MA123, Chapter 7 Word Problems (pp. 125-153) Chapter s Goal: In this chapter we study the two main types of word problems in Calculus. Optimization Problems. i.e., max - min problems Related Rates See animations and interactive applets of some of these at http://www.math.uky.edu/~pkoester/teaching/ Fall_2009/Math123/Notes () Chapter 7 October 31, 2009 1 / 64 Example 1: What is the largest possible product you can form from two nonnegative real numbers whose sum is 30? What are the unknowns? Two nonnegative real numbers, say x and y. What is given? Their sum is 30, the product is maximal, they are each greater than or equal to 0. Mathematize these givens: x + y = 30, x 0, and y 0 () Chapter 7 October 31, 2009 2 / 64
Example 1 (continued) xy is to be maximal. This suggests using calculus to find a maximum, but we can t yet since we have two variables. Elimate an unknown: y = 30 x so xy = x (30 x) = 30x x 2 The product now depends only on x. Also, y 0 = x 30 New problem: Find the maximum of P (x) = 30x x 2 on [0, 30]. This sort of problem can be solved by the techniques of Chapter 6. () Chapter 7 October 31, 2009 3 / 64 Example 1 (continued) Find the maximum of P (x) = 30x x 2 on [0, 30]. Endpoints: x = 0 and x = 30 Next, P (x) = 30 2x so P (x) = 0 implies x = 15. Critical Points: x = 15 Plug in endpoints and critical points: P (0) = 0, P (15) = 15 2 = 225, P (30) = 0 The maximum of P (x) is P (15) = 225. Relate back to our original problem: The largest possible product of x and y, whose sum is 30, occurs when x = y = 15. The product is 225. () Chapter 7 October 31, 2009 4 / 64
Guidelines for Optimization Problems: While there is no definite procedure to follow in solving optimization problems, these guidelines should help Read problem quickly: Get idea of what problem is asking What am I trying to find? What is given? Read problem more carefully: Pay more attention to details How does given information relate to the unknown? Define variables. Choose suggestive names for your varibles. Draw and label sketches. Determine if you want a maximum or a minimum. Determine if you want the extreme value or if you want the value of some quantity which gives the extreme value. () Chapter 7 October 31, 2009 5 / 64 Guidelines for Optimization Problems: Write down a general formula for what you are trying to optimize. Often your formula will involve several variables. If so: Write down equations relating the variables Solve the previous equations in terms of a single variable Substitute your equations from the last step into the general formula to get a function of a single variable Use methods from chapter 6 to find maximum or minimum. In Chapter 6 we were always given the interval on which function was defined Here we need to use the word problem to determine this interval Look for critical points, look for endpoints. You may have several possible points for max or min Check your point is max or min ( via First Derivative Test) () Chapter 7 October 31, 2009 6 / 64
Guidelines for Optimization Problems: REFLECT UPON YOUR SOLUTION You can learn a lot by looking back at your solution Does your answer seem reasonable? Is there an easier way to solve your problem? Can you solve variations of your problem with your solution? () Chapter 7 October 31, 2009 7 / 64 Second Derivative Test for Extrema The First Derivative Test is not always the most efficient way to determine if a given critical point is a max or a min. This next test is better, provided f (x) is easy to compute. Theorem Suppose f(x) has a critical point at x = c and f (x) exists for all x near c. If f (c) > 0 then f(c) is a local minimum of f(x). If f (c) < 0 then f(c) is a local maximum of f(x). If f (c) = 0 then we need to use the first derivative test to determine if maximum, minimum, or neither. () Chapter 7 October 31, 2009 8 / 64
Second Derivative Test for Extrema A curve bends up away from a local minimum. Thus, y = f(x) is concave up near a local minimum. Thus, f (c) > 0 at a local minimum. A curve bends down away from a local maximum. Thus, y = f(x) is concave down near a local maximum. Thus, f (c) < 0 at a local maximum. () Chapter 7 October 31, 2009 9 / 64 Example 2 Suppose the product of x and y is 26, and both x and y are positive. What is the minimum possible sum of x and y? What are the unknowns? Two positive real numbers x and y. What is given? Their product is 26, their sum is minimal, they are each greater than 0. Mathematize these givens: x > 0, y > 0, and xy = 26 () Chapter 7 October 31, 2009 10 / 64
