Chapter 3: Applications of Differentiation
Section 3.8: Optimization
Example 3.8.1
Of all rectangles with perimeter 100, find the one with maximal area.
Solution
Analysis
Figure 3.8.1(a) shows a labeled rectangle whose area is fw,h=w h, and whose perimeter is 2 w+2 h.
The constraint equation is gw,h≡2 w+2 h−100=0.
Solve the constraint equation for, say, h=50−w and write the area of the rectangle as w 50−w.
Maximize the objective function Fx=x 50−x
p1:=plottools[rectangle]([1,4],[5,1],style=line): p2:=plots:-textplot({[3,.7,typeset(w)],[.7,2.5,typeset(h)]},font=[Lucinda,18]): plots:-display(p1,p2,scaling=constrained, axes=none);
Figure 3.8.1(a) Labeled diagram of a rectangle
Graphical Solution
Figure 3.8.1(b) is a graph of the objective function Fx=x 50−x. The curve is a parabola, and the vertex of the parabola is the absolute maximum for the function F.
Using the Context Panel for the graph, set Probe Info to Nearest datum, and trace the graph with the cross-hair form of the probe.
Set the cross-hair on the vertex. Again in the Context Panel for the graph, in the Probe Info option, select Copy data.
Paste the contents of the Clipboard into the workspace:
25.11518285929648624.9867329089243
Figure 3.8.1(b) Graph of Fx=x 50−x
The density of pixels on the computer screen determines how accurately a graph can be "read." From the graphical approximation, it might be conjectured that the optimal dimensions of the rectangle are 25×25, for a maximal area of 625.
Numeric Solution
Numeric solution via the Context Panel
Form a sequence of the objective function and the constraint equation.
Context Panel: Optimization≻Maximize (local)
w h,2 w+2 h=100→maximize624.999999999999,h=25.0000000000000,w=25.0000000000000
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
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 Maximize
Click the Solve button. (See Figure 3.8.1(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 3.8.1(c) Solution by Optimization Assistant
Algebraic Solution
Control-drag Fx=…
Context Panel: Assign Function
Fx=x 50−x→assign as functionF
Type Fx and press the Enter key.
Context Panel: Expand≻Expand
Context Panel: Complete Square≻x
Fx
x50−x
= expand
−x2+50x
= complete square
−x−252+625
Upon completing the square in Fx, the equation y=−x−252+625 describes a parabola in vertex form. The coordinates of the vertex can be read from the equation: 25,625. Hence, the width of the rectangle is 25, its corresponding height is 50−25=25, and the maximal area is 625. The rectangle with fixed perimeter and maximal area is a square.
Analytic Solution
Write the equation for critical numbers. Press the Enter key.
Context Panel: Solve≻Solve
F′x=0
50−2x=0
→solve
x=25
Determine the maximal area.
F25 = 625
If w=25, then the constraint equation 2 w+2 h=100 determines that h=25, so the rectangle whose perimeter is fixed at 100, and whose area is a maximum, is a square.
The purist might demand that the Second-Derivative test be applied via the calculation F″25=−2, but the pragmatist will insist that Figure 3.8.1(b) is sufficient.
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