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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?

 Apr 5, 2015

Best Answer 

 #2
avatar+1038 
+10

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

 

This returns the solution for the question above.

solvec1,d,c2,c3(c1=π×d36c2=π×d34c3=(1(c1c2))){c1=0d=0c2=0c3=r12c1=r133×π6d=r13c2=r133×π4c3=13}

 

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

 

solvec1,d,c2,x(c1=2×π×0.5dc2=2×π×(0.5d+x)c2=c1+1){c1=r9×πd=r9c2=r9×π+1x=1(2×π)} 

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

 

This one is totally symbolic.

 

solvec2(c2×d2t2=c1×d1t1)c2=c1×d1×t2(d2×t1)

 Apr 5, 2015
 #1
avatar+130474 
+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 

 

  

 Apr 5, 2015
 #2
avatar+1038 
+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.

solvec1,d,c2,c3(c1=π×d36c2=π×d34c3=(1(c1c2))){c1=0d=0c2=0c3=r12c1=r133×π6d=r13c2=r133×π4c3=13}

 

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

 

solvec1,d,c2,x(c1=2×π×0.5dc2=2×π×(0.5d+x)c2=c1+1){c1=r9×πd=r9c2=r9×π+1x=1(2×π)} 

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

 

This one is totally symbolic.

 

solvec2(c2×d2t2=c1×d1t1)c2=c1×d1×t2(d2×t1)

Nauseated Apr 5, 2015
 #3
avatar+118703 
+5

Thanks for showing us this Nauseated  

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

 Apr 6, 2015

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