Example 2 (continued) x + y is to be minimal. This suggests using calculus to find a minimum, but we can t yet since we have two variables. Eliminate a variable: y = 26 x 1 New problem: Minimize the function on the interval (0, ). S(x) = x + 26 x 1 Endpoints: NONE () Chapter 7 October 31, 2009 11 / 64 Example 2 (continued) Next, S (x) = 1 26 x 2 So S (x) = 0 implies x = 26. (Recall x > 0.) Next, S (x) = 52 x 3 > 0 for all x > 0 S(x) is concave up on [0, ) so critical point gives a minimum. Relate back to the original problem: Minimal sum occurs when x = 26 and y = 26 26 = 26 Minimal sum 26 + 26 = 2 26 () Chapter 7 October 31, 2009 12 / 64
Example 3 A farmer builds a rectangular pen with three parallel partitions using 500 feet of fencing. What dimensions will maximize the total area of the pen? Let x denote the horizontal length and y denote the vertical length of fence. Unknowns: Dimensions of the fence which maximize area, xy Givens: 500 feet of fence to work with. Need to build four walls of length y and two walls of length x. Thus, 2x + 4y = 500, x > 0 and y > 0 () Chapter 7 October 31, 2009 13 / 64 Example 3 (continued) The condition 2x + 4y = 500 rearranges to x = 250 2y. x > 0 implies 250 2y > 0 so y < 125. New Problem: Maximize the function on [0, 125]. Find the critical points: and so y = 125 2 = 62.5 A(y) = (250 2y) y = 250y 2y 2 A (y) = 250 4y = 0 () Chapter 7 October 31, 2009 14 / 64
Example 3 (continued) Plug in critical points and endpoints to determine maximum: A(0) = 0, A(62.5) = 7812.5, A(125) = 0 so the maximum is at x = 62.5. Relate to original problem: Maximum area is obtained making vertical sides 62.5 feet long and horizontal sides 125 feet long. Maximal area is 7812.5 square feet. Remark: We need to build four vertical walls and only two horizontal walls, so we expect the vertical walls to be shorter. In fact, there are twice as many vertical walls as horizontal walls, so the vertical walls should have half the length of the horizontal walls. () Chapter 7 October 31, 2009 15 / 64 Example 4 A Norman window has the shape of a rectangle capped by a semicircle. What is the length of the base of a Norman window of maximal area if the perimeter of the window is 10? Let r be the radius of semicircle and h the height of rectangle. Base of rectangle is 2r. Perimeter is Two vertical walls + Base + Circumference Semi-circle 2h + 2r + π r So 10 = 2h + (2 + π) r () Chapter 7 October 31, 2009 16 / 64
Example 4 (continued) Solving for h: ( h = 5 1 + π ) 2 h 0 so (1 + π 2 ) r 5 so r 5 1 + π = 10 2 + π 2 Area of window is area of rectangle plus area of semicircle. Area of rectangle: (2r) h = 2r ( ( 5 1 + π ) 2 r ) r = 10r (2 + π) r 2 Area of semicircle: 1 2 π r2 () Chapter 7 October 31, 2009 17 / 64 Area of window: A(r) = (10r (2 + π) r 2 ) + 1 2 πr2 = 10r ( 2 + π ) 2 r 2 We need to maximize the function ( A(r) = 10r 2 + π ) 2 r 2 on the interval [0, 10 2+π ]. First find critical points: A (r) = 10 2 ( 2 + π ) 2 r = 0 so r = 5 2 + π 2 () Chapter 7 October 31, 2009 18 / 64
Example 4 (continued) Maximal area occurs at either a critical point or an endpoint. But look at second derivative: ( A (r) = 2 2 + π ) < 0 for all r 2 A(r) is concave down everywhere so critical point gives maximum. So maximum when r = 5 1.400 2+ π 2 What is the base of the maximal area window? The base is twice the radius, so base will be 10 2 + π 2 2.800 () Chapter 7 October 31, 2009 19 / 64 Example 5 Find the area of the largest rectangle with sides parallel to the coordinate axes that can be inscribed in a quarter circle of radius 10. Assume the center of the circle is located at the origin and one corner of the rectangle is at the origin and the opposite corner is on the quarter circle. Let x be the horizontal length of the rectangle and y the horizontal length. Right triangle, sides x and y, hypotenuse 10 x 2 + y 2 = 10 2 Solving for y y = 10 2 x 2, for x 10 () Chapter 7 October 31, 2009 20 / 64
