# Thread: Applications of maxima / minima: inscribed Cylinder

1. ## Applications of maxima / minima: inscribed Cylinder

a.)Estimate the dimensions of a right circular cylinder of greatest lateral surface area that can be incribed in a sphere with a radius of 6in. Confirm your estimates analytically.

b.)Estimate the dimensions of a right circular cylinder of Greatest Volume that can be inscribed in a sphere w/ radius 6in. Confirm your estimates.

I'll start this one . . .

a) Find the dimensions of a right circular cylinder of greatest lateral
surface area that can be inscribed in a sphere with a radius of 6 in.

The lateral surface area of a cylinder is: .S .= .2πrh
. . where r is the radius and h is the height.
Code:
              * * *
*     :     *
*-------+-------*
*|       :       |*
|       :       |
* |       :       | *
* |       *       | *
* |       | \ 6   | *
|      y|   \   |
*|       |  x  \ |*
*-------+------ *
*     |     *
* * *

We see that the radius is x and the height is 2y.

We also see that: .x² + y² .= .36 . . y .= .(36 - x²)^½

And we have: .S .= .2πrh .= .2πx(2y) .= .4πx(36 - x²)^½

Can you finish it now?

3. Originally Posted by ^_^Engineer_Adam^_^
a.)Estimate the dimensions of a right circular cylinder of greatest lateral surface area that can be incribed in a sphere with a radius of 6in. Confirm your estimates analytically.

b.)Estimate the dimensions of a right circular cylinder of Greatest Volume that can be inscribed in a sphere w/ radius 6in. Confirm your estimates.

Lateral Surface of right cylinder = 2pi(radius)(height)
Volume of right cylinder = pi(radius^2)(height)

Draw the figure.
Draw a circle. Draw the inscribed right circular cylinder vertically. Draw the vertical diameter of the circle. Draw a radius to any of the four corners of the "rectangle". A right triangle is formed, with these:
---hypotenuse = 6 in.
---vertical leg = (1/2)height of cylinder = h/2
---horizontal leg = radius of cylinder = r
---central angle, between hypotenuse and vertical leg = angle X

r = 6sinX

h/2 = 6cosX
h = 12cosX

-----------------------------------
1) Greatest lateral surface, A.

A = 2pi(r)h
A = 2pi(6sinX)(12cosX)
A = 36pi(2sinXcosX)
A = 36pi(sin(2X))
dA/dX = 36pi[cos(2X) *2]
dA/dX = 72pi[cos(2X)]
Set that to zero,
0 = 72pi[cos(2X)]
0 = cos(2X)
2X = arccos(0) = pi/2
X = pi/4 ----------------------for max A

Then,
r = 6sin(pi/4) = 6(sqrt(2) /2) = 3sqrt(2) = 4.24264 in.
h = 12cos(pi/4) = 12(sqrt(2) /2) = 6sqrt(2) = 8.48528 in.

Therefore, the inscribed right cylinder of greatest lateral area is 8.48528 inches in diameter and is 8.48528 inches also high. -----------answer.

-------------------------------------------------
b) Greatest Volume, V.

V = pi(r^2)h
V = pi[(6sinX)^2 ](12cosX)
V = pi[36sin^2(X)](12cosX)
V = 12*36pi[(1 -cos^2(X)]cosX
V = 12*36pi[cosX -cos^3(X)] -------------------(1)
dV/dX = 12*36pi[-sinX -3cos^2(X)*(-sinX)]
dV/dX = 12*36pi[(-sinX)[1 -3cos^2(X)]
Set that to zero,
(-sinX)[1 -3cos^2(X)] = 0

-sinX = 0
0 = sinX
X = arcsin(0) = 0 or pi -----------***

1 -3cos^2(X) = 0
cos^2(X) = 1/3
cosX = +,-sqrt(1/3) = +,-0.57735
X = arccos(+,-0.57735) = 0.9553 radians or 2.1863 radians. ----***

So there are 4 possible values of the angle X for max V. Plug them into (1). Only one of them gives a positive V---the X = 0.9553 radians only.

So,
r = 6sin(0.9553) = 4.8989 in. -----diameter = 2r = 9.7978 in.
h = 12cos(0.9553) = 6.9284 in

Therefore, the inscribed right cylinder of greatest volume is 9.7978 inches in diameter and is 6.9284 inches high. -----------answer.

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# find the dimensions of a right circular cylinder of maximum lateral surface which can be inscribed in a sphere of radius 4 inch.

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