathtt{c3}}=\left({\mathtt{1}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{c1}}}{{\mathtt{c2}}}}\right)\right)\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{c1}} = {\mathtt{0}}\\ {\mathtt{d}} = {\mathtt{0}}\\ {\mathtt{c2}} = {\mathtt{0}}\\ {\mathtt{c3}} = {\mathtt{r12}}\\ {\mathtt{c1}} = {\frac{{{\mathtt{r13}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{6}}}}\\ {\mathtt{d}} = {\mathtt{r13}}\\ {\mathtt{c2}} = {\frac{{{\mathtt{r13}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{4}}}}\\ {\mathtt{c3}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}}\\ \end{array} \right\}$$

The result is the same as CPhill’s solution.

Note that all the Cs & Ds are not affecting the result. This calculator has a high resistance to CDD

Here’s another one

$$\underset{\,\,\,\,{{\rightarrow {\mathtt{c1, d, c2, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{c1}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{0.5}}{d}\\ {\mathtt{c2}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}\left({\mathtt{0.5}}{d}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)\\ {\mathtt{c2}}={\mathtt{c1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{c1}} = {\mathtt{r9}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\\ {\mathtt{d}} = {\mathtt{r9}}\\ {\mathtt{c2}} = {\mathtt{r9}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\\ {\mathtt{x}} = {\frac{{\mathtt{1}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}}\\ \end{array} \right\}$$

^ - - - Does anyone know what it is for? It’s easy.

This one is totally symbolic.

$$\underset{\,\,\,\,{{\rightarrow {\mathtt{c2}}}}}{{solve}}{\left(\begin{array}{l}{\frac{{\mathtt{c2}}{\mathtt{\,\times\,}}{\mathtt{d2}}}{{\mathtt{t2}}}}={\frac{{\mathtt{c1}}{\mathtt{\,\times\,}}{\mathtt{d1}}}{{\mathtt{t1}}}}\end{array}\right)} \Rightarrow {\mathtt{c2}} = {\frac{{\mathtt{c1}}{\mathtt{\,\times\,}}{\mathtt{d1}}{\mathtt{\,\times\,}}{\mathtt{t2}}}{\left({\mathtt{d2}}{\mathtt{\,\times\,}}{\mathtt{t1}}\right)}}$$

Nauseated  Apr 5, 2015
#3
+91900
+5

Thanks for showing us this Nauseated

The power of the Web2 calc continues to amaze me  !!

Melody  Apr 6, 2015

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