Math 150A Ladder Around a Corner Laura Smith, Ph.D. We are to carry a stepladder down a hallway that is 9 feet wide. Unfortunately, at the end of the hallway there is a corner, where we must make a right-angled turn into a hallway that is 6 feet wide. Our goal is to determine the maximum length of a ladder that could be carried horizontally around this corner. In order to solve this problem, we must consider a ladder that touches the exterior wall in our hallway, the exterior wall in the other hallway, and the interior corner. Any ladder that cannot touch all three will be short enough to pass. Thus, the problem has turned into finding the shortest ladder that will touch all three. Let L be the length of a ladder that touches the exterior wall in the 9-foot hallway at point A, the exterior wall in the 6-foot hallway at point B, and the interior corner at point C (see Figure 1). Extending the walls of the hallways on a diagram, we let D be the point where the 6-foot hallway extension intersects the wall, and let E be the point where the 9foot hallway extension intersects the other wall (see Figure 1). Let x be the length of line segment AC, and let y be the length of line segment CB. We notice that the angle  between line segments DC and AD is the same as the angle between line segments EB and BC. Using the definitions of trigonometric functions, we can write the length of the ladder L as a function of this angle , 𝐿(𝜃) = 𝑥 + 𝑦 = 9 ∙ csc(𝜃) + 6 ∙ sec(𝜃). 𝜋 The domain of this function is 0 < 𝜃 < 2 . To find the minimal L, we find our absolute minimum using the first derivative test for absolute extreme values. Thus, we need to find the critical numbers by taking the derivative. We have 𝐿′ (𝜃) = −9 ∙ csc(𝜃) ∙ cot(𝜃) + 6 ∙ sec(𝜃) ∙ tan(𝜃). To find the critical numbers, we set our derivative equal to zero. This gives us 0= −9cos(𝜃) sin2 (𝜃) + 6sin(𝜃) cos2 (𝜃) = −9 cos3(𝜃)+6sin3 (𝜃) sin2 (𝜃)cos2 (𝜃) , which is zero when the numerator is equal to zero. Hence, −9 cos 3 (𝜃) + 6sin3 (𝜃) = 0, or 3 equivalently, 2 = 𝑡𝑎𝑛3 (𝜃). Thus, one critical number is 3 1/3 𝜃 = 𝑡𝑎𝑛−1 ((2) ). We then search for when the derivative is undefined, which occurs when either sine or cosine is equal to zero on the domain. However, there are no such values on our domain. We now examine where L’() is positive and negative on our domain. We have that 3 1/3 our derivative is negative for 0
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