Example 5 (continued) Objective: Maximize the area, xy. Substituting y, need to maximize the function on the interval [0, 10]. A(x) = x (10 2 x 2 ) 1/2 () Chapter 7 October 31, 2009 21 / 64 Example 5 (continued) Critical points: A (x) = (10 2 x 2 ) 1/2 + x ( ) 1 (10 2 x 2 ) 1/2 ( 2x) 2 = (10 2 x 2 ) 1/2 x 2 (10 2 x 2 ) 1/2 = 102 2x 2 102 x 2 Now, A (x) = 0 implies 10 2 2x 2 = 0 so x = 50. () Chapter 7 October 31, 2009 22 / 64
Example 5 (continued) Plug in critical points and endpoints A(0) = 0, A( 50) = 5, A(10) = 0 So maximal area is 50 and occurs when x = 50 and y = 50. Remark: In retrospect, this conforms to our intuition. The boundary curve is symmetric with respect to x and y so may expect maxima to occur at a symmetric point, x = y. () Chapter 7 October 31, 2009 23 / 64 Example 6 Let A be the point (0, 1) and B be the point (5, 3). Find the length of the shortest path that connects A and B if the path consists of two line segments and the path must touch the x axis. Let P = (x, 0). Minimize sum of distances from A to P and from P to B. Distance from A to P : (x 0)2 + (0 1) 2 = (x 2 + 1) 1/2 Distance from P to B: (5 x)2 + (3 0) 2 = (x 2 10x + 34) 1/2 () Chapter 7 October 31, 2009 24 / 64
Example 6 (continued) Minimize (x 2 + 1) 1/2 + (x 2 10x + 34) 1/2 Taking derivative and simplifying: 1 2 (x2 + 1) 1/2 2x + 1 2 (x2 10x + 34) 1/2 (2x 10) = x x2 + 1 + x 5 x2 10x + 34 = 0 So x x2 + 1 = x 5 x2 10x + 34 () Chapter 7 October 31, 2009 25 / 64 Example 6 (continued) Square both sides: Clear denominators: Multiply out: x 2 x 2 + 1 = x2 10x + 25 x 2 10x + 34 x 2 (x 2 10x + 34) = (x 2 10x + 25) (x 2 + 1) x 4 10x 3 + 34x 2 = x 4 10x 3 + 25x 2 + x 2 10x + 25 Cancel and collect like terms: 8x 2 + 10x 25 = 0 () Chapter 7 October 31, 2009 26 / 64
Example 6 (continued) Factor: (2x + 5)(4x 5) = 0 so x = 5/2 or x = 5/4. The minimum distance occurs at point (5/4, 0). The minimum distance is ((5/4) 2 + 1) 1/2 + ((5/4) 2 10 (5/4) + 34) 1/2 6.4 (plug x = 5/4 into total distance (x 2 + 1) 1/2 + (x 2 10x + 34) 1/2 ) Remark: Notice the two triangles in this case are similar, which means the angles of incidence and reflection are equal. Remark: What was the significance of the critical point x = 5/2? In fact this was not a critical point. We found the critical points by taking an algebraic equation and squaring that equation. x = 5/2 was an extraneous root. () Chapter 7 October 31, 2009 27 / 64 Example 6: Short and Elegant Solution Is there an easier way? YES!!! Shortest path from (0, 1) to (5, 3) that hits x-axis Shortest path from (0, 1) to (5, 3) that hits x-axis Any path from (0, 1) to (5, 3) hits x-axis Thus, shortest path from (0, 1) to (5, 3) is a line, so length of shortest path distance between (0, 1) and (5, 3) : (5 0)2 + ( 3 1) 2 = 25 + 16 = 41 6.4 () Chapter 7 October 31, 2009 28 / 64
Example 7 Find the area of the largest rectangle with one corner at the origin, the opposite corner in the first quadrant on the graph of the parabola f(x) = 9 x 2, and sides parallel to the axes. Let x be horizontal length of rectangle. The corner of rectangle on the parabola has coordinates (x, 9 x 2 ) so height of rectangle is 9 x 2. Have x 0 and y 0. But y 0 requires x 3 so 0 x 3. Thus, need to maximize A(x) = x(9 x 2 ) = 9x x 3 on [0, 3] () Chapter 7 October 31, 2009 29 / 64 Example 7 (continued) Find critical points: A (x) = 9 3x 2 = 0 = x 2 = 3 so x = 3. (Remember 0 x 3.) Next, A (x) = 6x 0 for all x 0 so A(x) is concave down for x > 0. Critical point must be a global maximum. Thus, the maximum at x = 3 and maximal area is A( 3) = 6 3 () Chapter 7 October 31, 2009 30 / 64
Example 8 Find the point P in the first quadrant that lies on the hyperbola y 2 x 2 = 6 and is closest to the point A(2, 0). Write the point P as (a, b). Closest point so need to minimize the distance. (a, b) lies on y 2 x 2 = 6 so b 2 a 2 = 6, so b = 6 + a 2. Distance between (a, b) and (2, 0) is (a 2)2 + (b 0) 2 = a 2 4a + 4 + 6 + a 2 = 2a 2 4a + 10 () Chapter 7 October 31, 2009 31 / 64 Example 8 (continued) We need to minimize y = 2a 2 4a + 10 However, it is enough to minimize f(a) = 2a 2 4a + 10 since y = x is increasing. Now, f (a) = 4a 4 = 0 = a = 1 Then b = 1 2 + 6 = 7. Minimal distance: 2 12 4 1 + 10 = 8 2.8284 () Chapter 7 October 31, 2009 32 / 64
