Expert: Alon Mandes - 1/9/2009

Question
I have a problem here which asks me to find the dimensions of the right-circular cylinder of greatest surface area that can be inscribed in a sphere of Radius R (note it is the surface area they are asking about, not the volume).



The answer is even given as

height = 2* (5-(5^1/2)/10)^1/2 R

radius = (5-(5^1/2)/10)^1/2 R

I asked another expert about this and he assisted me a little but then got frantically busy.

One way among several that I went about it was to set the Radius of the circle at R and the radius of the cylinder at r

The height of half the cylinder =h/2

So r^2 +(h/2)^2 = R^2 or r^2 = R^2 - (h/2)^2

and r = (R^2 - (h/2)^2)^1/2

I the took the surface formula

2pi(r^2) +2pi(r)(h)

Substituted in

2pi (R^2 - h^2/4) +2pi (R^2-(h/2)^2 (h)

and then simplified a little

2pi(R^2) -2pi(h^2/4) + 2=(R^2-(h/2)^2)^1/2 *(h)

I then differentiated (since R is a constant the first 2pi(R^2) drops out.

-pi(h) + (pi*(R^2-h^2/4)^-1/2 (-2h/4) +2pi(R^2-h^2/4)^1/2

And simplify

-pi(h) -pi(h)/2(R^2-(h^2)/4)^1/2 + 2pi(R^2-(h^2)/4)^1/2

set the derivative to zero and multiply by 1/pi so that all the pi's drop out.

and I get 0 = -h -h/(2(R^2-(h^2)/4)^1/2) + 2(R^2-(b^2)/4)^1/2

I then do some algebra and moving around, but I cannot seem to get rid of all of the square root variables to solve h for R and I certainly don't get anything like the answer in the text.

I also tried to solve this using trigonometric derivatives but to no avail.

Can you help?

Thanks,

Richard  

Answer
I agree that r=√[R²-(h²/4)] or r=(½)√[4R²-h²]. Our target function
is Sh=2*Πr²+2Πrh. So, S=2*Π*(¼)(4R²-h²)+2Πh*(½)√[4R²-h²].
S(h)=(½)Π(4R²-h²)+Πh*√[4R²-h²].
S'(h)=(½)Π*(-2h)+Π√[4R²-h²]-(Πh*2h)/2√(4R²-h²).
S'(h)=-Πh+Π√[4R²-h²]-(Πh²)/√(4R²-h²).
S'(h)=0 --> √[4R²-h²]=h + h²/√(4R²-h²).
We multiply boths sides by √[4R²-h²] :
4R²-h²=h√[4R²-h²]+h²
4R²-2h²=h√[4R²-h²].
We perform squaring operator for both sides :
(4R²-2h²)²=4R²h²-h^4
16R^4+4h^4-16R²h²=4R²h²+h^4
16R^4+5h^4-20R²h²=0
Let's set Γ=h² We get :
5Γ²-20R²Γ+16R^4=0
The solutions for this quadratic equation are :
Γ1=R²*(20+√80)/10 & Γ2=R²*(20-√80)/10.
We have h=√Γ , thus :
h=R* √[(5*4±2√2*√2*√5)/10]=R* √[4(5±√5)/10]=2R* √[5±√5]/√10.

Hope this can help.

Alon.