Optimization of cylinder inside sphere?

Question

A right circular cylinder is inscribed in a sphere of radius r. Find the largest possible surface area of such a cylinder.

So I have the formula for the volume of a sphere, I know the SA of a cylinder is 2(pi)r^2 + 2(pi)rh

can someone help me solve this? and also, how do i designate a "best answer"?

Answer 1

your sphere have radius r. so let you cylinder have radius R and height 2H.



r^2 = R^2 + H^2



surface area of cylinder, S(R)

= 2πR(R + 2H)

= 2πR[R + 2√(r^2 - R^2)]



S'(R) = 4πR + 4π√(r^2 - R^2) + 4πR(-2R)/2√(r^2 - R^2)

= 4πR + [4π(r^2 - R^2) - 4πR^2]/√(r^2 - R^2)

= 4π[R√(r^2 - R^2) + r^2 - 2R^2] / √(r^2 - R^2) = 0



R√(r^2 - R^2) + r^2 - 2R^2 = 0

R^2(r^2 - R^2) = (2R^2 - r^2)^2 = 4R^4 - 4(Rr)^2 + r^4

5R^4 - 5(Rr)^2 + r^4 = 0

R^2 = [5r^2 +- √{(5r^2)^2 - 4*5*r^4}] / 2*5

= [5r^2 +- r^2√5] / 10

0 < R < r yori,

R = + r√[0.5 +- 0.1√5]



R = 0.85065r or R = 0.52573r



S"(R) = 4π + {√(r^2 - R^2) * (-16πR) - (-8πR^2) * (-R) / √(r^2 - R^2)} / (r^2 - R^2)

= 4π + {(r^2 - R^2)(-16πR) - 8πR^3} / (r^2 - R^2)^1.5

= 4π + 8πR(R^2 - 2r^2) / (r^2 - R^2)^1.5



S"(0.85065r) < 0,

S(0.85065r) is max,

= 3.2360679775πr^2

= 3.236πr^2

= 10.166r^2

Answer 2

You need to get the SA in terms of one variable. Right now it's in terms of height and radius. The cylinder's radius, half its height and the sphere's radius always form a right triangle. Use pythagorean theorem to relate them. The sphere's radius is a constant so use the relationship to substitute into SA formula.

r-sphere radius

R -cylinder radius





SA = 2πR^2 + 2πR*2√(r^2-R^2)

SA' = 4πR -4R^2 *π/√(r^2-R^2)+4(π)*√(r^2-R^2)

=4π(R*√(r^2-R^2) -2R^2 +r^2)/√(r^2-R^2)



SA' = 0 =4π(R*√(r^2-R^2) -2R^2 +r^2)/√(r^2-R^2)

0 =R*√(r^2-R^2) -2R^2 +r^2

2R^2 - r^2=R*√(r^2-R^2)

4R^4 -4r^2 *R^2 + r^4=R^2(r^2-R^2)=R^2*r^2-R^4

5R^4 -5r^2 *R^2 + r^4 = 0

5R^4 -5r^2 *R^2 + r^4 = 0



quadratic formula

R^2 = (5r^2-√(25r^4 - 4(5)r^4))/10

R^2 = r^2 *(5-√5)/10, r^2 *(5+√5)/10

R = r * √((5-√5)/10, r * √((5+√5)/10) can't be negative



plugging in gives you

7.356481492r^2

and

10.16640739r^2



the second one is bigger so it is the answer

Answer 3

Assuming the pyramid must be a ordinary sq. pyramid, the biggest sq. pyramid interior a cube with section length d, could have top f = d and base section e = d (such that it shares a cube's section as its base), and its apex on the middle of the choice cube section, such that its volume V[pyramid] = (one million/3) e^2 f = d^3 / 3 = one-third the quantity V[cube] of the cube. subsequently, V[pyramid] is maximized as V[cube] is maximized. A cube interior a cylinder with radius b and top c, could have volume V[cube] such that: V[cube] = min(2b, c)^3 *** Eq. one million for the reason that c is inversely proportional Assuming the two caps of the cylinder are such that their around perimeters are parallel "small circles" of the sector, (it quite is, for the cylinders optimum radius b[max] = one million, c = 0; and additionally, for the cylinder's optimum top c[max] = 2, b = 0). permit element o be the middle of the sector, and permit element p be on the edge of a around cap. permit oq be a radius of the cylinder such that pq is perpendicular to the around cap. Then, op = the around radius a = one million, pq = 0.5 the cylinder's top = c/2, and oq = the cylinder's radius b opq is a top triangle with hypotenuse op. via the Pythagorean Theorem, c^2 / 4 + b^2 = one million => c = 2 sqrt(one million - b^2) *** Eq. 2 for the reason that 0 < b < one million and 0 < c < 2, we % basically evaluate the effective branches of the sqrts. Now, b is inversely proportional to c, and the two are monotonic, meaning that (via Eq. one million), V[cube] is maximized the place b = c. So our Pythagorean equation may be simplified to: b^2 / 4 + b^2 = one million => b = 2 sqrt(5) / 5 And, so V[pyramid, max] = V[cube, max] / 3 = b^3 / 3 = (2 sqrt(5) / 5)^3 / 3 = 8 sqrt(5) / seventy 5 *** answer ? 5.7% of the quantity of the sector --- because of the fact the inradius of a unit cube is one million/2 and a circumradius is sqrt(3)/2, i could think of packing the sector in the cube could maximize area utilization. the main awkward volume to fill seems to be the pyramid. So i could wager a maximal order could bypass like this (from innermost to outermost): pyramid, cylinder, cube, sphere i could wager that a minimum order could be cylinder, cube, sphere, pyramid