Chapter 3: Applications of Differentiation
Section 3.8: Optimization
Find the length of the longest ladder that can be carried horizontally around the corner of the passageway shown in Figure 3.8.14(a). (The horizontal and vertical segments, corridors of widths b and a, respectively, are at right angles to each other.)
Hint: The longest ladder that can be carried around the corner at point B is the shortest line segment from A to C that also passes through B.
Hint: Angles ABE and BCF are equal because they are corresponding interior angles of the parallel lines BE and CF.
Figure 3.8.14(a) Ladder in right-angled corridor
Figure 3.8.14(b) animates the passage of a ladder through the corridors. The accompanying slider controls the angle θ (see Figure 3.8.14(a)) measured in degrees. As the slider is moved, the corresponding value of L, the length of the ladder, is displayed. As θ→0° or θ→90°, the length of the ladder becomes arbitrarily large, but of course, such large ladders would not fit around the corner of the passageway. Hence, it is the length of the shortest segment that determines the length of the longest ladder that can be brought around the corner of the passageway.
Figure 3.8.14(c) is a graph of L against the angle θ, measured in degrees. For the particular values of a and b used in Figure 3.8.14(b), there appears to be a minimum value for L near θ≐44°.
Figure 3.8.14(b) Slider-controlled animation
Figure 3.8.14(c) Graph of Lθ
From Figure 3.8.14(a), L= AB+BC, so that Lθ=asinθ+bcosθ.
Define the objective function Lθ
Context Panel: Assign Function
Lθ=asinθ+bcosθ→assign as functionL
Find critical numbers
Write the equation L′θ=0 and press the Enter key.
Context Panel: Solve≻Obtain Solutions for≻θ
Context Panel: Conversions≻To List
Context Panel: Assign to a Name≻T
→solutions for theta
→assign to a name
Evaluate Lθ for the real value of θ=θ^
Write LT1 and press the Enter key.
Context Panel: Simplify≻Assuming Positive
Additional manipulations (performed by hand) put the length of the longest ladder into the form a2/3+b2/33/2.
An alternate solution can be constructed along the following lines. Let point B in Figure 3.8.14(a) be the origin of a Cartesian coordinate system. Then the equation of the green line representing the ladder is y=m x, where m, the slope of the line, is to be determined. The intersection of this line with y=b is the point b/m,b; the intersection with the vertical line x=−a is the point −a,−a m, where a is taken as positive.
The length of the segment AC is then Lm=b/m+a2+b+a m2. The equation L′m=0 has multiple solutions for m=m^ ; the appropriate real, positive solution gives Lm^=a2/3+b2/33/2.
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