Chapter 3: Applications of Differentiation
Section 3.8: Optimization
Example 3.8.5
Which point on the graph of fx=sinhx−x e−3 x, −1≤x≤3, is closest to the point 1,7?
Solution
Analysis
The slider attached to Figure 3.8.5(a) varies the location of the point x,fx, and shows the line segment between that point and 1,7.
As the point x,fx traverses the graph of f, the value of d, the distance between that point and 1,7 is displayed beneath the graph.
With care, and relative to the pixel density on the display device, it is possible to estimate that there are three relative minima, one of which is the absolute minimum.
These minima are approximately 7, 1.8, and 1.6; the corresponding critical numbers are approximately −0.023,−0.77, and 2.61. Hence, the absolute minimum of approximately 1.6 occurs near x=2.61.
x= = ⇒ d=
Figure 3.8.5(a) Distance from 1,7 to graph of f
The square of the distance between x,fx and 1,7 is given by
Fx=x−12+fx−72
Minimizing the square of the distance is equivalent, but computationally simpler, than minimizing the distance itself.
A graph of d=Fx is given in Figure 3.8.5(b). The two relative minima, and the relative maximum are clearly visible.
The implied constraints are contained in the finite domain, namely, −1≤x≤3.
Figure 3.8.5(b) Graph of d=Fx
Numeric Solution
Numeric solution via Context Panel
Form a sequence of the objective function (distance) and the constraint equation (curve).
Context Panel: Optimization≻ Minimize (local)
x−12+y−72,y=sinhx−x⋅exp−3⋅x→minimize1.62753460035659115,x=2.61042424734159,y=6.76462239075871
The return consists of a list with two objects. The first object is the optimal value of the objective function; the second, a list of the parameter values giving this extreme value.
Numeric Solution via the Optimization Assistant
Form a sequence of the objective function (distance)and the constraint equation (curve).
Context Panel: Optimization≻Optimization Assistant (The Optimization Assistant launches with the objective function and constraint equation inserted into the appropriate fields.)
In the Options section, select Minimize
Click the Solve button. (See Figure 3.8.5(c).)
Click the Quit button to exit the Optimization Assistant and have it write the Solution to the underlying worksheet.
Click to launch the Optimization Assistant with the data embedded.
Figure 3.8.5(c) Solution by Optimization Assistant
Analytic Solution
Define fx and the objective function Fx
Control-drag fx=… Context Panel: Assign Function
fx=sinhx−x ⅇ−3 x→assign as functionf
Control-drag Fx=… Context Panel: Assign Function
Fx=x−12+fx−72→assign as functionF
Obtain the critical numbers
Figure 3.8.5(d) contains a graph of F′x. The critical numbers for F are the zeros of F′, and those zeros are the x-intercepts in Figure 3.8.5(d).
From Figure 3.8.5(d), the critical numbers are near −1 and 0, and between 2 and 3.
module()
local f,F,P;
f := sinh(x)-x*exp(-3*x);
F := (x-1)^2+(f-7)^2;
P:=plot(diff(F,x),x=-1..3,-100..130,tickmarks=[4,2]); print(P);
end module:
Figure 3.8.5(d) Graph of F′x
Tools≻Load Package: Student Calculus 1
Loading Student:-Calculus1
Write the equation F′x=0 and press the Enter key.
Context Panel: Student Calculus1≻Solve≻Find Roots Complete Roots dialog as per Figure 3.8.5(e).
Context Panel: Assign to a Name≻c (Individual critical numbers can then be referenced as c1,c2,c3.)
Figure 3.8.5(e) Roots dialog
F′x=0
2x−2+2sinhx−xⅇ−3x−7coshx−ⅇ−3x+3xⅇ−3x=0
→roots
−0.7711222385,−0.02316600177,2.610424247
→assign to a name
c
Apply the Second-Derivative test
c1 = −0.7711222385
F″c1 = 2058.863556
c2 = −0.02316600177
F″c2 = −90.78527373
c3 = 2.610424247
F″c3 = 92.44081429
Calculate d at the critical numbers and at the endpoints
Context Panel: Evaluate and Display Inline
Fc1 = 1.771977164 Relative minimum
Fc2 = 7.072733480 Relative maximum
Fc3 = 1.627534600 Relative minimum
F−1. = 12.07708977 Absolute maximum
F3. = 3.620129089 Relative maximum
Useful Commands
The CriticalPoints command in the Student Calculus1 package will return all zeros of the first derivative in a specified interval. The numeric option allows it to find such zeros numerically.
The fsolve command computes numerically all zero of a polynomial equation. For a transcendental equation it returns just one solution, either in a specified interval, or at the end of a Newton iteration starting from a given initial value. This latter is the behavior of the "Solve Numerically from point" option in the Context Panel for Solve.
The Roots, CriticalPoints, and InflectionPoints commands in the Student Calculus1 package will return all zeros of the underlying equations. Unfortunately, Context Panel access to these more powerful and useful commands has not yet been provided.
CriticalPointsFx,x=−1..3,numeric = −0.7711222385,−0.02316600177,2.610424247
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