Additional Mathematics Differentation

Question

John has a rectangular piece of zinc with a perimeter with a perimeter of 32 cm. He wants to use that piece of zinc to build an open cylinder at both ends.

Find the length and the width, in cm, of the piece of zinc that makes the volume of the cylinder is maximum.​

Answer 1

Since we have to find the top circumference of the cylinder anyway, let us suppose it x.

Then the height of the cylinder, h = (32 - 2x) / 2 = 16 - x. ............. (1)

The radius, r = x/(2π) ..,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, (2). [since x = 2πr]

Area of the top, A = πr² = π. (x/(2π)² = x² / 4π .............................(3)

Volume of the cylinder, V = A.h = (x² / 4π) . (16 - x) ........................ [ Eq 3 multiplied by Eq 2)

= 4x²/π - x³/4π

Now we could optimize x for the maximum volume by differentiing the volume function of x ( as Alif said above).

dV/dx = 2x.4/π - 3x²/4π ................................................................... (4)

The top of the differentiation curve would be flat. Therefore, the value of Equation 4 at the top = 0.

Or, 2x.4/π - 3x²/4π = 0

or, x(8/π - 3x/4π) = 0 ................. [factoring out x]

So, either x = 0, or (8/π - 3x/4π) = 0, but x cannot be 0.

Therefore, (8/π - 3x/4π) = 0

or, 3x/4π) = 8/π

or, x = (8/π).(4π/3)

=32/3 cm = 10.67 cm...[2 dp}

Now the height, h = (16 - x) ...................................[Eq 1]

= (16 - 10.67) = 5.33 cm

We can have a check for the result given by Alif, namely 12 cm and 4 cm. The volume then would be

45.84 cc. But with the solution above the volume is 48.29 cc. Therefore, the answer in the answer sheet as mentioned by Alif must be wrong.

Answer 2

The perimeter of the rectangular piece of zinc can be written as 2(2πr + h) = 32 → 2πr + h = 16......(1)

Note: It's 2πr, because the length of the rectangle is the circumference of the cylinder, after the cylinder has been unrolled.

Then if you roll up the rectangular piece of zinc, to form a cylinder, the volume of the cylinder is V = πr²h......(2)

So, using equation (1), we can solve for h: h = 16 - 2πr and substitute this expression in for equation (2)

V = πr²(16 - 2r)

V = π(16r² - 2r³)

Now, you'll need a graphing calculator, to graph y = 16r² - 2r³ find where the maximum value is by going to 2nd trace → go to Maximum, and let the calculator do the work. (If you need more information on this, go to Youtube, and search up how to use a graphing calculator to find maximum and minimum values).

Once you know the x value (in this case that's the r value) that creates the maximum, the length of the rectangle is 2πr and the height is 16 - 2πr.

Hope this helps

Mr. K