Five Simple Steps to World Domination (or how to build the black box) Giving serious thought to your intended interpretation of …

Math 150 Topic7 Mystery Topic

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MATH 150 TOPIC 7 MYSTERY TOPIC

I How to Build a Function II Optimization III Word Problems Practice Problems

Math 150 Topic 7 Mystery Topic

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So why all the secrecy about Topic 7 maybe we thought if we said word problems you would avoid it entirely Now that youre here, stay awhile We cant promise to cure all of those word problem ills, but we can help This section is quite extensive actually, we went overboard However, if you go over each example carefully and work as many practice problems as you can stand, you should be well-prepared when these problems arise in class I How to build a Function

A In calculus, word problems require constructing a function in algebra, word problems usually involve equations Calculus I is restricted to the study of single variable functions So that is our goal: turning a word problem into a function of one variable Example: Express the volume of an open box formed when 1 squares are removed from the corners of a rectangular sheet whose length is 8 more than its width Solution: Let x and x 8 represent the width and length of the rectangular sheet
x8 x8 x6 x-2

-
1 1

-
x6

1 x-2

The figures
above show a rectangular sheet, with corners removed, folded into an open box Since v lwh, a complete list of formulas appears on pg 13 vx x 6x - 21 or vx x2 4x - 12

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This means that for any x 2, vx indicates its volume This restriction on the domain of V x is based on the box having x - 2 as a side Functions formed from word problems also called mathematical models often have domain restrictions based on quantities being measured See, that wasnt so bad maybe those nasty rumors about word problems arent all true B Some problems appear to need a second variable Dont worry, well just use algebra to get a function of one variable

Example: An open box with a square base has a volume of 48 in3 Express its surface area SA as a function of x x

y x

Solution: Lets start by writing an equation that measures surface area in terms of x and y SA x2 4xy [Sum of the areas of the 5 surfaces no top] Note, this is not yet a function of one variable However, since 48 V x2y 48, we know that y 2 By substitution, x 48 SAx x2 4x x2 192 x2 x The function to be found is called the objective function The fixed condition in this case V 48 is often
referred to as a constraint We dont want the surface area of just any open box with a square base The box must also have a volume of 48 in3 Note Anytime the objective contains 2 variables, a constraint must be determined and substitution must be used to express the objective function in terms of one variable

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C Some functions are built using the Pythagorean Theorem Example: A rectangle is inscribed in a semi-circle of radius 2 Express the area A of the rectangle as a function of x

x, y

Solution: We begin with A lw 2xy and proceed as before This time the constraint comes from the Pythagorean Theorem Since x2 y 2 22, y x, y
2 x y

4 - x2

Substitution then yields the objective function Ax 2x 4 - x2, 0 x 2

Example: A right circular cone with height h is inscribed in a sphere of radius 1 Express the volume of the cone as a function of h, ie find V h

1 r

Solution: First, suppose we use the figure above with h 1 The 1 volume of a cone is given by V r2 h To get a function in terms of 3 2 h, we must substitute for r Once again we use a right triangle to get a constraint: r2 h - 12 1 This leads to r2 2h - h2 After
substitution, 1 V h 2h2 - h3 The preceding 3 argument is based on a picture with h 1 You get the same answer if h 1 Try it and see

1 h
h-1

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D If you havent left for a chat room, lets try setting up a few more functions, this time using proportions If you prefer to try a few practice problems before going on we suggest 7175 Example:
A 10 4 E D B h x C

Given the figure with dimensions as shown, express h as a function of x

Solution: Since ACE and BCD are similar, a relationship of their corresponding sides may be expressed as a proportion Separating the two triangles may help:
A 10 B

and
E 4x C

h D x C

Thus, BD AE h 10 DC EC x 4x Cross multiplying and solving for h yields h Example:
4

10x 10x or hx 4x 4x

r h

A water tank in the form of a right circular cone has a height of 15 ft and a radius of 4 ft The tank is filled to a height of h ft with a radius of r ft Express the volume of the water as a function of r

