Chapter 3: Applications of Differentiation
Section 3.8: Optimization
Example 3.8.12
An 8 × 10 rectangular sheet of paper is oriented with the long sides vertical. The upper left corner is placed along the right edge, and the paper folded. Where should the upper left corner be located so that the crease so formed has minimal length?
Solution
Analysis
In Figure 3.8.12(a) the upper left corner is folded over to the right edge. The top edge is divided into two segments, 8−x and x. The length of the crease is L. The arrow helps the eye focus on the displaced corner. The slider controls the value of x∈0,3, and for each such x, the corresponding value of L is displayed. To within the accuracy of the pixel density of the display, the minimum of L (10.39230) occurs at x=2. For x>3, the bottom of the crease falls on the bottom edge of the sheet of paper. At x=4, the paper has been folded in half vertically.
x= = ⇒ L=
Figure 3.8.12(a) Dynamic view of creased paper problem
Figure 3.8.212(b) shows the paper with a generic crease. The blue and red triangles are congruent because the red triangle came, by the fold, from the region in blue. Consequently,
PQ=PB=L2−8−x2
QA=AB=8−x
RB=8−x2−x2
PS=8
The green and yellow triangles are similar because they are right triangles in which ∠θ and ∠ψ are complements. Consequently,
ABPB=RBPS
use plots,plottools in ex3825:=module() local p1,p2,p3,p4,p5,p6,AA,BB,CC,dd,PP,QQ; export fig; AA:=[0,10-6*sqrt(2)]; BB:=[0,10]; PP:=[6,10]; CC:=[8,10]; QQ:=[8,10-4*sqrt(2)]; dd:=[8,10-6*sqrt(2)]; p1:=rectangle(BB,[8,0],style=line); p2:=polygon([AA,BB,PP],color=blue, transparency=.7); p3:=polygon([AA,PP,QQ],color=red,transparency=.5); p4:=polygon([PP,CC,QQ],color=yellow,transparency=.5); p5:=polygon([AA,dd,QQ],color=green,transparency=.5); p6:=textplot({[3.7,6,typeset(L)], [2.6,10.4,typeset(8-x)], [7,10.4,typeset(x)], [1.8,1.8,typeset(theta)], [7.7,6,typeset(theta)], [7.7,3.8,typeset(psi)], [6.6,9.6,typeset(psi)], [-.3,1.5,typeset(P)], [-.3,10,typeset(Q)], [6,10.4,typeset(A)], [8.3,10,typeset(R)], [8.3,4.3,typeset(B)], [8.3,1.5,typeset(S)]},font=[default,12]); fig:=display(p1,p2,p3,p4,p5,p6,scaling=constrained,axes=none); end module; end use:
ex3825:-fig;
Figure 3.8.12(b) Schematic of creased paper
or
8−xL2−8−x2=8−x2−x28
from which it follows that Lx=8−x3/24−x.
Numeric and Graphical Solutions
Define the objective function Lx
Control-drag Lx=… Context Panel: Assign Function
Lx=8−x3/24−x→assign as functionL
Numerical and graphical estimates of the minimum
Type Lx and press the Enter key.
Context Panel: Optimization≻Minimize (local)
Lx
8−x3/24−x
→minimize
10.3923048454132641,x=1.99999999864265
The graph of Lx in Figure 3.8.12(c) could be examined for an estimate of the minimum value for L, but because of the shallow shape of the curve near the minimum, it would be difficult to obtain an accurate answer.
According to the graph, x=4 is a vertical asymptote that corresponds to the paper being folded in half vertically. If the paper were infinitely tall, then L would indeed be unbounded, but of course, the model does not permit L to be greater than the height of the paper.
Figure 3.8.12(c) Graph of Lx
Analytic Solution
Obtain the critical number
Write L′x=0 and press the Enter key.
Context Panel: Solve≻Solve (If x=8, the upper left corner will not be on the right edge of the paper.)
L′x=0
−328−x4−x+128−x3/24−x3/2=0
→solve
x=2,x=8
Apply the Second-Derivative test
Since L″2>0, x=2 is a minimum.
L″2 = 1462
Obtain the minimum length
Context Panel: Approximate≻10 (digits)
L2 = 362≐10.39230484
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