Related Rates and Implicit Derivatives The derivative measures the rate of change of a quantity. But rate of change with respect to what? The area of a circle, A, depends on its radius, r. But suppose the radius is increasing with time, t Thus, the area of the circle will also increase with respect to time. What is A? This symbol is ambiguous. It could mean the rate of change of the area as the radius changes, or it could be the rate of change of the area with respect to time. () Chapter 7 October 31, 2009 33 / 64 Related Rates and Implicit Derivatives Here it is better to use Leibniz notation for derivatives: da = rate of change of area with respect to time da dr = rate of change of area with respect to radius dr = rate of change of radius with respect to time This notation clearly specifies both variables, the independent and the dependent. () Chapter 7 October 31, 2009 34 / 64
Related Rates and Implicit Derivatives The derivatives da, are related by the chain rule: da dr, da = da dr dr Thus, if we know any two of these derivatives, then we will be able to find the third. In this case, A = πr 2 and so da dr = 2πr2 But knowing da would require also knowing dr. A related rates problem requires us to find the derivative of some quantity by relating that derivative to a given (or related) quantity. () Chapter 7 October 31, 2009 35 / 64 dr Example 9 Consider the area of a circle A = π r 2 and assume that r depends on t. Find a formula for da. Compute derivative with respect to t on both sides of area equation: d (A) = d ( ) π r 2 Left hand side is da Right hand side is Thus, π d da ( r 2 ) = π 2r dr = 2π r dr () Chapter 7 October 31, 2009 36 / 64
Example 9 (continued) da = 2π r dr The 2πr gives the rate of change of the area with respect to the radius (that is, it is da dr ) () Chapter 7 October 31, 2009 37 / 64 Guidelines for Related Rates Problems: While there is no definite procedure to follow in solving related rates problems, these guidelines should help Read problem quickly: Get idea of what problem is asking Ask yourself What am I trying to find? What is given? Read problem more carefully: Pay more attention to details. Ask yourself How do the givens relate to the unknown? Define variables. Choose suggestive names for your varibles. Draw and label sketches. In most problems, all quantities are implicit functions of time. Decide which given quantities are constant and which givens are variable. DO NOT plug in numeric values for variables at this point. () Chapter 7 October 31, 2009 38 / 64
Guidelines for Related Rates Problems: Write down each given derivative. These will have units blah per blah, like feet per second, or miles per hour, or dollars per unit. Write down the derivative you want to find. Find an equation relating all of the variables. DO NOT plug in numeric values for any quantity that can change. Use the chain rule to find the derivative implicitly using the equation in the last step. Plug in all given values AFTER taking derivatives and solve for the desired derivative. () Chapter 7 October 31, 2009 39 / 64 Guidelines for Related Rates Problems: REFLECT UPON YOUR SOLUTION You can learn a lot by looking back at your solution Does your answer seem reasonable? Does the derivative have the correct ± sign? Is there an easier way to solve your problem? Can you solve variations of your problem with your solution? () Chapter 7 October 31, 2009 40 / 64
Example 10 Boyle s Law states that when a sample gas is compressed at a constant temperature, the pressure and volume satisfy the equation P V = c where c is a constant. Suppose that at a certain instant the volume is 600 cubic centimeters, the pressure is 150 kpa, and the pressure is increasing at a rate of 20 kpa/ min. At what rate is the volume decreasing at this instant? Want dv. c is constant so dc = 0. On the other hand, dc = d dp (P V ) = V + P dv so 0 = dp V + P dv () Chapter 7 October 31, 2009 41 / 64 Example 9 (continued) Given: V = 600, P = 150, and dp = 20 Plugging in gives 20 600 + 150 dv = 0 Solving for unknown dv dv 20 600 = = 80. 150 Volume changing at rate of 80 cubic centimeters per minute. Negative sign means volume is decreasing at 80cm 3 /min () Chapter 7 October 31, 2009 42 / 64
