Functions and Graphs

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Functions and Graphs

• Chapter 2

2
2.7

• Mathematical Models Constructing Functions

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Example 1

• The price p, in dollars, and the quantity x sold
  of a certain product obey the demand equation
• P -(1/3)x 100 0x300
• Express the revenue R as a function of x.
• Pricequantity Revenue
• (-(1/3)x 100)x R(x)
• 2. What is the revenue if 100 units are sold?
• X is units sold, so plug in 100 for x.
• (-(1/3)(100) 100)100 6666.67
  
  
  
  

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Example 1

• Graph the revenue function on a calculator.
• Standard zoom doesnt work, zoom fit doesnt
  quite work either, so hit the window button and
  change xmin and xmax to equal the domain. The
  lowest y should be 0 and since one output was
  6666.67 lets try 10000 for ymax.

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Example 1

• What quantity x maximizes revenue? What is the
  maximum revenue? Use the max function on the
  calculator.
• 150 products maximize revenue. The maximum
  revenue is 7500.

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Example 1

• What price should the company charge to maximize
  revenue?
• Revenue/sold price so
• Maxrevenue/soldformax priceformax
• 7500/150 50.00

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Example 2

• The price p, in dollars, and the quantity x sold
  of a certain product obey the demand equation
• x -20p 500 0p25
• Express the revenue R as a function of x.
• Pricequantity Revenue
• -20p x-500
• P (-1/20)x 25
• (-(1/20)x 25)x R(x)
• 2. What is the revenue if 20 units are sold?
• X is units sold, so plug in 20 for x.
• (-(1/20)(20) 25)20 480
  
  
  
  

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Example 2

• Graph the revenue function on a calculator.
• If p 0, then x500, so x ranges from 0 to 500.
  Lets try 10000 again for y.

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Example 2

• What quantity x maximizes revenue? What is the
  maximum revenue?
• 250 products maximize
  revenue. Max revenue is 3125.

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Example 2

• What price should the company charge to maximize
  revenue?
• Revenue/sold price
• 3125/250 12.50

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Example 3

• Beth has 3000 feet of fencing available to
  enclose a rectangular field. One side of the
  field lies along a river, so only three sides
  require fencing.
• Express the area A of the rectangle as a function
  of x, where x is the length of the side parallel
  to the river. See next slide for picture.
• X((3000-x)/2) A(x)

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Example 3
3000


X
3000 - x

(3000-x)/2
X
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Example 3

• Beth has 3000 feet of fencing available to
  enclose a rectangular field. One side of the
  field lies along a river, so only three sides
  require fencing.
• 2. Graph A(x) on a calculator. For what value
  of x is the area largest?

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Example 4

• Let P (x,y) be a point on the graph of y x2
  8.
• Express the distance d from P to the point (0,-1)
  as a function of x.
• P (x,x2 8)
• Use distance formula
• Sqrt(((x2-8) - -1)2 (x-0)2)
• D Sqrt((x2-7)2 (x)2)

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Example 4

• What is d if x0?
• D Sqrt((02-7)2 (0)2) 7
• What is d if x -1?
• D Sqrt(((-1)2-7)2 (-1)2) sqrt(37)

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Example 4

• 4. Graph d(x) on the calculator.
• For what values of x is d smallest?
• 2.54 and -2.54

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Example 5

• A wire of length x is bent into the shape of a
  square.
• Express the perimeter of the square as a function
  of x.
• Express the area of the square as a function of
  x.
• Length of a side of the square x/4.
• The perimeter would be 4(x/4) x.
• The area would be (x/4)2 x2/16

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Example 6

• A wire 10 meters long is to be cut into two
  pieces. One piece will be shaped as an
  equilateral triangle, and the other piece will be
  shaped as a circle.
• 1. Express the total area A enclosed by the
  pieces of wire as a function of the length x of a
  side of the equilateral triangle.

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Example 6
X
3x
10-3x
10
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Example 6

• Circumference of the circle is 10-3x, so
• 2pr 10-3x
• R (10-3x)/(2p)
• Area of the circle pr2 p((10-3x)/(2 p)
  (10-3x)/2
• Area of triangle .5bh .5x(.5xsqrt(3))
  .25sqrt(3)x2
• A(x) (10-3x)/2 .25sqrt(3)x2

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Example 6

• What value of x gives you the smallest value for
  A?
• A(x) (10-3x)/2 .25sqrt(3)x2
• Graph the equation and use the minimum function
  to find this value.
• x 1.732meters as
• a length of a side of
• the equilateral
  triangle will
• yield the smallest
  area.

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Example 7

• Two cars leave an intersection at the same time.
  One is headed south at a constant speed of 30mph,
  and the other is headed west at a constant speed
  of 40mph. Express the distance between the cars
  as a function of time. At t0, the cars leave the
  intersection.
• Ratetime distance
• dw distance traveled by the car headed west.
• dS distance traveled by the car headed south.
• Note the triangle is right.

dW
dS
D
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Example 7

• ds 30t
• dw 40t
• The distance between the cars would be the
  hypotenuse of the right triangle, so
• Sqrt((30t)2 (40t)2) D