Question 1205006
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This can be done on a regular calculator, but I'll turn to <a href="https://www.geogebra.org/">GeoGebra</a> instead.


Here is the particular GeoGebra workbook that I created for this problem. 
<a href="https://www.geogebra.org/calculator/gwhxa25b">https://www.geogebra.org/calculator/gwhxa25b</a>
Normally I use GeoGebra to make some kind of graph. 
However, the graph part is blank. Focus on the column of items on the left side.
I'll explain the inputs below. Each input is highlighted in <font color=blue>blue</font>



<font color=blue>R = 0.635</font> and <font color=blue>r = 0.6</font> represent the radius values of the conical frustum. 


<font color=blue>h = 1.2</font> is the height or depth of the ladle.
So far everything is in meters, so I'll convert the diameters "12.5cm and 17.5cm" to meters
12.5 cm = 0.125 m
17.5 cm = 0.175 m
Divide by 100 to convert from cm to meters.



The formula <font color=blue>Vf = (1/3)*pi*h*(R^2+R*r+r^2)</font> represents the volume of the conical frustum. 
It is the formula your teacher gave you. 
GeoGebra will replace the letters h, R and r with their proper values. 
The result of the calculation is roughly Vf = 1.43788 when rounding to 5 decimal places.


So far we've just taken care of the conical frustum only.


Recall that the volume of a full sphere is (4/3)*pi*(radius)^3


The larger sphere has volume (4/3)*pi*(external radius)^3
The smaller sphere has volume (4/3)*pi*(internal radius)^3


The volume of a spherical outer shell, ignoring the hollow empty space, is:
shell volume = (larger sphere volume) - (smaller sphere volume)
shell volume = (4/3)*pi*(external radius)^3 - (4/3)*pi*(internal radius)^3
shell volume = (4/3)*pi*( (external radius)^3 - (internal radius)^3 )
shell volume = (4/3)*pi*( (eR)^3 - (iR)^3 )


where,
eR = external radius
iR = internal radius


In this case:
eR = 0.175/2 = 0.0875
iR = 0.125/2 = 0.0625
both are in meters. 
Keep in mind the 0.175 m and 0.125 m represent diameters. 
Cut those in half to get their corresponding radius values.


The formula (4/3)*pi*( (eR)^3 - (iR)^3 ) represents a full spherical shell. But we want a hemispherical shell instead. We'll cut that in half to get (2/3)*pi*( (eR)^3 - (iR)^3 )


So that's where the formula <font color=blue>Vh = (2/3)*pi*( (eR)^3 - (iR)^3 )</font> comes from.



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Here's a quick recap of the inputs to type in (type them in the order mentioned)
<font color=blue>R = 0.635</font>
<font color=blue>r = 0.6</font>
<font color=blue>h = 1.2</font>
<font color=blue>Vf = (1/3)*pi*h*(R^2+R*r+r^2)</font>
That takes care of the frustum
Then for the hemispheres we have these inputs
<font color=blue>eR = 0.0875</font>
<font color=blue>iR = 0.0625</font>
<font color=blue>Vh = (2/3)*pi*( (eR)^3 - (iR)^3 )</font>


GeoGebra should display these approximate values
Vf = 1.43788
Vh = 0.00089
which represent the frustum volume and hemisphere volume respectively.


The last calculation to make is to type in <font color=blue>ratio = Vf/Vh</font> and the result should be roughly 1612.41248
Round this down to the nearest whole number and it shows we can make <font color=red size=4>1612</font> hemispheres.
This assumes that there is no waste. 


Again this could have been done with any calculator, but I used GeoGebra so we could name the inputs and use them at a later time. Likely other calculators have this capability as well. 


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Answer: <font color=red size=4>1612</font>
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