KCC's Quizzes AQQ243 about maximizing a cylinder volume

We can assume the side of rectangle as x and y, so perimeter will be 2(x+y) =p so we can calculate y in term of x and p. So y is (p/2-x) and volume of cylinder is pi*r^2*h. So V is equal pi*(x^2*p/2 -x^3). For maxima we need to take derivative and check if second derivative is negative or positive. 

So dV/dx= pi(2*p*x/2-3x^2) and dV/dx=0.

After calculations will get p=3x and second derivate is negative means local maxima. 

So max volume will be when x=p/3 and y=p/6.

V= pi*(p/3)^2*p/6

Getting rid of π, the “*” volume and perimeter are:

V* =r²h
P* = 2r² +2rh

Lagrange gives us:

• 2 r h = λ (4r+2h)
• r² = λ 2r
• P*₀ = 2r² + 2rh

The second equation leads to  λ = r/2

Into the first leads to 2r=h

So,  P* = 6r²,   or  P_given = 6 π r²,  so r = sqrt( P₀/(6 π))

Note: the first two equations come from the derivate of V with respect to r (and h) which should equal to lambda times of the derivate of P with respect of the same parameter. (Because the volume is hopefully optimal and that the constraint has to be tangent to it since, if not, it would not intercept the equation to optimize or would intercept at two different points, so would not be optimal, since the volume could then be reduced (increased) while satisfying the constraint).

I also think that this is related to Archimedes' proof about it is the same exact constant π  which is involved in formulas for the circle and for the sphere.  

p=4*r+2*h V=r^2*pi*h

h=(p-4*r)/2

V=r^2*pi*(p-4*r)/2=pi/2*(r^2*p-4*r^3)

dV/dr=pi/2*(2*r*p-12*r^2)=0 => r(2*p-12r)=0 r1=0 r2=p/6

d^2V/dr^2=pi/2*(2*p-24*r):

pi*p > 0 for r1 => Minimum

pi/2*(-2p) < 0 => Maximum

V is max for r=h=p/6

p=2(2r+h)     2r + h = p/2   h = p/2 - 2r

v= Pi * r^2 *  h

v = pi * r^2 * (p/2 - 2r)  =  pi * r^2 * p/2  - 2 pi * r^3  

dv/dr  =    pi r p  -  6 pi r^2

for max, dv/dr = 0

so, pi r p = 6 pi r^2

p = 6r

r = p/6

h = p/2 - 2r  =  p/2 - 2p/6 = p/2 - p/3

h = 3p/6 - 2p/6 =  p/6 

so we also get r = h = p/6

Reading the other answers, I see that I misread p as the total surface (it has length ^ 2  unit in my formulation), instead of the p defined in the formulation of the problem, that is p is the perimeter (it has unit of length ^1 )  f the generating rectangle.

Using the same Lagrange multipliers technique, but this time, with the right definition of P, for this problem: 

V=πr²h
P=2(2r+h)

Lagrange: (we have just one contraint, so one lamba)

• 2πrh = 4 λ  , or    πrh = 2λ
• πr²= 2 λ
• P₀ = 2(2r+h)

 (1) and (2) :  2πrh = πr²     so   r=h  and then P₀ = 6r = 6h  and V follows.

Answer is r=h=p/6.

You can write p=4*r + 2*h, so h= (p -4*r)/2. 

V= r*r*h= (r*r*p)/2 - 2* r*r*r. (We can get rid of PI). Deriving by r, you get dV/dr= r*(p-6*r) which is zero for r=p/6 and since by further deriving by r you can prove that the second derivative is negative you demonstrate that r=p/6 is a maximum. So you get maximum volume for r=p/6 and h=p/6. 

Wow, I didn't know that Lagrange method... It is indeed very efficient since you obtain the right answer, congratulation vanderghast !

Congratulation Sona22 ! I suppose x for you is the diameter and y the height of the cylinder