Functions as Real World Models

Functions as Real World Models


Many of the processes studied in the physical and social sciences involves understanding how one quantity is related to another quantity. Finding the function that describes the dependence of one quantity to another is called modeling. Modeling real world problems especially those that require optimization is one of the important applications of the study of functions.


Example 2.4.1.

Emma and Brandt drive away from a campground at right angles to each other. Emma’s speed is 65 kph and Brandon’s is 55 kph.

a) Express the distance between the cars as a function of time.b) Find the domain of the function.

camp ground


Emma’s direction

Illustration:

Brandon’s direction

Distance between Emma and Brandt at time t

timeratedistance


Solution: a. Suppose 1 hr has gone by. At that time, Emma has traveled 65 km and Brandt has traveled 55 km. We can use the Pythagorean theorem to find the distance between them. This distance would be the length of the hypotenuse of a triangle with legs measuring 65 km and 55 km.


Solution:(continuation)

After 2 hours, the triangle’s legs would measure 130 km and 110 km. Observe that the distances will always be changing. We make a drawing and let t be the time in hours that Emma and Brandt have been driving since leaving the campground.


Illustration:

After t hours, Emma has traveled 65t km and Brandt 55t km.



Using the Pythagorean theorem:

Because distance must be nonnegative, we need consider only the positive square root when solving for d(t):

Thus, d(t) = 85.15t, t ≥ 0.

,tttd 222 5565

22 5565 tttd


Solution: b. Since the time traveled, t, must be nonnegative, the domain is the set of nonnegative real numbers .),0[


Example 2.4.2

A rectangular field is to be fenced along the bank of a river, and no fence is required along the river. The material for the fence costs PhP8 per running foot for the two ends and PhP12 for running foot for the side parallel to the river; PhP3600 worth of fence is to be used.

a. Let x be the length of an end; express the number of square feet in the area of the field as a function of xb. What is the domain of the resulting function?


Illustration:

x

Given:

The material for the fence costs PhP8 per running foot for the two ends and PhP12 for running foot for the side parallel to the river; PhP3600 worth of fence is to be used.


Solution

a. Let y be the length of the side of the field parallel to the river and A square feet be the area of the field.

Then because the cost of the material for each end is PhP8 per running foot and the length of an end is x feet , the total cost of the fence for each end is 8x pesos. Similarly, the total cost of the fence for the third side is 12y pesos. We then have

8x + 8x + 12y =3600. (1)



To express A in terms of a single variable, we first solve equation (1) for y in terms of x.

We substitute this value of y in the equation , yielding as a function of x, and

xy

xy

3

4300

16360012

)x(x)x(A3

4300


Solution:

b. Both x and y must be nonnegative. The smallest value that x can assume is 0. The smallest value that y can assume is 0, and when y=0, we obtain from equation (1) x=225. Thus is 225 is the largest. And [0,225] is the domain of A as a function of x.


Example 2.4.3 A buko pie store can produce buko pie at a cost of PhP95 per piece. It is estimated that if the selling price of the buko pie is x pesos, then the number of buko pie that are sold each day is 1000-x.

a. Express the daily profit of the store as a function of x.b. Use the result in a) to determine the daily profit , given that the selling price is PhP160.


Solution:

a. The profit P(x) can be obtained by subtracting the total cost C(x) from the total revenue, R(x).

The total revenue is the product of the selling price and the number of buko pie sold in a day. So R(x)= x(100-x) . On the other hand, the total cost is the product of cost per buko pie and the number of buko pie sold in a day. Equivalently, C(x) = 95(1000-x).



Hence, P(x) = R(x) – C(x) P(x) = x(1000 - x) – 95(1000 - x)

P(x) = (1000 - x)(x - 95)


Solution:

b. P(160) = (1000 -160)(160 - 95)P(160) = (840)(65) = PhP54600

Before we proceed with more examples let us define the following relationships.


Definition 1. Directly Proportional

A variable y is said to be directly proportional to a variable x if

y = kx, where k is a nonzero constant. More generally, a variable y is said to be directly proportional to the nth power of x if y = kxn

where k is a nonzero constant.


Example 2.4.4.

The approximate weight of a person’s muscles is directly proportional to his or her body weight.

a. Express the number of kilograms in the approximate muscle weight of a person as a function of the person’s body weight, given that a person weighs 68 kg has muscles weighing approximately 27 kg.b. Find the approximate muscle weight of a person weighing 50 kg.


Solution:

a. Let x kg be the approximate muscle weight of a person having a body weight of kg.

Then f(x) = kx. Because a person of body weight 68 kg has muscles weighing approximately 27 kg, x = 68 and f(x) = 27. Then 27 = k(68)

Thus, 68

27k

xxf68

27)(


Solution:

b. Since the approximate muscle

weight of a person weighing 50 kg is

xxf68

27)(

kg851950f

5068

2750f

.)(

)()(


Definition 2. Inversely Proportional

A variable y is said to be inversely proportional to a variable x if

where k is a nonzero constant. More generally, a variable y is said to be inversely proportional to the nth power of x if


x

ky

nx

ky


Example 2.4.5

For an electric cable of fixed length, the resistance is inversely proportional to the square of the diameter of the cable.a. Given that a cable having the fixed length is ½ cm in diameter and has a resistance of 1 ohm, express the number of ohms in the resistance as a function of the number of centimeters in the diameter.b. What is the resistance of a cable having the fixed length and a diameter of 2/3 cm?


Solution:

a. Let f(x)ohms be the resistance of a cable having the fixed length with x cm in diameter. Then . If a cable having the fixed length is ½ cm in diameter and has a resistance of 1 ohm,

then we obtain . Then k = ¼.

Thus,

2)(

x

kxf

221

1k

24

1)(

xxf


Solution :

b. If then .

Therefore, the resistance of a cable having a fixed length and a diameter of 2/3 cm is 9/16 ohm.

3

2x

16

932f )/(


Definition 3. Jointly Proportional

A variable z is said to be jointly proportional to variable x and y if z = kxy where k is a nonzero constant.

More generally, a variable z is said to be jointly proportional to the nth power of x and the mth power of y if


mn ykxz

TIME TO THINK

1. A right circular cylinder of height h and radius r is inscribed in a right circular cone with a height of 10 ft and a base with radius 6 ft.

a. Express the height h of the cylinder as a function of r. b. Express the volume V of the cylinder as a function of r. c. Express the volume V of the cylinder as a function of h.

TIME TO THINK

2. An open-top box with a square base is to be constructed from two materials, one for the bottom and one for the sides. The volume of a box is to be 9 cubic feet. The cost of the material for the bottom is Php4 per square foot, and the cost of the material for the sides is Php3 per square foot.

a. Determine a model for the cost of the box as a function of its height h. What is the domain of the function?b. Which will be the most expensive to construct, a box with a height of 1 foot, 2 feet, or 3 feet?

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Functions as Real World Models