# It follows that sec θ 3 or sec θ 1 we reject the

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It follows that sec θ = 3 or sec θ = 1. We reject the latter as before, and find that sec θ = 3, so θ 1 . 23095 (radians). The resulting minimum volume of the cone is 8 3 πR 3 , twice the volume of the sphere! r Θ Θ h R 3.7.82: Let L be the length of the crease. Then the right triangle of which L is the hypotenuse has sides L cos θ and L sin θ . Now 20 = L sin θ + L sin θ cos 2 θ , so L = L ( θ ) = 20 (sin θ )(1 + cos 2 θ ) , 0 < θ π 4 . Next, dL = 0 when (cos θ )(1 + cos 2 θ ) = (sin θ )(2 sin 2 θ ); (cos θ )(2 cos 2 θ ) = 4 sin 2 θ cos θ ; 203
so cos θ = 0 (which is impossible given the domain of L ) or cos 2 θ = 2 sin 2 θ = 2 2 cos 2 θ ; cos 2 θ = 2 3 . This implies that cos θ = 1 3 6 and sin θ = 1 3 3. Because L + as θ 0 + , we have a minimum either at the horizontal tangent just found or at the endpoint θ = 1 4 π . The value of L at 1 4 π is 20 2 28 . 28 and at the horizontal tangent we have L = 15 3 25 . 98. So the shortest crease is obtained when cos θ = 1 3 6; that is, for θ approximately 35 15 52 . The bottom of the crease should be one-quarter of the way across the page from the lower left-hand corner. 3.7.83: Set up coordinates so the diameter is on the x -axis and the equation of the circle is x 2 + y 2 = 1; let ( x , y ) denote the northwest corner of the trapezoid. The chord from (1 , 0) to ( x, y ) forms a right triangle with hypotenuse 2, side z opposite angle θ , and side w ; moreover, z = 2 sin θ and w = 2 cos θ . It follows that y = w sin θ = 2 sin θ cos θ and x = 1 w cos θ = cos 2 θ. Now A = y (1 x ) = (2 sin θ cos θ )(1 cos 2 θ ) = 4 sin θ cos θ sin 2 θ, and therefore A = A ( θ ) = 4 sin 3 θ cos θ, π 4 θ π 2 . A ( θ ) = 12 sin 2 θ cos 2 θ 4 sin 4 θ = (4 sin 2 θ )(3 cos 2 θ sin 2 θ ) . To solve A ( θ ) = 0, we note that sin θ = 0, so we must have 3 cos 2 θ = sin 2 θ ; that is tan 2 θ = 3. It follows that θ = 1 3 π . The value of A here exceeds its value at the endpoints, so we have found the maximum value of the area—it is 3 4 3. 3.7.84: Let θ = α/ 2 (see Fig. 3.7.18 of the text) and denote the radius of the circular log by r . Using the technique of the solution of Problem 82, we find that the area of the hexagon is A = A ( θ ) = 8 r 2 sin 3 θ cos θ, 0 θ π 2 . After some simplifications we also find that dA = 8 r 2 (sin 2 θ )(4 cos 2 θ 1) . 204
Now dA/dθ = 0 when sin θ = 0 and when cos θ = 1 2 . When sin θ = 0, A = 0; also, A (0) = 0 = A ( 1 2 π ) . Therefore A is maximal when cos θ = 1 2 : θ = 1 3 π . When this happens, we find that α = 2 3 π and that β = π θ = 2 3 π . Therefore the figure of maximal area is a regular hexagon. 3.7.85: The area in question is the area of the sector minus the area of the triangle in Fig. 3.7.19 and turns out to be A = 1 2 r 2 θ r 2 cos θ 2 sin θ 2 = 1 2 r 2 ( θ sin θ ) = s 2 ( θ sin θ ) 2 θ 2 because s = . Now dA = s 2 (2 sin θ θ cos θ θ ) 2 θ 3 , so dA/dθ = 0 when θ (1 + cos θ ) = 2 sin θ . Let θ = 2 x ; note that 0 < x π because 0 < θ 2 π . So the condition that dA/dθ = 0 becomes x = sin θ 1 + cos θ = tan x.