CALCULUS I HOMEWORK 9.. Taking the derivative, f (x) = 1 1 x 2.

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1 CALCULUS I HOMEWORK 9 Due November 30. Let x > 0. Prove that x + x 2. Solution: Let f(x) = x + x. Taking the derivative, Setting f (x) = 0 gives f (x) = x 2. = x 2. If x > 0, the only critical point is at x =. On (0, ), we have f < 0 and the function f is decreasing. On (, ), we have f > 0 and the function f is increasing. Therefore x = is the absolute minimum of f on (0, ). We can evaluate f() = + = 2. Therefore f(x) = x + x Compute the antiderivative (a) f(x) = e 3x + x 3 (b) f(x) = x 4 + 3x 2 + 2x (c) f(x) = x + x (d) f(x) = 2 cos x sin 2x F (x) = 3 e3x x C. F (x) = x5 5 + x3 + x 2 + C. F (x) = ln x + 2x 2 + C.

2 2 CALCULUS I HOMEWORK 9 F (x) = 2 sin x + cos 2x + C (a) The lord of a castle decides to build a city wall. The wall is to be m thick and 5 m high. The lord has 000 m 3 worth of stone to build this fortification. What are the lengths of the sides of the wall which maximize the enclosed area? Solution: Let x be the length of one side of the wall (measured from outside the castle), and y be the length of the other side (also measured from outside the castle). The total volume of the wall is 000 = ()(5)(y)+()(5)(y)+()(5)(x 2)+()(5)(x 2) = 0y +0(x 2). Note that we cannot set 000 = 0x+0y as this would be overcounting the volume of the wall in the corners. The wall has thickness, so we subtracted (2)() from two of the sides. We can solve for y in terms of x. The area enclosed is given by 00 = y + x 2, y = 02 x. A = xy = x(02 x) = x x. Taking the derivative, da = 2x Setting the derivative to zero, we see that the critical point is at x = 5. This is a maximum since da da > 0 for x < 5 and < 0 for x > 5. Hence the optimal dimensions are to have both sides of length 5m. What this means is if you took a tape measure and measured the length of one of the walls from outside the castle, the length would be 5 meters. If you measured from inside the castle, the length would be 5 = 49 meters. (b) Several years after the construction of the wall, a large army of aggressors attack the castle. A catapult is installed 50 m from the city wall. If a projectile is catapulted towards the wall at an angle θ at time t = 0, its height in meters is given by { (25 sin θ) t 5t 2 if (25 sin θ) t 5t h(t) = otherwise The projectile s distance from the catapult in meters is d(t) = (25 cos θ) t.

3 CALCULUS I HOMEWORK 9 3 (Note: if you have taken physics, these equations come from the equations of motion of a projectile with initial velocity 25 m/s with angle θ from the horizontal, taking the gravitational acceleration to be g = 0m/s 2.) Find the angle θ which maximizes the height of the projectile when the projectile reaches the wall. Is it possible for the catapult to launch a projectile over the city wall? Solution: The projectile reaches the wall when d = 50, and we can solve for the time of impact 50 = (25 cos θ) t, t = 2 cos θ. The height of the projectile at the wall is thus ( ) ( ) h(θ) = (25 sin θ) 5 = 50 tan θ 20(cos θ) 2. cos θ cos θ Taking the derivative dh dθ = 50(cos θ) 2 40(cos θ) 3 sin θ. Setting the derivative to zero gives the following equation for the critical angle 50(cos θ) 2 = 40(cos θ) 3 sin θ. Simplification yields, sin θ cos θ = 5 4. Therefore, the critical angle is θ = arctan(5/4), which is approximately radians, or 5.3 degrees. At this angle, the projectile reaches the wall at height h = 50 tan arctan(5/4) 20(cos arctan(5/4)) 2 =.25. The wall only has height 5m, so the projectile is flying over the city wall. (c) The wizard who lives in the castle climbs to the top of the castle and retaliates against the attack. The wizard has 500 mana and can blast a fireball at a cost of mana or conjure a lightning strike at a cost of 6 mana. Mathematically, this means that if the wizard blasts x fireballs and conjures y lightning strikes, then we must have x + 6y 500. A fireball causes 2 units of damage. The first lightning strike causes 320 units of damage, but each strike is only (3/4) as strong as the previous one due to the depletion of the electric field. For example, a first lightning strike would deal 320 units of damage, a second strike would deal 240 units of the damage, a third strike would deal 80 units of damage, and so on.

4 4 CALCULUS I HOMEWORK 9 To maximize damage, how many fireballs should the wizard propel into the attacking army, and how many lightning strikes should he conjure? The following formula for the geometric series may be useful: if 0 < r <, then n r i = + r + r r n = rn+ r. i=0 Solution: Let x denote the number of fireballs, and y denote the number of lightning strikes. The damage dealt is then { ) 2 ( } 3 y D = 2x + (320) + (320) + (320) + + (320) 4) 4 4) Using summation notation, this can be expressed as y ) i. D = 2x + (320) 4 Using the formula for geometric series, i=0 D = 2x + (320) (3/4)y (3/4). Simplifying D = 2x + (280)( (3/4) y ). Supposing for the moment that x and y are continuous variables, we may solve x + 6y = 500. We will need to address the fact that x and y must be integers after finding the critical point, but for now we solve x = 500 6y. Therefore, the damage as a function of y is D = 2(500 6y) + (280)( (3/4) y ). Taking the derivative dd d = dy dy (3/4)y = ln(3/4)(3/4) y. Setting dd dy = 0 Solving for y gives 2 = 280 ln(3/4)(3/4) y, (3/4) y = ln(3/4). y = ( ln(3/4) ln ).9 ln(3/4) Testing points on either side of.9, we see that dd > 0 on (,.9) and dd dy < 0 on (.9, ). Hence.9 is the absolute maximum. dy

5 CALCULUS I HOMEWORK 9 5 Since y needs to be an integer, the above analysis tells us that the maximum will occur when we are closest to y =.9. To be safe, we check both y = and y = 2. When y =, the largest value that x can be is x = 500 6() = 434 When y = 2, the largest value that x can be is x = 500 6(2) = 428 We check x = 434, y = into the damage function D = 2(434) + (280)( (3/4) ) We check x = 428, y = 2 into the damage function D = 2(428) + (280)( (3/4) 2 ) The maximum damage occurs when the wizard shoots 428 fireballs and generates 2 lightning strikes.