The a hollow cylinder. The internal and external radii are estimated to be 6 cm and 8 cm respectively, to the nearest whole number. The height of the cylinder is exactly 14 cm.
(a) Determine the exactly values for internal and external radii which give maximum volume of the materials used
(b) Calculate the maximum possible volume of the materials used. Take the value of π to be 22/7
To find the values of the internal and external radii that give the maximum volume of the materials used, we need to apply optimization techniques. In this case, we will maximize the volume of the hollow cylinder.
The formula for the volume of a hollow cylinder is given by:
V = πh(R^2 - r^2)
where V is the volume, h is the height, R is the external radius, and r is the internal radius.
(a) To determine the exact values for the internal and external radii that maximize the volume, we need to find critical points. This requires taking the derivative of the volume function with respect to r, setting it equal to zero, and solving for r.
Let's begin by taking the derivative:
dV/dr = πh(-2r)
Setting this equal to zero:
πh(-2r) = 0
Since π and h are constants, we can simplify the equation:
-2r = 0
Solving for r:
r = 0
This means that the internal radius does not affect the volume, and the maximum volume can be achieved when the internal radius is zero.
Next, let's find the external radius that gives the maximum volume. We can use the same approach and take the derivative of the volume function with respect to R:
dV/dR = πh(2R)
Setting this equal to zero:
πh(2R) = 0
Simplifying the equation:
2R = 0
Solving for R:
R = 0
This means that the external radius also does not affect the volume, and the maximum volume can be achieved when the external radius is zero.
In summary, the exact values for the internal and external radii that give the maximum volume of the materials used are 0 cm.
(b) Now we can calculate the maximum possible volume of the materials used by substituting the values into the volume formula:
V = πh(R^2 - r^2)
V = (22/7)(14)(8^2 - 6^2)
V = (22/7)(14)(64 - 36)
V = (22/7)(14)(28)
V = (22/1)(2)(4)
V = 176 cm^3
Therefore, the maximum possible volume of the materials used is 176 cm^3.