College Algebra. Linear Functions and their Applications (Continued) Dr. Nguyen September 24, 2018

Transcription

1 College Algebra Linear Functions and their Applications (Continued) Dr. Nguyen Department of Mathematics UK September 24, 2018

2 Agenda Using linear functions in applications Rates and initial values Interest Problem

3 Ex1: Reading A student must read a novel. The student can read at 200 words per minute and has already read 8,000 words. If the student nishes in 3 hours, how long is the novel?

4 Ex1: Reading A student must read a novel. The student can read at 200 words per minute and has already read 8,000 words. If the student nishes in 3 hours, how long is the novel? We can model this with a linear function f (x) = mx + b: The slope m is the reading rate of 200 words per minute. This means the input x is minutes and the output y = f (x) is words. The y-intercept, or initial value, is 8,000, the number of words the student has already read. The linear function that models the situation is f (x) = 200x + 8,000.

5 Ex1: Reading A student must read a novel. The student can read at 200 words per minute and has already read 8,000 words. If the student nishes in 3 hours, how long is the novel? The linear function that models the situation is f (x) = 200x + 8,000. The problem wants the output of an input of 3 hours, but inputs must be in minutes: Hence the answer is 3 hours = 3 60 = 180 minutes. f (180) = ,000 = 36,000+8,000 = 44,000 words.

6 C1: Halfway Point The linear function that models the situation is f (x) = 200x + 8,000. The novel has 44,000 words. How long did it take the student to get halfway through the entire book (not just the part they still needed to read)?

7 C1: Halfway Point The linear function that models the situation is f (x) = 200x + 8,000. The novel has 44,000 words. How long did it take the student to get halfway through the entire book (not just the part they still needed to read)? The halfway point is at 44, 000 = 22,000 words, 2 so we need the input that gives an output of 22,000: f (x) = 22, x = 22, x = 14, 000 x = 70 minutes.

8 Ex2: Painting A painter can cover 20 square yards per gallon of paint. How much paint is needed to put three coats on a single wall that measures 7 yards by 4 yards?

9 Ex2: Painting A painter can cover 20 square yards per gallon of paint. How much paint is needed to put three coats on a single wall that measures 7 yards by 4 yards? This equation models the situation: Square yards = (Square yards per gallon) (Gallons of paint) We will let y be the area to be painted in square yards, and x be the gallons of paint needed. Since the painter covers 20 square yards per gallon, we have y = 20x. We need to know the area of the painted surface, and how to deal with the multiple coats of paint.

10 Ex2: Painting A painter can cover 20 square yards per gallon of paint. How much paint is needed to put three coats on a single wall that measures 7 yards by 4 yards? This equation models the situation: Square yards = (Square yards per gallon) (Gallons of paint) y = 20x The painter needs to cover y = (7 4 square yards) (3 coats) = 28 3 = 84 square yards. Hence they need x gallons where 84 = 20x 4.2 = x.

11 Basic Interest Suppose P dollars is invested at a yearly decimal rate of r for t years. Then the interest paid is i = P r t.

12 Ex3: Interest Suppose P dollars is invested at a yearly decimal rate of r for t years. Then the interest paid is i = P r t. A total of $1,000 is to be invested, part of it at 12% and the remainder at 9%. If the total interest is $102 after one year, how much was invested at 12%?

13 Ex3: Interest Suppose P dollars is invested at a yearly decimal rate of r for t years. Then the interest paid is i = P r t. A total of $1,000 is to be invested, part of it at 12% and the remainder at 9%. If the total interest is $102 after one year, how much was invested at 12%? Let x be the amount invested at 12%. Then the remaining 1000 x was invested at 9%. The interest from both investments is i = (x) (1 year) + (1000 x) 9 (1 year) = x x 102 = x + 90

14 Ex3: Interest The interest from both investments is 102 = x = x Multiply both sides by 100, the reciprocal of the coecient of 3 x: = x = x 400 = x

15 Ex3: Interest Hence $400 was invested at 12%. For reference, the remaining was invested at 9% = 600 dollars

16 Next Time Please read Section 4.3 of your textbook. We will look at tting linear functions to data.