Chapter 3: Applications of Differentiation
Section 3.8: Optimization
Example 3.8.3
One acre (43560 sq ft) is to be enclosed with a rectangular fence that costs $5.00 per ft. Additionally, the enclosed area is to be subdivided into three equal rectangles with fencing that costs $3.50 per ft. What is the minimal cost of the construction, and what are its dimensions?
Solution
Analysis
Figure 3.8.3(a) shows a labeled rectangle subdivided into three equal-area rectangles. The expensive outer fencing is drawn in blue; the cheaper inner fencing, in red.
The objective function is the cost, given by C=56 u+2 v+722 v, in dollars.
The constraint is 3 u v=43560, so that the area enclosed is one acre. Of course, there are the implied constraints that u and v are nonnegative.
Solve the constraint for, say, v=14520/u and write the objective function to be minimized as Fx=30 x+17 14520/x.
p1:=plottools[rectangle]([0,1],[3,0],style=line,color=blue): p2:=seq(plot([[x,0],[x,1]], style=line,color=red),x=1..2 ): p3:=plots:-textplot([seq([z-0.5,-0.1,typeset(u)],z=1..3),[2.9,0.5,typeset(v)]],font=[Lucinda,18]): plots:-display(p1,p2,p3,scaling=constrained, axes=none,view=[0..3,-.2..1]);
Figure 3.8.3(a) Labeled diagram of the enclosure
Graphical Solution
Figure 3.8.3(b) is a graph of the objective function Fx=30 x+17 14520/x.
Given the scale of the graph, a graphical estimate of the minimum point will be difficult; the critical number seems to be approximately 100, and the minimum cost, slightly more than $5,000.
Of course, a more accurate approximation might be obtained if the scale of the graph were modified. That is left to the reader's discretion.
Figure 3.8.3(b) Graph of Fx
Numeric Solution
Control-drag a sequence of the objective function and the constraint equation.
Context Panel: Optimization≻Optimization Assistant
To the right of "Iteration Limit," change "default" to 200.
Press the Solve button to obtain the solution displayed in Figure 3.8.3(c).
Press the Quit button to write the solution to the underlying worksheet.
Click to launch the Optimization Assistant with the data embedded.
Figure 3.8.3(c) Solution by Optimization Assistant
Analytic Solution
Define the objective function Fx
Control-drag Fx=… Context Panel: Assign Function
Fx=30 x+1714520/x→assign as functionF
Obtain the critical number
Write the equation for the critical number. Press the Enter key.
Context Panel: Solve≻Solve
F′x=0
30−246840x2=0
→solve
x=2217,x=−2217
Perform the Second-Derivative test
Evaluate F″ at the positive critical number. Context Panel: Evaluate and Display Inline
F″2217 = 3018717
Obtain the minimal cost and the minimizing dimensions
Evaluate the cost function: Context Panel: Evaluate and Display Inline Context Panel: Approximate≻10 (digits)
F2217 = 132017→at 10 digits5442.499426
Select the positive critical number. Context Panel: Approximate≻5 (digits)
2217→at 5 digits90.708
Evaluate v=14520/u with u=2217: Context Panel: Evaluate and Display Inline Context Panel: Approximate≻5 (digits)
145202217 = 6601717→at 5 digits160.08
Although there are two critical numbers, the positive value is chosen because the relevant variable represents a dimension.
Since F″ at the critical number is positive, the extremum is a minimum.
The minimum cost is $5442.50, and the minimizing dimensions are u=2217, v=660/17. The outer fence encloses a rectangle whose dimensions are 3 u=6617 by v=660/17.
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