SOLUTION: The question is as under : " The sum of the surface areas of a sphere and a cube is given. Show that when the sum of their volumes is least, the diameter of the sphere is equal

Algebra ->  Proofs -> SOLUTION: The question is as under : " The sum of the surface areas of a sphere and a cube is given. Show that when the sum of their volumes is least, the diameter of the sphere is equal       Log On


   

Question 890514: The question is as under :
" The sum of the surface areas of a sphere and a cube is given. Show that when the sum of their volumes is least, the diameter of the sphere is equal to the edge of the cube."
The above question is related to Maxima and Minima chapter of Differential calculus.
Please send step by step solution.
Regards,
Khoka123.
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
You are to prove it in general for any given value of the total surface
area,  So we just call it a constant A.

Since the problem mentions diameter rather than radius, let's
convert the standard formula for the volume of a sphere so that
it is in terms of the diameter D instead of the radius, so we
substitute D%2F2 for r in the usual formula:



And also for the surface area of a sphere

A=4pi%2Ar%5E2=4pi%2A%28D%2F2%29%5E2=4pi%2A%28D%5E2%2F4%29=pi%2AD%5E2





A-pi%2AD%5E2%22%22=%22%226x%5E2

%28A-pi%2AD%5E2%29%2F6%22%22=%22%22x%5E2

sqrt%28%28A-pi%2AD%5E2%29%2F6%29%22%22=%22%22x

%28%28A-pi%2AD%5E2%29%2F6%29%5E%281%2F2%29%22%22=%22%22x

6%5E%28-1%2F2%29%28A-pi%2AD%5E2%29%5E%281%2F2%29%22%22=%22%22x

6%5E%28-3%2F2%29%28A-pi%2AD%5E2%29%5E%283%2F2%29%7D%7D%7B%7B%7B%22%22=%22%22x%5E3

Substitute in:









Put that equal to zero:


Multiply both sides by 2 to clear the fraction:



Add the exponents of 6



Divide by pi%2AD  (we aren't interested in when D=0)





Square both sides:




Multiply both sides by 6 to clear the fraction:

matrix%281%2C5%2C%0D%0A%0D%0A6D%5E2%2C%22%22%2C+%22%22=%22%22%2C%22%22%2C+A-pi%2AD%5E2%29

matrix%281%2C5%2C%0D%0A%0D%0A6D%5E2%2Bpi%2AD%5E2%2C%22%22%2C+%22%22=%22%22%2C%22%22%2C+A%29
matrix%281%2C5%2C%0D%0A%0D%0AD%5E2%286%2Bpi%29%2C%22%22%2C+%22%22=%22%22%2C%22%22%2C+A%29
matrix%281%2C5%2C%0D%0A%0D%0AD%2C%22%22%2C+%22%22=%22%22%2C%22%22%2C+A%2F%286%2Bpi%29%29

Take positive square roots of both sides:



Now we only need to show that x also equals that.

sqrt%28%28A-pi%2AD%5E2%29%2F6%29%22%22=%22%22x

We substitute for D:

sqrt%28%28A-pi%2A%28sqrt%28A%2F%286%2Bpi%29%29%29%5E2%29%2F6%29%22%22=%22%22x

sqrt%28%28A-pi%2A%28A%2F%286%2Bpi%29%29%29%2F6%29%22%22=%22%22x

sqrt%28%28A-%28pi%2AA%2F%286%2Bpi%29%29%29%2F6%29%22%22=%22%22x

Multiply top and bottom under the radical by %286%2Bpi%29

sqrt%28%28A%286%2Bpi%29-pi%2AA%29%2F%286%286%2Bpi%29%29%29%22%22=%22%22x

sqrt%28%286A%2Bpi%2AA-pi%2AA%29%2F%286%286%2Bpi%29%29%29%22%22=%22%22x

sqrt%28%286A%29%2F%286%286%2Bpi%29%29%29%22%22=%22%22x

sqrt%28A%2F%286%2Bpi%29%29%22%22=%22%22x

So x = r and the proposition is true.

Edwin