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1 Solution: The problem starts with V r2 h and again requires sub3 stitution, this time for h From the similar triangles, 4 15 15 h r r h 4 1 So V r r2 3 15 r 4 5
r3 4

Now is probably a good time for a break Youve seen enough to try practice problems 71-79 Just dont forget about us Come back when youre ready for more II Optimization Optimization is first introduced in algebra, provided the function is quadratic Since the graph of any quadratic is a parabola, the vertex will indicate the least minimum or greatest maximum output of the function Finding such a value is referred to as optimization Lets start with a function in quadratic form, f x ax2 bxc The graph of f is a parabola opening up or down with vertex identified by h, k

h, k

a 0, k is max value a 0, k is min value Note: k f h To optimize a quadratic function; 1 Change f x ax2 bx c into f x ax - h2 k by completing the square; or

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b 2 Use h - and then evaluate f h 2a Note: h is the input but the output f h is the optimum value Example: Find the optimal value of f x -x2 6x 10 Solution: Since f is quadratic and concave down with a -1, the optimum value is a maximum Method 1: Completing the Square f x -x - 6x 10 -x2 - 6x 9 10 9 -x - 32 19
2

62 2 Because -9 9 0, f x has changed but is still equivalent to its
original form from

Thus f is concave with vertex at 3, 19 This means f has a maximum value of 19 when x 3 Method 2: h h- b 2a

-6 3; since f 3 -9 18 10 19, the results are the same -2

If you want to practice optimization before continuing, go to problem 710

Now that we can optimize, lets try an application In calculus, position functions are of special interest Changes in position relative to time lead to the discussion of velocity and derivatives For now, we will limit ourselves to the relationship between time and position Example: A ball is thrown straight upward from ground level The function st 64t - 16t2 gives the height of the ball in feet after t seconds a What is the maximum height reached by the ball? b When does the ball return to the ground?

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Solution: a By completing the square, st -16t - 22 64 Conclusion: after 2 sec, the ball reaches a maximum height of 64 ft st -16t2 - 4t -16t2 - 4t 4 64 -16t - 22 64

b On the ground, the height of the ball is 0 The problem requires finding the input t whose output is 0 st 0 64t - 16t2 16tt - 4 t 0 and t 4 This means the ball returns to the ground after 4 seconds Here
is a graph of st relating time and height st 64 48

2, 64

Note: the motion of the ball is straight up and down The parabola is the graph of the relationship between time and height It does not represent the path of ball t

Up to now weve only optimized quadratic functions All kinds of functions can be optimized as you will learn in calculus III Word Problems Now all the pieces should be in place Are you ready to try a real word problem?
Yes I guess If you insist Please dont make me

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Example: Find two numbers whose product is maximal if the sum of twice the first and second is 40 This type of optimization, where a precondition constraint must be maintained, is common in calculus Weve already done several examples like this Solution: Suppose x and y represent the two numbers Their product which is the objectiveis given by P xy The constraint 2x y 40 will be used for substitution to form the product function P x P x x40 - 2x 40x - 2×2 This time use x - y 40 - 210 20 b to get x 10 From the constraint, 2a

Do you realize what youve found? Of the infinite ordered pairs that satisfy the linear constraint, x 10 and y 20 yield
the largest product in this case 200 Example: Suppose you have 40 yards of fencing to enclose a rectangular garden one side of which is against a house What dimensions will yield a maximum area? A xy Ax x40 - 2x 40x - 2×2

Solution:

garden x y

using 2x y 40 y 40 - 2x

Since Ax is quadratic and concave down, the existence of a maximum is guaranteed Thus, dimensions of 10 by 20 will yield a maximum area To verify this result find dimensions and area when x 99 and x 101 Notice how the area peaks when x 10 Note: Have you noticed this example and the previous one are essentially the same? There is one difference The product function P x will accept any real x, but Ax is only feasible for 0 x 20

