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# KCC's Quizzes AQQ243 about maximizing a cylinder volume

A cylinder V of radius r and height h is formed by the rotation of a rectangle of perimeter p around the vertical axis as shown in the picture. Question:

For a given fix perimeter p, what should be the radius r and the height h (versus p) to maximize the volume V?

Try to be among the first ones giving the correct answers!

Good luck!

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• ADI North America will be on summer shutdown starting August 24, 2023; perhaps another community member can assist you until our return on September 5th.
• 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  Pgiven = 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)

 V P differentiate with respect to r 2πrh 4 differentiate with respect to h πr2 2

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.