206 Calculus and Structures

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1 06 Calculus and Structures

2 CHAPTER 4 CURVE SKETCHING AND MAX-MIN II Calculus and Structures 07 Copright

3 Chapter 4 CURVE SKETCHING AND MAX-MIN II 4. INTRODUCTION In Chapter, we developed a procedure for graphing polnomial functions. In this Chapter we take this one step further and epand the procedure to include functions of an kind. In Chapter we found the maimum and minimum for word problems that involved onl polnomials. The same ideas can be applied to functions that are more comple than polnomials. In order to etend these results we will make use of our new abilit to differentiate functions using the product, quotient, and chain rules. 4. CURVE SKETCHING Eample : Sketch the graph of, 4 a) Find a simple point: = 0, = 4, and place this point on Fig. b) Find the behavior of the function for large and small values of : For large and small values of, 4 4 f ( ) lim 0, as or using the language of limits, f ( ) 0. Place marks on the graph in Fig. to illustrate this behavior. We sa that f() has a horizontal asmpote at = 0 as + ( 0,4 ) X c) Find critical points. Rewrite Eq. as, Fig. 08 Calculus and Structures

4 Section 4. 4( ) () Use the chain rule to differentiate Eq. and set the result equal to 0. d d ( ) 0 or = 0. Therefore, = 0, = 4 is the onl critical point. Place a short horizontal line through this point in Fig.. Place an mark at = 0 on the -ais. This divides the - ais as follows: d) Determine the regions over which the function increases and decreases: Let =, d d 0. Let = -, 0. The curve increases until it hits the horizontal line at (0,4) d d and then decreases. It is also asmptotic to = 0 for large and small values of. e) Putting together this information, the resulting curve is shown in Fig.. M ( 0,4 ) Fig. 4 Check: The function, f ( ) is even. Notice that the graph of t his function is smmetric with respect to the -ais. Remark: We will show in Eample 6 of Section 6. that the area under this curve over the interval [0,] equals. Eample Sketch the graph of, (3) Calculus and Structures 09

5 Chapter 4 CURVE SKETCHING AND MAX-MIN II a) Find a eas point: = 0, = 0 and place this point on Fig. 3 b) Find the behavior of t he function for large and small values of. For large and small values of onl the leading terms in the numerator and denominator matter so that Eq. 3 can be rewritten as Eq. 4, lim ( f ) (4) Therefore, f ( ). The line = is called a horizontal asmptote. To find whether the curve approaches = above or below it, place a large value of into Eq. 4, i.e., when = 5, = 5/4 so the curve approaches the horizontal asmptote from above it. Places marks above = for large values of as shown in Fig. X - ( 0,0 ) Fig. 3 b) This graph brings a new element into the graphing procedure. Rewriting Eq. 3, ( )( ) ou will notice that the denominator equals 0 when =. Therefore the graph has what is known as vertical asmptotes at. These vertical asmptotes are vertical lines as shown in Fig.. To see how the graph of the function behaves as approaches these vertical lines first let approach from above, i.e., let = 3/ in Eq. 3. We see that = 9/5. Since this result is positive approaches, i.e., as +,. Also as - let = / in Eq. 3. We see that = -/3 a negative number so that. Likewise, we find that as,. This is illustrated in Fig. 3 b marks placed along the asmptotes. c) Find the critical points: d d ( ) 0 (5) 0 Calculus and Structures

6 Section 4. = 0, = 0 Place the critical point on the -ais in Fig. 3. The critical point divides the -ais as follows: d) Find where the curve increases and decreases: From Eq. 4, d d for < 0, 0, and for > 0, 0. Mark the -ais appropriatel with d d + s and s in Fig. 3. Therefore, the graph increases from the mark at =, = 0 until it connects with the = - - asmptote, then it rises from the = - + asmptote to the critical point. Then it falls to the = asmptote and falls again from the = + asmptote until it connects with the mark as =, = 0. Putting this all together we get Fig X Check: Notice that the graph of the even function, respect to the -ais. f Fig. 4 ), is smmetric with ( Eample 3 Sketch the graph of, e for 0 (5) An eas point is = 0 when = 0. Place this point in Fig. 5. Find the behavior of the function for large values of : Referring back to Eample of Chapter 9, using L Hopital s Rule, e 0. Place a mark above the -ais for large in Fig. 5. lim Calculus and Structures