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Lets summarize the process used in the last several examples: a Identify the quantity to be optimized b Assign variables and write the quantity to be optimized in terms of the variables c Using constraints, rewrite the quantities to be optimized in terms of one variable; ie derive the objective function d Using algebra if the function is quadratic or calculus, find the optimum values

Example: In the figure below, AC lies along the shoreline
of a lake, and the distance from A to C is 10 miles P represents a small island 4 miles from A Leaving from P , a well-conditioned triathlete swims along the hypotenuse P B, and after reaching shore, runs to C
A x B 10 C

a Express the total distance traveled from P to C as the function of x b If the swimming rate is 3 mph and the running rate is 5 mph, express the total time it takes to get from P to C as a function of x

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10-x

Solution:
4

x2 16

a Dx

x2 16 10 - x distance 1 1 2 x 16 10 - x , T x rate 3 5

b Using time

Lets examine these functions more thoroughly 1 Both functions are defined over 0 x 10 2 Below is a table of total distance and total time for specific values of x input 0 1 2 4 6 10 14 131 125 117 112 108 output 33 32 309 308 32 36

0 x 10 Dx in miles T x in hours

3 Conclusions: Distance decreases as point B moves closer to point C Thats understandable since the shortest distance between 2 points is a straight line in this case that line is P C, the hypotenuse of AP C Initially, T x also decreases, but near x 4 we get a surprise: T x starts to increase and

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Mystery Topic

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continues to do so for the remainder of the interval That means that near x 4, T x has a minimal value Thus, if our athlete is interested in getting from P to C in the least amount of time, the task would be to minimize T x Since this function is not quadratic, the parabola method will not apply It is calculus that provides an easy method for minimizing this function Try this problem again after you study derivatives Now you can continue to practice problems on this topic or, if necessary, go over this entire section again

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Formulas from Geometry Rectangular Solid
h w

Volume lwh Surface area 2lw wh lh Surface area is the sum of the areas of all its surfaces

Cylinder
r

V r2 h SA 2r2 2rh

Cone
r

1 V r2 h 3

Trapezoid b1 h b2 Sphere 4 V r3 3 SA 4r2 1 A hb1 b2 2

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PRACTICE PROBLEMS for Topic 7 Word Problems 71 A square piece of tin with side x is to be used to form a box without a top by cutting off a 1 inch square from each corner and then folding up the sides Express the volume as a function of x
Return to Review Topic Answer

72 An 8 inch square piece of
tin is to be used to form a box without a top by cutting off squares of size x from each corner Express the volume as a function of x Note: Although problems 71 and 72 appear similar, they are different because of what x represents
Answer

73 A rectangle is inscribed in a semicircle of radius r r is a constant Express the area of the rectangle as a function of x

x, y

Return to Review Topic

Answer

74 ABCD is an isosceles trapezoid Express the area of the trapezoid as a function of its height D 3 A h 8 C 3 B
Answer

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75 A right circular cylinder has a volume of 20 cm3 Express its surface area as a function of h r h

Return to Review Topic

Answer

B 76 a Express r in terms of h b Express the SHADED area as a function of h

10 p r h

Answers

77 In this figure, S represents the length of a shadow cast by a six-foot person standing x feet from a light source The light is 20 feet above the ground Express S as a function of x

20 6 x
Return to Review Topic

s
Answer

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78 Starting at point A, a runner goes to point P and then to point D Express the runners total distance as a function of x D A 5
miles x P 15 miles
Answer

8 miles

79 A 10 foot long board is leaning against a wall so that its base is 6 feet from the wall When the base of the board is pulled y feet further from the wall, the top of the board drops x feet Express y as a function of x
x

y
Return to Review Topic

6
Answer

710 Find the maximum or minimum value of each function and state the x-value at which this occurs a f x -x2 10x - 4 b f x 2×2 10x - 4 c st 160t - 16t2
Return to Review Topic Answers

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711 A rancher has 500 feet of fence to enclose 2 adjacent rectangular grazing areas with the same dimensions What dimensions would yield maximum area?
Answer

