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

Jun 25, 2016

$\left(1 + \sqrt{2}\right) \pi$

#### Explanation:

The problem can be stated as a maximization/minimization.

Find maximum/minimum of

$S \left(x , y\right) = 2 \pi x y + 2 \pi {x}^{2}$

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

${g}_{1} \left(x , y\right) = {x}^{2} + {y}^{2} - {r}^{2} = 0$
${g}_{2} \left(x , y\right) = y - {s}^{2} = 0$

The lagrangian is

$L \left(x , y , {\lambda}_{1} , {\lambda}_{2} , s\right) = S \left(x , y\right) + {\lambda}_{1} {g}_{1} \left(x , y\right) + {\lambda}_{2} {g}_{2} \left(x , y , s\right)$

being analytic, the stationary conditions are given by

$\nabla L \left(x , y , {\lambda}_{1} , {\lambda}_{2} , s\right) = \vec{0}$

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 ${g}_{1} \left(x , y\right)$ so, the stationary points qualification is made over

${f}_{{g}_{1}} \left(x\right) = 2 \pi {x}^{2} + 2 \pi x \sqrt{{r}^{2} - {x}^{2}}$

having a minimum $\frac{{d}^{2} {f}_{{g}_{1}}}{{\mathrm{dx}}^{2}} > 0$ 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 $\frac{{d}^{2} {f}_{{g}_{1}}}{{\mathrm{dx}}^{2}} < 0$ 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

$S \left(\frac{r}{2} \sqrt{2 + \sqrt{2}} , \frac{r}{2} \sqrt{2 - \sqrt{2}}\right) = \left(1 + \sqrt{2}\right) \pi {r}^{2}$

If $r = 1$ then $S \left(\frac{1}{2} \sqrt{2 + \sqrt{2}} , \frac{1}{2} \sqrt{2 - \sqrt{2}}\right) = \left(1 + \sqrt{2}\right) \pi$

Jun 25, 2016

$\pi \left(\sqrt{2} + 1\right)$ 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

$S \left(a\right) = 2 \pi {\sin}^{2} a + 2 \pi \sin a \cos a$.

For the maximum of S, $S ' = 0 \mathmr{and} S ' ' < 0$.

$S ' = 2 \pi \left(2 \sin a \cos a + {\cos}^{2} a - {\sin}^{2} a\right)$

$= 2 \pi \left(\sin 2 a + \cos 2 a\right) = 0$, when

$\sin 2 a = - \cos 2 a$. So, $2 a = \frac{3 \pi}{4} \mathmr{and} a = \frac{3 \pi}{8}$..

As minimum is 0, this a gives the maximum.

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

$= \pi \left(1 - \cos \left(\frac{3 \pi}{4}\right) + \sin \left(\frac{3 \pi}{4}\right)\right)$.

$= \pi \left(1 + \cos \left(\frac{\pi}{4}\right) + \sin \left(\frac{\pi}{4}\right)\right)$

$= \pi \left(\sqrt{2} + 1\right)$.