Thickness- Crust

[Guest]

A loaf of bread, which is a half sphere (hemisphere) has a diameter of 12 inches. The crust is baked to thickness T, so the remainder is half the contents of the loaf. Determine thickness of crust.


Volume of hemisphere - 2/3 * pi * r^3

2/3 * 3.14 * 6^3 = 452.155 = 452

How do I proceed from here ?

[Guest]

Is the crust only over the curved top of the hemisphere?
Has the flat base no crust because it is sitting on the baking tray?

If that is the case then a simple answer......

The half of a half is a quarter sphere....and.... Let R = radius of remaining hemisphere...

1/3 * 3.14 * 6^3 = 2/3 * 3.14 * R^3

divide across by ( 1/3 * 3.14) gives....

6^3 = 2 * R^3

take cube root of each side

6 = (cuberoot 2) * R

then .... R = 6 / (cuberoot 2)

R = 6 / 1.26

R = 4.76

So crust = 6 - 4.76 = 1.24 inches

[Guest]

"Is the crust only over the curved top of the hemisphere? - The problem does not indicate.
Has the flat base no crust because it is sitting on the baking tray?" - The problem does not indicate.

[Guest]

Now considering a crust on the base of the loaf as well.......

Assume the crust on the base to approx. a circular disk the thickness of the crust......keep it simple...

1/3 * 3.14 * 6^3 = 2/3 * 3.14 * R^3 - (3.14 * R^2 * (6 - R)

as was done on last post, divide across by ( 1/3 * 3.14) gives....

6^3 = 2 * R^3 - 3 * R^2 * (6 - R)

6^3 = 2 * R^3 +3 * R^3 - 18 * R^2

216 = 5 * R^3 - 18 * R^2

5 * R^3 - 18 * R^2 - 216 = 0

Cubic equation .....solve by graph and/or trial methods.....

R = 5.2 inches

So crust equals 6 - 5.2 = 0.8 inches

We now have a slightly thinner crust spread over all of the loaf.

[Guest]

" 2/3 * 3.14 * R^3 - (3.14 * R^2 * (6 - R)

6^3 = 2 * R^3 - 3 * R^2 * (6 - R)

6^3 = 2 * R^3 +3 * R^3 - 18 * R^2

216 = 5 * R^3 - 18 * R^"

I am not clear on these steps.

[Guest]

" 2/3 * 3.14 * R^3 - (3.14 * R^2 * (6 - R)

2/3 * 3.14 * R^3.....this is the volume of the hemisphere of radius R

(3.14 * R^2 * (6 - R))....this is the volume of the circular disk of crust on bottom...it should have had another ) on the end

It is crust so subtract it from the hemisphere of good loaf.......1st line

6^3 = 2 * R^3 - 3 * R^2 * (6 - R).....this is after dividing across by ( 1/3 * 3.14)

6^3 = 2 * R^3 +3 * R^3 - 18 * R^2....this is after multiplying out the brackets above

216 = 5 * R^3 - 18 * R^2"......this is rearranging the above...adding 3 + 2 to get 5 and 6^3 = 6x6x6 to get 216

I am not clear on these steps.

[Guest]

"6^3 = 2 * R^3 - 3 * R^2 * (6 - R).....this is after dividing across by ( 1/3 * 3.14)

6^3 = 2 * R^3 +3 * R^3 - 18 * R^2....this is after multiplying out the brackets above"

I still don't understand these steps.

One question:

If the crust is 0.8 inches, what is the contents of the loaf ?

[Guest]

your question....
If the crust is 0.8 inches, what is the contents of the loaf ?

You must read the question.....
The question tells us that crust is T which we have found to be 0.8 inches, if we consider the base crusted as well.
and the question tells us that the remainder is half of the original loaf contents.....copy on line below....
"The crust is baked to thickness T, so the remainder is half the contents of the loaf."

the original loaf was a hemisphere of dia 12 inches or radius 6 inches. So the remainder of the loaf under the crust is the volume of a half of a hemisphere of radius 6 inches.
A hemisphere volume is 2/3*pi*6^3.....
The volume of half a hemisphere is 1/3*pi*6^3 and this is what we set equal a hemisphere of radius R minus a disk from the bottom for the crust on the base. And we calculated the Radius R of this hemisphere (what is left) with the base cut off it.

In figures...your question...If the crust is 0.8 inches, what is the contents of the loaf ? I assume you mean the bit left no crust.
1/3*pi*6^3 = 226.08 cubic inches and is a hemisphere with a slice off the base. its radius 5.2 inches.