712 The position function st 80t - 16t2 gives the distance in feet above the ground reached by an object in t seconds a How high is the object after one second? b In how many seconds will the object be at a height of 50 feet? c How long will it take the object to reach maximum height? d What is the maximum height? e How long does it take for the object to hit the ground?
Answers

The last 3 problems require trigonometry You may want to complete your trig review before attempting these 15

713

Leaning
against a wall, a 15 foot ladder reaches a height of x feet

a Express the distance from the ladder to the base of the wall in terms of x If is the angle formed by the ladder and the ground, b Express x as a function of c Express as a function of x
Answers

Math 150 Topic 7 Mystery Topic

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A x B 10-x C

714

Using the last example in this section, start at point C, run to B, and then turn and swim out to point P If represents the amount of turn, express the total distance as a function of

Answer

715 A rain gutter is to be constructed of aluminum sheets 10 inches wide by folding up a three inch length from each edge refer to the figure h 4 3 in Express the area of the gutter as a function of Refer to 74
Answer

Math 150 T7-Mystery Topic Review Answers

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ANSWERS to PRACTICE PROBLEMS Topic 7 Mystery Topic

71
x-2

1 x-2

vx x - 22 or vx x2 - 4x 4

Return to Problem

72
8-2x

x 8-2x

vx x8 - 2×2 or vx 64x - 32×2 4×3

Return to Problem

73 If x2 y 2 r2 , y
Return to Problem

r2 - x2 Since A xy, Ax x r2 - x2

8 74 3
9-h2

h 8

3
9-h2

b1 8, b2 8 2 9 - h2 1 A h8 8 2 9 - h2 2 Ah 8h h 9 - h2

Return to Problem

75 SAh is the
objective; v 20 is the constraint v r2 h 20 r2 20 , r h 20 h Thus SA 2r2 2rh 2 SA
Return to Problem

20 h

2h 20 h

20 h

40 2h h

Math 150 T7-Mystery Topic Review Answers

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76 Constraint relationship is a proportion 5 r 5 r h 7h 7 1 2r h 2 1 10 hh 27 5 Ah h2 7 A
Return to Problem

6 20 xs s 20s 6x 6s 3 s x 7
Return to Problem

A 78 5 miles
x2 25

D
15-x2 64

8 miles 15 - x

x P

15 miles T x
Return to Problem

x2 25

15 - x2 64, x 15

Math 150 T7-Mystery Topic Review Answers

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79 Initially 10 10 - 6y

8-x

6 so 6 y2 8 - x2 100 6 y2 100 - 8 - x2 y 6y
Return to Problem

100 - 8 - x2 - 6; x 8 100 - 8 - x2

710

by completing the square f x -x - 52 21

from x -

b 2a

x 5, f 5 21

a 0, concave down Max value of 21 at x 5 b 5 f x 2 x 2 5 5 33 x- ,f - - 2 2 2 5 33 at x - a 0, concave up Min value of 2 2 - t 5, s5 400
2

33 2

st 16t - s2 400

a 0, concave down Max value of 400 at t 5
Return to Problem

Math 150 T7-Mystery Topic Review Answers

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711

4x 3y 500 500 - 4x or y , 3 Substituting into

y x

A 2xy 1000x 8×2 yields Ax - 3 3 Since a 0, we get a max at x - b 125 250 or and y
2a 2 3 125 The problem didnt ask for max area even though A is easily found 2

Return to Problem

712

a s1 64 feet b Solve st 50 for t The quadratic in t has 2 solutions 10 5 2 t approximately 73 and 427 4 5 c st -16 t - 100 2 max ht occurs at t 25 seconds d s25 100 feet e Solve st 0, t 5 seconds
Return to Problem
2

Math 150 T7-Mystery Topic Review Answers

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713

a d

225 - x2 x x 15 sin 15 x ,x0 15

b sin

c arcsin
Return to Problem

B 16×2

10-x

714

16 x2 10 - x

To get D we need a trig relationship to allow substitution for x : x cot x 4 cot Therefore, 4 D 16 16 cot2 10 - 4 cot