6 - 1 © 2012 Person Education, Inc.. All rights reserved. Chapter 6 Applications of the Derivative.

6 - 7 © 2012 Person Education, Inc.. All rights reserved. Your Turn 1 Find the absolute extrema of the function on the interval [0,8]. Solution: First look for critical numbers in the interval (0,8). Continued

6 - 8 © 2012 Person Education, Inc.. All rights reserved. Your Turn 1 continued Notice that at x = 2/5 and 2/5 is in the interval (0,8). The derivative is undefined at x = 0 but the function is defined there, so 0 is also a critical number. Evaluate the function at the critical numbers and the endpoints. Compare. Conclusions? x - ValueExtrema Candidates Value of the Function 00 2/50.977 Absolute Maximum 8-84 Absolute Minimum

6 - 9 © 2012 Person Education, Inc.. All rights reserved. Your Turn 2 Find the locations and values of the absolute extrema, if they exist, for the function Solution: In this example, the extreme value theorem does not apply since the domain is an open interval (−∞,∞), which has no endpoints. Begin as before by finding any critical numbers. x = 0 or x = − 4 or x = 1. Evaluate the function at the critical numbers. Continued

6 - 10 © 2012 Person Education, Inc.. All rights reserved. Your Turn 2 continued For an open interval, rather than evaluating the function at the endpoints, we evaluate the limit of the function when the endpoints are approached. Because the negative x 4 term dominates the other terms as x becomes large, f approaches negative infinity, so it has no absolute minimum. The absolute maximum 148, occurs at x = − 4. x - ValueExtrema Candidates Value of the Function -4148 Absolute Maximum 0 20 1 23

6 - 11 © 2012 Person Education, Inc.. All rights reserved. Practical example (textbook p.309) As the number of workers (hours of labor) varies, the output varies. The maximum output occurs at 320 hours of labor (that’s 8 employees). When is output per hour best? Ans: when the slope is the “most positive”, that’s the point where the tangent slope is maximum. This happens around 270 hours of labor (see graph). Graph of production output versus hours at work in a business.

6 - 14 © 2012 Person Education, Inc.. All rights reserved. Your Turn 1 Find two nonnegative numbers x and y for which x + 3y = 30, such that x 2 y is maximized. Solution:  Step 1: reading and understanding the problem  Step 2: does not apply in this example; there is nothing to draw.  Step 3: we determine what is to be maximized and assign a name to that quantity. Here x 2 y, is to be maximized, so let’s call it the function M M = x 2 y. we must express M in terms of just one variable, which can be done using the equation x + 3y = 30 and solving for either x or y. Solving for y gives: So now:

6 - 15 © 2012 Person Education, Inc.. All rights reserved. Your Turn 1 Continued  Step 4: Find the domain of the function. Because of the nonnegativity requirement, x must be at least 0. Since y must also be at least 0, this requires: So the domain of x is: [0,30].  Step 5: ﬁnd the critical points for M by finding and then solving

6 - 16 © 2012 Person Education, Inc.. All rights reserved. Your Turn 1 Continued x = 0 or x = 20  Step 6: find M for the critical numbers and endpoint of the domain. Conclusion: the maximum value of the function occurs when x = 20, y = 10/3 and the maximum value is 1333.3 x 0 20 30 M’ + - M

6 - 17 © 2012 Person Education, Inc.. All rights reserved. Your Turn 2 A math professor participating in the sport of orienteering must get to a specific tree in the woods as fast as possible. He can get there by traveling east along the trail for 300 m and then north through the woods for 800 m. He can run 160 m per minute along the trail but only 40 m per minute through the woods. Running directly through the woods toward the tree minimizes the distance, but he will be going slowly the whole time. He could instead run 300 m along the trail before entering the woods, maximizing the total distance but minimizing the time in the woods. Perhaps the fastest route is a combination, as shown. Find the path that will get him to the tree in the minimum time.