Example 11 A train is travelling over a bridge at 30 miles per hour. A man on the train is walking toward the rear of the train at 2 miles per hour. How fast is the man travelling across the bridge in miles per hour? Speed of man relative to bridge is speed of train relative to bridge minus speed of man relative to bridge. Man s speed relative to bridge is then 30 2 = 28 miles per hour () Chapter 7 October 31, 2009 43 / 64 Example 12 Two trains leave stations 100 miles apart on parallel tracks one hundred feet apart. (the first track is one hundred feet north of the second track). The trains travel toward each other, one going east and the other going west, each traveling at 47 miles per hour. How fast is the distance between the trains changing just as they pass each other? Let D be the distance between the two trains. Distance will be minimal as the trains pass each other. Therefore, distance has a critical point when the trains pass. Therefore, dd = 0 when the trains are passing each other. () Chapter 7 October 31, 2009 44 / 64
Example 13 Two trains leave a station at the same time. One travels north on a track at 30 miles per hour. The second travels east on a track at 46 miles per hour. How fast are they travelling away from each other when the northbound train is 60 miles from the station? x is distance travelled by Eastbound train y is distance travelled by Northbound train D = x 2 + y 2 distance between trains Given: dx Unknown: dd = 46 mph, dy = 30 mph after northbound travels 60 miles. () Chapter 7 October 31, 2009 45 / 64 Example 13 (continued) Could take d on both sides of D = x 2 + y 2. Easier to take d if we square both sides: D 2 = x 2 + y 2 = d ( D 2 ) = d ( x 2 ) + d ( y 2 ) 2D dd = 2xdx + 2y dy Replacing D = x 2 + y 2 gives = dd = x dx + y dy D dd = x dx + y dy x2 + y 2 () Chapter 7 October 31, 2009 46 / 64
Example 13 (continued) Northbound travels 60 miles in 2 hours Eastbound thus travelled 46 2 = 92 miles Plug in x = 60, y = 92, dx dd = 30, dy = 46 : = 60 30 + 46 92 602 + 92 2 54.92 miles per hour () Chapter 7 October 31, 2009 47 / 64 Example 14 Two trains leave a station at noon. One travels north at 30 miles per hour. The other travels east at 80 miles per hour. At 1 PM the northbound train stops for one half hour at a station while the eastbound train continues without stopping. At 1 : 30 PM the northbound train continues north at 30 miles per hour. How fast are the trains traveling away from each other at 2 PM? x is distance travelled by Eastbound train y is distance travelled by Northbound train D = x 2 + y 2 distance between trains () Chapter 7 October 31, 2009 48 / 64
Example 14 (continued) As in previous problem Given: dx = 80, dy = 30 Unknown: dd at 2:00. At 2:00 dd = x dx + y dy x2 + y 2 x = 2 hours 80 mph = 160 miles y = 1.5 hours 30 mph = 45 miles (Northbound stopped for half hour) () Chapter 7 October 31, 2009 49 / 64 Example 14 (continued) Plug in all givens: dd = 160 80 + 45 30 1602 + 45 2 85.13 mph () Chapter 7 October 31, 2009 50 / 64
Example 15 A ladder 10 feet long rests against a vertical wall. The bottom of ladder slides away from wall at rate of 1 foot / sec. How fast is top of ladder sliding down the wall when bottom of ladder is 6 feet from wall? Let x denote horizontal distance between ladder and wall Let y denote vertical distance between ladder and ground Then 100 = x 2 + y 2 () Chapter 7 October 31, 2009 51 / 64 Example 15 (continued) Taking d on both sides and using chain rule: so Given: But So x = 6, 0 = 2 x dx + 2 y dy dy = x y and dx dx = 1 x 2 + y 2 = 100 = y = 8 dy = 6 8 1 = 3 4 feet per second () Chapter 7 October 31, 2009 52 / 64
