Chapter 3: Applications of Differentiation
Section 3.8: Optimization
The strength of fields such as gravity or electromagnetism is inversely proportional to distance from the source. If two such sources are separated by a distance of L units, determine where on the line connecting the sources will the minimum strength be detected? Assume that the strength of one source is k times that of the other.
Let one source at x=0 on a number line have strength 1; and the other at x=L, have strength k.
Except for a proportionality constant, the combined strength at x is given by
Figure 3.8.10(a) is a representative graph of Sx with L=10 and k=5.
Figure 3.8.10(a) Field strength, L=10,k=5
Without loss of generality, L can be set to 1 so that Sx=1x2+k1−x2.
Define the objective function Sx
Context Panel: Assign Function
Sx=1x2+k1−x2→assign as functionS
Obtain the critical number
Write the equation S′x=0 and press the Enter key.
Context Panel: Solve≻Obtain Solutions for≻x
→solutions for x
Control-drag the first (and only real) solution.
Context Panel: Assign to a Name≻Xmin
k2/3k+1−k1/3k+1+1k+1→assign to a nameXmin
Apply the Second-Derivative test
S″Xmin = 6k2/3k+1−k1/3k+1+1k+14+6⁢k1−k2/3k+1+k1/3k+1−1k+14
The even powers in the denominators of S″Xmin imply that the expression is positive, and hence that Xmin is indeed a relative minimum. Figure 3.8.10(a) is invoked as proof that Xmin is a global minimum.
The graph of Xmin as a function of the relative strength k in Figure 3.8.10(b) suggests computing
limk→0Xmin = 1 and limk→∞Xmin = 0
As the strength of the source on the right decreases, the minimum of S moves toward x=1.
As the strength of the source on the right increases, the minimum of S moves toward x=0.
Figure 3.8.10(b) Graph of Xmin as a function of k
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