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# Calculus/Application of Derivatives

Question
Show that the height 'h ' of a right circular cylinder of maximum total surface area , including
the two ends , that can be inscribed in a sphere of radius 'r ' is given byhsquare =2rsquare{1-1/5}

Thanks!

Show that the height 'h ' of a right circular cylinder of maximum total surface area , including the two ends , that can be inscribed in a sphere of radius 'r ' is given byhsquare =2rsquare{1-1/5}
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I am not sure how you get that, but my work looks like this:

If the radius of the sphere is r,

Let  2h = the height of the cylinder.

Then h = the height of a right triangle in a quadrant.
and  the base, R, is the radius of the cylinder.

And  R^2 + h^2 = r^2  is the 'constraint'.

Surface area is:
Side : Circumference * height = 2 pi R h.
End*2 = 2 pi R^2

Now R^2 = r^2 - h^2
and R = sqrt(r^2 - h^2)

Total A = 2 pi(r^2 - h^2) + 2 pi sqrt(r^2 - h^2) h

Ignore the 2pi factor:
r^2 - h^2 + sqrt(r^2 - h^2) h

r^2 - h^2 + sqrt(r^2h^2 - h^4)

Now you can start your maximizing as a function of h.

dA/dh = - 2h + 1/2[2r^2 h - 4h^3]/sqrt(r^2h^2 - h^4)]

dA/dh = - 2h + [r^2 h - 2h^3]/sqrt(r^2h^2 - h^4)]

Set that = 0:

[r^2 h - 2h^3]/sqrt(r^2h^2 - h^4) = 2h

r^2 h - 2h^3 = 2h sqrt(r^2h^2 - h^4)

r^2 h - 2h^3 = 2h^2 sqrt(r^2 - h^2)

r^2 - 2h^2 = 2h sqrt(r^2 - h^2)

Square both sides:

r4 + 4h4 - 4r2h2 = 4h2(r2 - h2)

r4 + 4h4 - 4r2h2 = 4h2r2 - 4h4

r4 + 8h4 = 8h2r2

Call  h^2 = H, for convenience,  r^2 = S

S^2 + 8H^2 = 8HS

8H^2 - 8HS + S^2 = 0

Use the QF: (ignore the S)
8 +- sqrt(64 - 32)
H = ------------------
16

8 +- sqrt(32)
H = ------------------
16

8 +- 4sqrt(2)
H = -------------
16

2 +- sqrt(2)
H = ------------
4

Calculus

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