# Math Help - Optimisation of area

1. ## Optimisation of area

Hi guys, heres another great optimisation problem Just having a bit of trouble getting the right equations etc any help appreciated. Also as is stated in the brackets a net of the shape is required. As usual any help is appreciated. Thanx.

Question: Consider a rectangle of perimeter 60Cm. Form a cylinder by revolving this rectangle about one of its edges. What dimensions of the rectangle will result in a cylinder of maximum volume? (Include a net of the final shape)

2. 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.

3. Thanx for that mate

4. Hello, mathsB!

I have no idea what a "net" is . . .

Consider a rectangle of perimeter 60Cm.
Form a cylinder by revolving this rectangle about one of its edges.
What dimensions of the rectangle will result in a cylinder of maximum volume?
(Include a net of the final shape)
We have a rectangle with dimensions $L \times W$.
Code:
      * - - - - - *
|           |
W |           |
|           |
* - - - - - *
L
The perimeter is 60 cm: . $2L + 2W \:=\:60\quad\Rightarrow\quad W \:=\:30 - L$ .[1]

Revolve the rectangle about a vertical side.
The cylinder has radius $L$ and height $W$.
. . It volume is: . $V \:=\:\pi r^2h \:=\:\pi L^2W$. . [2]

Substitute [1] into [2]: . $V \;=\;\pi L^2(30-L)$

Now maximize this function.
. . $\text{(I got: }L = 20,\;W = 10)$

5. Originally Posted by Soroban

I have no idea what a "net" is . . .

...
Hello,

only for the record: I've attached a "net" of a cube and a "net" of perpendicular circular cylinder. Maybe you now can provide me with the correct English expression