How to use this applet:
Instructions:
The switch shapes button toggles between the cylinder and cone shapes.
At any time you can press the lock volume button to fix the volume at set value and you can then use either the increment/decrement height or width buttons to examine how the surface areas alter with respect to the height/width.
The lock radius and lock height buttons can be used to freeze the height or width and investiage how the volume and surface area changes if just one of the height or width parameters are altered.
The increment/decrement height and width buttons may be used to alter the height and width of the shape when the volume is locked.
Notes on the maths used in the applet:
This applet shows how the surface areas of two shapes, a cone and a cylinder, can have a
minimum value when their volume is fixed and their height and radii are altered. It allows
you to select either the cylinder or cone, select values for its radius and height and
consequently its volume. You can then click the "Lock volume" button which will fix the
volume at its current value. Now you can use the buttons to alter height or radius and
observe how the surface area changes with respect to the height and the radius and the ratio
of height to radius.
You can see that for given volume, the surface area is a minimum for a given ratio of height
to width, and that when the problem is repeated for a different volume, the same ratio of
height to radius gives the minimum surface area.
So what is this ratio of height to radius?
As can be seen in this program, it is very hard (impossible?) to get an absolutely accurate
indication of some of the conditions that give a minimum value of surface area.
It is differential calculus that solves this problem for us, by providing us with a tool
which can instantly achieve the same, or better, results that would take a lot of testing to
achieve.
For example, for a cylinder of fixed volume, V, height H, and radius R. (This is a similar
situation to a recent LCH exam question). For what ratio of height to radius will the
surface area of the cylinder be a minimum?
1) V (Volume) = pi*r*r*h. (from a simple formula for volume, p9, maths tables).
2) SA (surface area)= (2*pi*rh) + 2(pi*r*r).
3) h(height) = V/(pi*r*r) (height in terms of volume, pi, and radius).
4) SA = (2*pi*r*(V/(pi*r*r))) + 2(pi*r*r). (substitute in for h from 3)
5) SA = ((2*V)/r) + 2(pi*r*r) (tidy up equatation)
6) d(SA)/dr = (-2*V)/(r*r) + (4 *pi *r) (differentiation both sides wrt r)
7) 0 = (-2*V)/(r*r) + (4 *pi *r) (for minimum value of SA, put d(SA)/dr = 0.
8) (2*V)/(r*r) = (4 * pi * r)
9) 2V = 4 * pi * r*r*r
10) V/(2pi) = r*r*r
11) (pi*r*r*h)/(2pi)= r*r*r (from initial value of volume)
12) h = 2r (cancel parts common to both sides)
This shows that the surface area of a cylinder is a minimum when the height is twice the
radius. If you enter values into the cylinder program you can see that this result is
verified to within the programs limits of accuracy.
This is meant to show just what max/min differentiation problems are actually about.
Use of the cone to verify the result obtained for a min value of its surface area for fixed
volume is left to you as an exercise. Be aware that surface area refers to the total surface area, not the curved surface area, in this program.
