The problem seeks the dimension of the cylindrical tin can that is capable of holding 14 cubic inches of liquid, and has the smallest surface area. Before we can determine the dimension, let us consider the volume, V, of this cylindrical tin can, which is the product of the area of the circular base with radius r, and the height h of the cylinder, and the surface area of the tin can, A, which is the area of the circular base, the circular top, and around the sides.
V= πr2h (1)
14= πr2h
h= 14πr2 (2)
A=2πr2+2πrh (3)
According to (2), the value of h can be written in terms of r.
A(r)=2πr2+28πr (4)
However, we have to consider the constraints as stated in the problem. In other words, we have to find the minimum value of A(r)=2πr2+28πr over [1, 2].
The first derivative test is a method that involves the determination whether the intervals in the function is increasing or decreasing. Take note that an interval (a, b) is considered increasing if f’(x) over (a, b) is positive, and decreasing if f’(x) over (a, b) is negative. Now, critical values are potential values within (a, b) where the direction of the graph changes, and are referred to as the relative extreme points. To demonstrate, let us go back to the surface area problem.
A(r)=2πr2+28πr
A'r=4πr-28πr2 (5)
0=4πr-28πr2,r≠0
r=37π2 (6)
The critical points are r=0, 37π2 , both of which are outside [1, 2]. Thus, the only points of interest are r = 1, and r = 2.
Since the function is increasing over [1, 2], we can deduce that the smallest value of A(r) over [1, 2] is when r is 1, and the maximum value is when r is 2.
On the other hand, the second derivative test evaluates the concavity, which is the turning behavior of a graph over certain intervals. If within an interval (a, b) of a differentiable function f, there exists critical values where f’(c) = 0, then the graph is concaved up if f’’(c) positive, and concaved down if the f’’(c) is negative. To demonstrate, let us go back to our problem, but this time, let us consider within the interval [0, 2].
A''r=4π+56πr3 (7)
Therefore, the graph has a relative minimum at r=∛(7/π^2 ). Take note that we considered the interval [0, 2] because there is no critical value within [1, 2].
Finally, the closed interval method uses the endpoints and all critical values within the closed interval to determine the absolute extreme points. The absolute minimum/maximum refers to the point within the endpoints and critical values that gives the lowest/highest value. To demonstrate, let’s go back to the problem. Since there is no critical value within [1, 2], we will consider only the points 1 and 2.
In short, the smallest surface area can be achieved when the radius is 1 inch, and from (2), the height is 14.4563 inches.