Example 16 A cylindrical tank with circular base parallel to the ground is being filled at rate of 4 cubic feet per minute. The radius of the tank is 2 feet. How fast is level of water in tank rising when tank is half full? Your answer should be in feet per minute. Volume of cylinder with radius r and height h is V = π r 2 h Radius is constant, r = 2 feet, so volume depends only on height V = 4 πh so dv = 4 π dh () Chapter 7 October 31, 2009 53 / 64 Example 16 (continued) Given: so so dv = 4 cubic feet per minute dh = 1 π 4 = 4 π dh feet per minute Remark: We never used fact that tank was half full. This is because the rate of change of height depended only of rate of change of volume. () Chapter 7 October 31, 2009 54 / 64
Example 17 A conical salt spreader is spreading salt at rate of 3 cubic feet per minute. Diameter of cone is 4 feet and height is 5 feet. How fast is height of salt in spreader decreasing when height of salt in spreader (measure from the vertex up) is 3 feet? Give your answer in feet per minute. (It will be a positive number since we are measuring the rate at which it is decreasing.) Volume of cone with height h and radius r is V = π 3 r2 h Unlike the previous problem, here both h and r will be changing. () Chapter 7 October 31, 2009 55 / 64 Example 17 (continued) Salt is being spread at rate of 3 cubic feet per minute, so dv = 3 Initially, radius is 2 feet and the height is 5 feet. Let h be height and r be radius at a generic point in time. The ratio h/r is constant (Similar Triangles), and initially ratio is 5 2, so r = 2 5 h = 0.4 h () Chapter 7 October 31, 2009 56 / 64
Example 17 (continued) Now, So ( π ) ( ) 0.16 π V = (0.4h) 2 h = 3 3 dv = But h = 3 and dv ( ) 0.16 π = 3 so 3 3h 2 dh 3 = 0.16 3 2 π dh h 3 = 0.16 π h2 dh dh = 3 0.16 π 3 2 = 1 0.48 π 0.6631 feet per second () Chapter 7 October 31, 2009 57 / 64 Example 18 It is estimated that the annual advertising revenue received be a certain newspaper will be R(x) = 0.5 x 2 + 3 x + 160 where x is thousands of newspapers in circulation and R(x) is measured in thousands of dollars. The circulation is currently 10, 000 papers and is increasing at a rate of 2, 000 papers per year. At what rate will the annual advertising revenue be increasing with respect to time 2 years from now? Know dr dx = x + 3 () Chapter 7 October 31, 2009 58 / 64
Example 18 (continued) So dr = dr dx dx = (x + 3) dx Current circulation is 10, 000. Circulation increasing by 2000 papers per year so in two years circulation will be 10, 000 + 2 2, 000 = 14, 000 papers Thus, two years from now x = 14. Also, dx = 2. So dr = (x + 3) dx = (14 + 3) 2 = 34 Thus, revenue is increasing at rate of $34, 000 per year. () Chapter 7 October 31, 2009 59 / 64 Example 18 (continued) Remark: We could have done this without using related rates. Let t be the number of years from the present day. Now, x(t) = 10 + 2t and so R is a function of t: R(t) = 0.5 (10 + 2t) 2 + 3 (10 + 2t) + 160 = 240 + 26t + 2t 2 So so plugging in t = 2 gives dr = 26 + 4t dr = 26 + 4 2 = 34 () Chapter 7 October 31, 2009 60 / 64
Example 19 A stock is increasing in value at rate of $10 per share per year. An investor is buying shares of the stock at rate of 26 shares per year. How fast is the value of the investor s stock growing when the stock price is $50 per share and investor owns 100 shares? Let s be number of shares owned and let p be price per share. The total value, V, of investor s portfolio is V = p s and so rate of change of value of investor s portfolio is dv = dp s + p ds () Chapter 7 October 31, 2009 61 / 64 Example 19 (continued) Given ds = 26, dp = 10 s = 100, p = 50 so dv = 10 100 + 50 26 = 2300 dollars per year Value of the investor s portfolio is increasing at rate of $2300 per year. () Chapter 7 October 31, 2009 62 / 64
Example 20 Suppose the demand function q for a certain product is given by q = 4000e 0.01 p where p is the price of the product and q in the number of units sold at price p. If the product is currently selling for $100 per unit and quantity supplied is decreasing at rate of 80 units per week, find the rate at which the price of the product is changing. (Give this answer in dollars per week) Given: p = 100, and dq = 80 () Chapter 7 October 31, 2009 63 / 64 Example 20 (continued) Take d on both sides: dq = 4000 ( 0.01) dp e 0.01 p Plug in p = 100 and dq = 80 80 = 4000( 0.01) dp e 0.01 100 1 dp = 40e so dp = 2e 5.44 dollars per week The price is decreasing at a rate of $5.44 per week. () Chapter 7 October 31, 2009 64 / 64