Math 1a Worksheet 25 (11/19) Optimization Using Calculus II SOLUTIONS Reader Beware! These are meant to be a sketch of the solutions. They do not necessarily represent the type of complete solutions you are expected to give on homeworks and exams. If you have a question or spot a mistake, please do let me know! 1. To get the best view of the Statue of Liberty, you should be at the position where θ is at a maximum. If the statue stands 92 meters high, including the pedestal, which is 46 meters high, how far from the base should you be? [Hint: Find a formula for θ in terms of your distance from the base. Use this function to maximize θ, noting that 0 < θ < π/2.] We let x be the distance from the base of the statue to the observer. Notice that as x changes, both the angle θ and the angle of the smaller triangle (with height 46) change. We label this angle α. 92 - 46 - Θ α x We are looking for a formula for θ. Using trig we know that tan(α + θ) = 92 . x To reduce this to an equation involving only x and θ, we can use the relation tan α = 46 x and the trig identity tan(α + θ) = tan α + tan θ . 1 − tan α tan θ This gives us 46/x + tan θ 92 = 1 − (46/x) tan θ x which we can solve for tan θ to get tan θ = x2 46x + 4232 The domain for the function θ is 0 < x < ∞. We can either solve for θ by taking the arctangent of both sides, or use implicit differentiation. Let’s try the second way: d d 46x tan θ = dx ( x2 +4232 ) dx dθ sec2 θ dx = dθ dx 46(x2 +4232)−2x(46x) (x2 +4232)2 2 = (cos2 θ)(46) (x4232−x 2 +4232)2 dθ This is defined everywhere, so the only critical points will be where dx is zero. Since 0 < θ < π/2, the cosine will never be equal to zero. We can see that the √ √ derivative is positive on (0, 4232) and negative on ( 4232, ∞). By the first derivative √ test for absolute extrema, this means that θ has an absolute maximum when x = 4232 ≈ 65 meters. The angle θ(x) is ≈ .34 radians, or 19 degrees. Therefore, you should stand about 65 meters from the statue to maximize your viewing angle θ. 2. A holiday ornament is to be constructed by inscribing a right circular cone of brightly colored material in a transparent spherical ball of radius 2 inches. What is the maximum possible volume of such a cone? To think about this, we want to find the dimensions of the cone: it’s radius and height. Here’s a two-dimensional slice of the ornament: 2 x 2 y (Note: the drawing isn’t perfect; the outer curve should be a perfect circle.) We want to maximize the volume of the cone: 1 V (r, h) = πr2 h 3 Using the notation from the drawing, we see that the radius r of the cone is labeled as y and the height h of the cone is labeled as x + 2 1 V (x, y) = πy 2 (x + 2) 3 To get a function of one variable, we use the relationship of x and y: they are√the legs of a right triangle with hypotenuse 2 (the radius of the sphere), and so y = 4 − x2 . We now have 1 1 √ V (x) = π( 4 − x2 )2 (x + 2) = π(4 − x2 )(x + 2) 3 3 Notice that the domain of x is [−2, 2] (though the endpoints give a cone with no volume, so they certainly won’t give the absolute maximum.) To find the critical points, we look at V 0 (x) = − 13 π(3x2 + 4x − 4) which is defined everywhere and zero at x = −2 and x = 23 . Using the Closed Interval Method, we check all critical points and the endpoints x = ±2: x −2 2 3 2 V (x) 0 1 π(256/27) ≈ 9.93 3 0 Clearly, the absolute maximum volume occurs when x = 23 . In this case, the height √ is 38 and the radius is 4 3 2 . Note: There are many other ways to solve this problem, for example, you can represent the height and radius of the cone in terms of the angle θ, representing half the top angle of the triangle. 3. Do Dogs Know Calculus?1 A math professor is playing fetch with his Corgi dog at the shore. He stands 15 m away from his dog on the shore and throws the ball 2 meters into the water. The dog runs part way down the beach then jumps in the water and swims to the ball. If the dog can run at a speed of 6 m/s and swim at a speed of 1 m/s, and the dog wants to get to the ball the fastest he can, where does he jump in the water? 2.0 1.5 1.0 0.5 2 4 6 8 10 12 14 The function we want to minimize is the time for the dog to reach the ball. If the dog jumps into the water x meters from the professor, we have √ 15 − x 4 + x2 T (x) = + . 6 1 The domain is [0, 15] and we’re looking for extreme values, so we consider √ 2x 6x − 4 + x2 1 √ = T 0 (x) = − + √ 6 2 4 + x2 6 4 + x2 q 4 It’s defined everywhere and is 0 at x = ± 35 . q 4 We can use the first derivative test to that T at x = 35 ≈ .34 has a local minimum q q 4 4 since T 0 (x) < 0 for x < 35 and T 0 (x) > 0 for x > 35 . In fact, it gives an absolute minimum (no need to check the endpoints) since q the graph is decreasing from T (0) √ 4 ) must be the absolute minimum. until T ( 435 and increasing afterwards, so T ( 35 q 4 The time it takes for the dog to reach the ball is T ( 35 ) ≈ 2.55 seconds. 1 This problem is adapted from an article by Professor Tim Pennings in the College Mathematics Journal 34(May):178-182, 2003.