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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

Best Answer 

 #2
avatar+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.

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{c1, d, c2, c3}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{c1}}={\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{d}}}^{{\mathtt{3}}}}{{\mathtt{6}}}}\\
{\mathtt{c2}}={\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{d}}}^{{\mathtt{3}}}}{{\mathtt{4}}}}\\
{\mathtt{c3}}=\left({\mathtt{1}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{c1}}}{{\mathtt{c2}}}}\right)\right)\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{c1}} = {\mathtt{0}}\\
{\mathtt{d}} = {\mathtt{0}}\\
{\mathtt{c2}} = {\mathtt{0}}\\
{\mathtt{c3}} = {\mathtt{r12}}\\
{\mathtt{c1}} = {\frac{{{\mathtt{r13}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{6}}}}\\
{\mathtt{d}} = {\mathtt{r13}}\\
{\mathtt{c2}} = {\frac{{{\mathtt{r13}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{4}}}}\\
{\mathtt{c3}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}}\\
\end{array} \right\}$$

 

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

 

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{c1, d, c2, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{c1}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{0.5}}{d}\\
{\mathtt{c2}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}\left({\mathtt{0.5}}{d}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)\\
{\mathtt{c2}}={\mathtt{c1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{c1}} = {\mathtt{r9}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\\
{\mathtt{d}} = {\mathtt{r9}}\\
{\mathtt{c2}} = {\mathtt{r9}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\\
{\mathtt{x}} = {\frac{{\mathtt{1}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}}\\
\end{array} \right\}$$
 

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

 

This one is totally symbolic.

 

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{c2}}}}}{{solve}}{\left(\begin{array}{l}{\frac{{\mathtt{c2}}{\mathtt{\,\times\,}}{\mathtt{d2}}}{{\mathtt{t2}}}}={\frac{{\mathtt{c1}}{\mathtt{\,\times\,}}{\mathtt{d1}}}{{\mathtt{t1}}}}\end{array}\right)} \Rightarrow {\mathtt{c2}} = {\frac{{\mathtt{c1}}{\mathtt{\,\times\,}}{\mathtt{d1}}{\mathtt{\,\times\,}}{\mathtt{t2}}}{\left({\mathtt{d2}}{\mathtt{\,\times\,}}{\mathtt{t1}}\right)}}$$

Nauseated  Apr 5, 2015
Sort: 

3+0 Answers

 #1
avatar+81083 
+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
avatar+1037 
+10
Best Answer

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

 

This returns the solution for the question above.

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{c1, d, c2, c3}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{c1}}={\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{d}}}^{{\mathtt{3}}}}{{\mathtt{6}}}}\\
{\mathtt{c2}}={\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{d}}}^{{\mathtt{3}}}}{{\mathtt{4}}}}\\
{\mathtt{c3}}=\left({\mathtt{1}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{c1}}}{{\mathtt{c2}}}}\right)\right)\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{c1}} = {\mathtt{0}}\\
{\mathtt{d}} = {\mathtt{0}}\\
{\mathtt{c2}} = {\mathtt{0}}\\
{\mathtt{c3}} = {\mathtt{r12}}\\
{\mathtt{c1}} = {\frac{{{\mathtt{r13}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{6}}}}\\
{\mathtt{d}} = {\mathtt{r13}}\\
{\mathtt{c2}} = {\frac{{{\mathtt{r13}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{4}}}}\\
{\mathtt{c3}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}}\\
\end{array} \right\}$$

 

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

 

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{c1, d, c2, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{c1}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{0.5}}{d}\\
{\mathtt{c2}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}\left({\mathtt{0.5}}{d}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)\\
{\mathtt{c2}}={\mathtt{c1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{c1}} = {\mathtt{r9}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\\
{\mathtt{d}} = {\mathtt{r9}}\\
{\mathtt{c2}} = {\mathtt{r9}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\\
{\mathtt{x}} = {\frac{{\mathtt{1}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}}\\
\end{array} \right\}$$
 

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

 

This one is totally symbolic.

 

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{c2}}}}}{{solve}}{\left(\begin{array}{l}{\frac{{\mathtt{c2}}{\mathtt{\,\times\,}}{\mathtt{d2}}}{{\mathtt{t2}}}}={\frac{{\mathtt{c1}}{\mathtt{\,\times\,}}{\mathtt{d1}}}{{\mathtt{t1}}}}\end{array}\right)} \Rightarrow {\mathtt{c2}} = {\frac{{\mathtt{c1}}{\mathtt{\,\times\,}}{\mathtt{d1}}{\mathtt{\,\times\,}}{\mathtt{t2}}}{\left({\mathtt{d2}}{\mathtt{\,\times\,}}{\mathtt{t1}}\right)}}$$

Nauseated  Apr 5, 2015
 #3
avatar+91505 
+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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