7 Chapter 4 CURVE SKETCHING AND MAX-MIN II (, /e) Find the critical Using the product rule, X Fig. 5 d d e e e ( ) (6) d 0 when = and = /e d Place a short horizontal line at (, /e) to mark the critical point. And place a mark on the -ais in Fig. 5 at =. The -ais divides into the segments: d d From Eq. 6, 0 for > and 0 for 0 < <. Increasing and decreasing d d intervals of the function are indicated b + and signs. Locate the inflection points: d d e ( e e ) 0 d e ( - ) 0 d d Therefore = and = /e is the location of an inflection point. Since 0 to the left of d d = and 0 to the right of =, the curve is concave down up to = and then d concave up thereafter. From all of this evidence, we are able to sketch the graph of Eq. 5 in Fig. 6. Calculus and Structures

8 Section 4. (, 3/e) Problem : Fig. 6 Sketch the graph of the following functions. Label critical points and determine if the are relative maima or relative minima. Find all vertical and horizontal asmptotes. a. b. c. f ( ) ( ) f ( ) f ( ) / ABSOLUTE MAXIMA AND MINIMA If we wish to find the absolute maimum and minimum values of a function, f(), over and closed interval [a,b] (including its endpoints) or an open interval (a,b) (not including its endpoints), then there is no need to draw a graph. We use the procedure introduced in Chapter. This is illustrated b the following eamples. Find the absolute maimum and minimum values of f() over the indicated intervals Eample 4 4 f ( ), [-, ] Find critical points: According to Eample there is one critical point at (0, 4) Evaluate the function at the endpoints: Make a table: = when = - and = 4/5 when = - 0 f() 4 4/5 Therefore the absolute maimum is at = 0 and the absolute minimum is at =. Calculus and Structures 3

9 Chapter 4 CURVE SKETCHING AND MAX-MIN II Eample 5 f ( ) e, [0, ) Find critical points: According to Eample 3, there is one critical point at (, /e) Check endpoints: = 0 when = 0 and Make a table lim e 0. Using L hospital s rule 0 f() 0 /e 0 The absolute minimum is at = 0 and. The absolute maimum is at =. Eample 6 / 3 f ( ) ( ), [0, 9] Find critical points: / 3 / 3 f '( ) ( ) / 3( ) ( ) / 3 5 / 3 = 0 / 3 / 3 ( ) ( ) f () = 0 when = 3/5 Note that the derivative is infinite when = however, the curve is continuous at that point, so the curve has a cusp at =. Check endpoints: f(0) = 0 and f(9) = 36 Make a table: 0 3/5 9 3 f() 0 + (3/5) ( / 5) / 0 36 The minimum value is at = 0 and the ma is at = 9 Problems Find the absolute maimum and minimum values of the following functions over the given intervals. 4 Calculus and Structures

10 Section 4.3 t a) g(t) = te, t 0 b) f ( ) ln, > 0 c) t f ( t), for all t t d) f ( ) e sin, [0, ] 4.4 WORD PROBLEMS Eample 7 Find a point on the parabola = that is closest to the point (,4). What is the closest distance? Entr phase: Sketch graph and label point as shown if Fig. 7 (, 4) (, ) = Let D be the distance from (,4) to (,) on the curve. Fig. 7 Find (,) such that D is a minimum. Compute D min. Attack phase: D ( ) ( 4) (7) where = or = ½ (8) Replacing into Eq. 7, D ( ) ( 4), is in the interval (-, ) Calculus and Structures 5

11 Chapter 4 CURVE SKETCHING AND MAX-MIN II D and D both have minima at the same value of so we shall compute the minimum of D since the math is easier.. We will call z = D. z D ( ) ( Y 4) Find the critical point of D:: Using the chain rule, dz d ( ) ( 4) The critical point is : = As z So there is no maimum. The minimum occurs at the critical point. Therefore, from Eq. 8 when =, = and D = 5 Review phase: The answer appears reasonable b looking at the sketch in Fig. 7. Eample 8: A wire 00 inches long is cut into two pieces and each piece is bent into a square. Where should the wire be cut so that the sum of the areas of the two squares is a maimum and a minimum? Entr Phase: The information of the problem is placed in the diagram in Fig Let be the position of the cut 4 ( 00 ) 4 Fig.8 Find such that A = A + A is a min. Attack Phase: (00 ) A A A, Calculus and Structures

