Chapter 3: Applications of Differentiation
Section 3.8: Optimization
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Example 3.8.3
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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?
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Solution
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Analysis
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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.
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The objective function is the cost, given by
, in dollars.
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The constraint is , so that the area enclosed is one acre. Of course, there are the implied constraints that and are nonnegative.
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Solve the constraint for, say, and write the objective function to be minimized as .
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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]);
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Figure 3.8.3(a) Labeled diagram of the enclosure
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Graphical Solution
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Figure 3.8.3(b) is a graph of the objective function .
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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.
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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.
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Figure 3.8.3(b) Graph of
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Numeric Solution
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Control-drag a sequence of the objective function and the constraint equation.
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Context Panel:
Optimization≻Optimization Assistant
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To the right of "Iteration Limit," change "default" to 200.
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Press the Solve button to obtain the solution displayed in Figure 3.8.3(c).
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Press the Quit button to write the solution to the underlying worksheet.
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Click
to launch the Optimization Assistant with the data embedded.
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Figure 3.8.3(c) Solution by Optimization Assistant
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Analytic Solution
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Define the objective function
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Control-drag
Context Panel: Assign Function
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Obtain the critical number
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Write the equation for the critical number.
Press the Enter key.
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Context Panel: Solve≻Solve
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Perform the Second-Derivative test
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Evaluate at the positive critical number.
Context Panel: Evaluate and Display Inline
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=
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Obtain the minimal cost and the minimizing dimensions
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Evaluate the cost function:
Context Panel: Evaluate and Display Inline
Context Panel: Approximate≻10 (digits)
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=
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Select the positive critical number.
Context Panel: Approximate≻5 (digits)
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Evaluate with :
Context Panel: Evaluate and Display Inline
Context Panel: Approximate≻5 (digits)
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=
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Although there are two critical numbers, the positive value is chosen because the relevant variable represents a dimension.
Since at the critical number is positive, the extremum is a minimum.
The minimum cost is $5442.50, and the minimizing dimensions are , . The outer fence encloses a rectangle whose dimensions are by .
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