6 - 18 © 2012 Person Education, Inc.. All rights reserved. Your Turn 2 Continued Solution : Step 1:We are trying to minimize the total amount of time, which is the sum of the time on the trail plus the time through the woods. We must express this time as a function of x. Since: Time = Distance/ Speed, the total time is Notice in this equation that 0 ≤ x ≤ 300.

6 - 19 © 2012 Person Education, Inc.. All rights reserved. Your Turn 2 Continued xT(x) 021.88 20721.24 Absolute Minimum 30021.36 Conclusion: We see from the table that the time is minimized when x = 207 m. That is, when the professor goes 93 m along the trail and then heads into the woods.

6 - 24 © 2012 Person Education, Inc.. All rights reserved. Implicit Differentiation So far we worked with functions where one variable is expressed in terms of another variable—for example: y = or y = x sin x (in general: y = f (x). ) Some functions, however, are defined implicitly by a relation between x and y, example: x 3 + y 3 = 6xy We say that f is a function defined implicitly - Equation above means: x 3 + [f (x)] 3 = 6x f (x)

6 - 27 © 2012 Person Education, Inc.. All rights reserved. Your Turn 1 Solution: Differentiate with respect to x on both sides of the equation. Now differentiate each term on the left side of the equation. Think of xy as the product (x)(y) and use the product rule and the chain rule. Continued

6 - 31 © 2012 Person Education, Inc.. All rights reserved. Your Turn 3 The graph of is called the devil’s curve. Find the equation of the tangent line at the point (1, 1). Solution: We can calculate by implicit differentiation. Continued

6 - 32 © 2012 Person Education, Inc.. All rights reserved. Your Turn 3 Continued To find the slope of the tangent line at the point (1, 1), let x = 1 and y = 1 The slope is The equation of the tangent line is then found by using the point-slope form of the equation of a line.

6 - 36 © 2012 Person Education, Inc.. All rights reserved. Your Turn 1 Solution: We start by taking the derivative of the relationship, using the product and chain rules. Keep in mind that both x and y are functions of t. The result is Continued

6 - 38 © 2012 Person Education, Inc.. All rights reserved. Application problem: A 25-ft ladder is placed against a building. The base of the ladder is slipping away from the building at a rate of 3 ft per minute. Find the rate at which the top of the ladder is sliding down the building when the bottom of the ladder is 7 ft from the base of the building. Continued

6 - 39 © 2012 Person Education, Inc.. All rights reserved. Solution: Solution: Starting with Step 1, let y be the height of the top of the ladder above the ground, and let x be the distance of the base of the ladder from the base of the building. We are trying to find dy/dt when x = 7. To perform Step 2, use the Pythagorean theorem to write Both x and y are functions of time t (in minutes) after the moment that the ladder starts slipping. According to Step 3, take the derivative of both sides of the equation with respect to time, getting Continued

6 - 40 © 2012 Person Education, Inc.. All rights reserved. To complete Step 4, we need to find the values of x, y, and dx/dt. Once we find these, we can substitute them into Equation to find dy/dt. Since the base is sliding at the rate of 3 ft per minute, dx/dt = 3. Also, the base of the ladder is 7 ft from the base of the building, so x =7. Use this to find y. Continued

6 - 42 © 2012 Person Education, Inc.. All rights reserved. Problem 1 An oil rig leak spreads in a circular pattern with a radius increasing at the rate of 30m/hr. How fast is the area of the oil increasing when it has a radius of 100m? Application problems:

6 - 43 © 2012 Person Education, Inc.. All rights reserved. Problem 2 Sand falls from an overhead bin and forms a cone with a radius that remains always 3 times the height. If the sand is falling at the rate of 120 ft 3 /min, how fast is the height of the sand pile changing when it is 10 ft high?

6 - 44 © 2012 Person Education, Inc.. All rights reserved. Problem 3 Two planes approach an airport, one flying due West at 120 mi/hr and the other flying due North at 150mi/hr. Assuming they stay at the same constant elevation, how fast is the distance between them changing when the westbound plane is 180 mi from the airport, and the northbound one is at 225 mi from the airport?

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6 - 1 © 2012 Person Education, Inc.. All rights reserved. Chapter 6 Applications of the Derivative.

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