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

Question

Charles Arocha · Accepted Answer

# (1 + sqrt[2]) pi#

The problem can be stated as a maximization/minimization.

Find maximum/minimum of

#S(x,y) = 2pi x y+2 pi x^2#

subjected to

#x^2+y^2=r^2#
#y > 0#

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

#g_1(x,y) = x^2+y^2-r^2 = 0#
#g_2(x,y) = y-s^2=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)#

being analytic, the stationary conditions are given by

#grad L(x,y,lambda_1,lambda_2,s)=vec 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#

Elijah Abram · Answer

#pi(sqrt 2+1)# areal units.

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)#.

Emilia Aguilar · Answer

undefined To find the maximum possible total surface area of a cylinder inscribed in a hemisphere of radius 1, we can use calculus. Let $ r $ be the radius of the cylinder and $ h $ be its height.

The total surface area $ A $ of the cylinder is given by:

$$ A = 2\pi rh + 2\pi r^2 $$

Given that the cylinder is inscribed in a hemisphere, we have the constraint that $ r \leq 1 $ and $ h \leq 2 $.

To maximize $ A $, we can express $ h $ in terms of $ r $ using the Pythagorean theorem:

$$ h = 2\sqrt{1 - r^2} $$

Now, substitute this expression for $ h $ into the formula for $ A $:

$$ A(r) = 2\pi r(2\sqrt{1 - r^2}) + 2\pi r^2 $$

To find the maximum surface area, differentiate $ A $ with respect to $ r $, set the derivative equal to zero, and solve for $ r $.

$$ \frac{dA}{dr} = 2\pi (2\sqrt{1 - r^2} - \frac{2r^2}{\sqrt{1 - r^2}}) + 4\pi r $$

Setting $ \frac{dA}{dr} = 0 $, we find:

$$ 2\sqrt{1 - r^2} - \frac{2r^2}{\sqrt{1 - r^2}} + 2r = 0 $$

Solving this equation yields the value of $ r $. After obtaining $ r $, substitute it back into the expression for $ h $ to find the corresponding value of $ h $. Then calculate the total surface area $ A $ using the formula mentioned earlier.