To find: "What dimensions of the rectangle will result in a cylinder of maximum volume?"

So the dimensions of the rectangle when the first derivative of the volume of the cylinder is zero.

There are two possibilities: one is when the rectangle is revolved about one of its horizontal edges, and the other is when the rectangle is revolved about one of its vertical edges---if the 4 edges were either horizontal or vertical.

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1) When revolved about a horizontal edge.

Let vertical edges be x cm each, so the horizontal edges are (30-x) cm each.

Volume of generated cylinder, V1 = "pi(r^2)L", where r=x, and L = (30-x) here.

So,

V1= pi(x^2)(30-x)

V1 = pi(30x^2 -x^3)

Differentiate both sides with respect to x,

dV1/dx = pi[60x -3x^2]

Set that to zero,

0 = 60x -3x^2

Divide both sides by 3x,

0 = 20 -x

x = 20 cm ----------vertical side or edge.

(30-x) = 10 cm ----horizontal edge.

Therefore, the rectangle must be 20cm by 10cm. -------answer.

(max V1 = pi(20^2)(10) = 4000pi cu.cm.)

The net of the generated cylinder?

The surface area of the cylinder is a rectangle that is (10cm wide) by (2pi*20 cm long), or a rectangle 20cm by 40pi cm.

----If the net in question refers only to this suface area , then the net is a rectangle 10cm by 40pi cm.

----If the net in question includes also the 20cm by 10cm original rectangle,then the net is composed of two rectangles, side by side at their respective 10cm sides, a 10-by-20 and a 10-by-40pi. Or it is a longer rectangle that is 10cm by (20 +40pi)cm.

----If the net in question includes also the circular planes at each vertical side of the cylinder, then include two circles of radius 20cm each.

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1) When revolved about a vertical edge.

Let again vertical edges be x cm each, so the horizontal edges are (30-x) cm each.

r = (30-x), and L=x here.

So, V2 = pi[(30-x)^2](x)

V2 = pi[900 -60x +x^2](x)

V2 = pi[900x -60x^2 +x^3]

Differentiate both sides with respect to x,

dV2/dx = pi[900 -120x +3x^2]

Set that to zero,

0 = 900 -120x +3x^2

Divide both sides by 3,

x^2 -40x +300 = 0

Factor that,

(x-10)(x-30) = 0

x = 10 or 30

When x=30, the horizontal edge (30-x) becomes zero, so no good.

Therefore, x = 10 cm ------vertical edge of the rectangle.

(30-x) = 20 cm -----------horizontal edge rectangle.

Therefore, the rectangle must be 10cm by 20cm. -----answer.

(max V2 = pi(20^2)(10) = 4000pi cu.cm. also, same as max V1)

The net of the generated cylinder?

The same as in V1.