12 Section 4.4 da (00 ) Find the critical point: 0 d 6 6 After some algebra, Check the endpoints: = 50, A = 5. When = 0 and = 00, A = (5) Make a table: A (5) (5) (5) The Absolute Ma occurs at = 0 and = 00. The absolute min occurs at = 50. Review: The answer makes sense because b smmetr both squares should have the same area. Eample 9a: A bird is released from point A, an island which is 5 km, from the nearest point B. The bird flies along a straight shoreline to point C and then continues fling along the shoreline to its nesting area D. If it uses.4 times as much energ to fl over water than it does over land, what path will minimize t he ependiture of energ. Points B and D are km apart Entr Phase: Fig. 9 illustrates the given information A 5km B C D 0 Fig.9 Calculus and Structures 7

13 Chapter 4 CURVE SKETCHING AND MAX-MIN II Find: such that Energ E is a minimum. Attack Phase E (.4) 5 (.0)( ), 0 Find the critical point: Using t he chain rule, de d = 5 After some algebra, Check the endpoints: = 5 or = When = 0, E = (.4)(5) + = 9, when =, E = (.4)(3) = 8. Make a table: E The least energ path occurs for = 5.03 with and energ of 6.9 units Problem 3 In Eample 9a, Let W be the energ per km over water L be the energ per km over land i) If the bird lands at = 4, what is the ratio W/L? ii) If the ratio W/L 0, i.e., W<<L, where does the bird land? iii) If the ratio W/L is large ( ), i.e., L<<W, where does the bird land? For this problem ou can use the energ formula, E W 5 L( ) 8 Calculus and Structures

14 Section 4.4 where W and L are treated like constants. Problem 4: Fig. 0 shows a one-mile-square cit park. A local power compan needs to run a power line from the northwest corner A of the park to the southeast corner B. To preserve the beaut of the park, onl underground lines ma be run through the park, onl overhead lines are permissible along the boundar of the park. The power compan plans to construct an overhead line a distance along the west edge of the park, then from the southern end of this line continue with a straight power lie to point B. If overhead lines cost $40 thousand per mile and underground lines cost $00 thousand per mile, how should the power compan construct the line to minimize cost? $40/ mi $00/ mi Problem 5 A 00 foot length of wire is cut at point and the left part of the wire is bent into a square while the right length of wire is bent into a circle. Where should the cut be made so that the sum of the areas of square and circle is a minimum? A maimum? Problem 6 A Norman window has the shape of a semicircle placed on top of a rectangle. If the perimeter, P, of the window is 30 ft.., find the dimensions and such that the window lets in the most light, i.e., the area A = Area of rectangle + area of semicircle is a maimum. See Fig.. Fig. Calculus and Structures 9

15 Chapter 4 CURVE SKETCHING AND MAX-MIN II Problem 7 During a flu outbreak in a school of 763 children, the number of infected children, I, was epressed in terms of the number of susceptible (but still health) children, S, b the epression S I = 9 ln ( ) S What is the maimum number of infected children? Problem 8: An apple tree produces, on average, 400 kg, of fruit each season. However, if more than 00 trees are planted per km, crowding reduces the ield b kg. for each tree over 00. a) Epress the total ield from one square kilometer as a function of the number of trees on it. Graph this function. b) How man trees should a farmer plant on each square kilometer to maimize ield? Problem 9 For a parabola, =, find (,) on the parabola such that the square of the distance from (0,) to (,) is a minimum. See Fig.. (0,) (, ) = Fig. 0 Calculus and Structures

16 Section 4.4 Problem 0 For the function = /, find the point, ) on this curve such that the length of the ( 0 0 tangent line AB to the curve at, ) is a minimum. See Fig. 3 ( 0 0 = / A (, ) 0 0 B Fig. 3 Problem For the rectangle with area 64 square units shown in Fig 4, line AB is drawn from verte A to the midpoint of the opposite side. Find the values of and such that the length of AB is a minimum. A A=64 B Fig. 4 Calculus and Structures

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