Find the maximum possible total surface area of a cylinder inscribed in a hemisphere of radius 1?

2 Answers
Jun 25, 2016

(1 + sqrt[2]) pi(1+2)π

Explanation:

The problem can be stated as a maximization/minimization.

Find maximum/minimum of

S(x,y) = 2pi x y+2 pi x^2S(x,y)=2πxy+2πx2

subjected to

x^2+y^2=r^2x2+y2=r2
y > 0y>0

Introducing a slack variable ss to transform the inequality restriction into an equality equivalent restriction we have

g_1(x,y) = x^2+y^2-r^2 = 0g1(x,y)=x2+y2r2=0
g_2(x,y) = y-s^2=0g2(x,y)=ys2=0

The lagrangian is

L(x,y,lambda_1,lambda_2,s) = S(x,y)+lambda_1 g_1(x,y) + lambda_2 g_2(x,y,s)L(x,y,λ1,λ2,s)=S(x,y)+λ1g1(x,y)+λ2g2(x,y,s)

being analytic, the stationary conditions are given by

grad L(x,y,lambda_1,lambda_2,s)=vec 0L(x,y,λ1,λ2,s)=0

giving

{ (2 lambda_1 x + 4 pi x + 2 pi y = 0), (lambda_2 + 2 pi x + 2 lambda_1 y = 0), ( x^2 + y^2 -r^2= 0), (y -s^2= 0), (2 lambda_2 s = 0) :}

Solving for x,y,lambda_1,lambda_2,s we obtain

(x= -r, y= 0, lambda_1= -2pi, lambda_2= 2pi r, s= 0), (x = r, y = 0, lambda_1= -2pi, lambda_2 = -2pi r, s = 0), (x= -r/2 sqrt[2 -sqrt[2]], y=r/2 sqrt[2 + sqrt[2]], lambda_1 =(sqrt[2]-1)pi, lambda_2 = 0, s ne 0), (x=r/2 sqrt[2 + sqrt[2]], y= r/2 sqrt[2 - sqrt[2]], lambda_1 =-(sqrt[2]+1)pi, lambda_2 = 0, s ne 0)

As we kow the active restriction is g_1(x,y) so, the stationary points qualification is made over

f_{g_1}(x)=2 pi x^2 + 2 pi x sqrt[r^2 - x^2]

having a minimum (d^2f_{g_1})/(dx^2)>0 at

(x= -r/2 sqrt[2 -sqrt[2]], y=r/2 sqrt[2 + sqrt[2]], lambda_1 =(sqrt[2]-1)pi, lambda_2 = 0, s ne 0),

and maximum (d^2f_{g_1})/(dx^2)<0 at

(x=r/2 sqrt[2 + sqrt[2]], y= r/2 sqrt[2 - sqrt[2]], lambda_1 =-(sqrt[2]+1)pi, lambda_2 = 0, s ne 0)

The maximum value being

S(r/2 sqrt[2 + sqrt[2]], r/2 sqrt[2 - sqrt[2]]) = (1 + sqrt[2]) pi r^2

If r = 1 then S(1/2 sqrt[2 + sqrt[2]], 1/2 sqrt[2 - sqrt[2]]) = (1 + sqrt[2]) pi

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Jun 25, 2016

pi(sqrt 2+1) areal units.

Explanation:

Let a be the inclination to the axis of the cylinder, of the radii of the

hemisphere, to the circular top of the cylinder..Then the radius of the

cylinder is sin a, the height is cos a and the surface area is

S(a)=2pisin^2 a+2pisin a cos a.

For the maximum of S, S'=0 and S''<0.

S'=2pi(2 sin a cos a+cos^2a-sin^2a)

=2pi(sin 2a+cos 2a)=0, when

sin 2a = -cos 2a. So, 2a = (3pi)/4 and a = (3pi)/8..

As minimum is 0, this a gives the maximum.

Thus, max S = 2pi(sin^2 (3pi/8))+pisin (3pi/4))

= pi(1-cos((3pi)/4)+sin ((3pi)/4)).

=pi(1+cos(pi/4)+sin(pi/4))

=pi(sqrt 2 + 1).