Some Algebra Word Problems – Part II

(edited 5/31/2017)

Some Algebra Word Problems – Part II

Source:  Blitzer, Robert  Introductory & Intermediate Algebra for College Students 3^rd Edition Pearson, Prentice Hall:  Upper Saddle River, New Jersey 2009  ISBN-13: 978-0-13-602895-6

Problems and diagrams are rewritten.  No claims of profit are made.

Here are five more problems.  For the previous installment, click this link:  http://edspi31415.blogspot.com/2017/05/some-algebra-word-problems.html

What size is the Garden?

Problem:  

The length of a rectangular garden is 5 feet greater than its width.  The garden’s area is 300 square feet.  Find the dimensions of the garden.  [pg. 452, number 82 – Blitzer]

Discussion:

The area of a rectangle is:

Area = Length * Width

In this case, Width = x, Length = x + 5, and Area = 300, with width and length in feet, area in square feet.

Then:

(x + 5) * x = 300

x^2 + 5 * x = 300

What we have is a quadratic equation.  Subtract 300 from both sides yields

x^2 + 5*x – 300 = 0

We can use the quadratic equation or try to factor.  I’m going to factor:

(x – 15) * (x + 20) = 0

Which leads to x = 15 and x = -20.   In this case, negative lengths do not make sense, hence the solution we are looking for x = 15.

To conclude, the Width is 15 feet, while the Length is 20 feet (15 + 5). 

Trees and Triangles

Problem: 

A person who is 5 feet tall is standing 80 feet from a tree.  At the time, the tree casts an 86-foot shadow, while the person casts a 6-foot shadow.  What is the tree’s height?  [pg. 527, number 32 – Blitzer]

Discussion:

We are working with similar triangles.  Please see the diagram.  The ratio of height/length is:

x/86 = 5/6

x = 86 * 5/6

x = 71.666666

The height of the tree is 71.666666 feet. 

* I previously had the incorrect answer of 70.83333.  I thank Gianfranco Cazzaro for alerting me on my mistake.  The answer is now correct.  - EWS 5/31/2017

From Bangkok to Phnom Penh

Problem:

On a map, Bangkok are given coordinates (-115, 170) and Phnom Penh has the coordinates (65, 70).  Units are in miles.  How long will it take a plane flying 400 miles an hour (mph) on a direct flight between Bangkok and Phnom Penh? Round off the answer to the nearest tenth of an hour.  Assume the flight takes place in a straight line.  [pg. 750, number 121 – Blitzer]

Discussion:

We can use the following two formulas to help us solve the problem:

Distance between points (x1, y1) and (x2, y2):

Distance = √( (x1 – x2)^2 + (y1 – y2)^2 )

Relationship between Distance, Speed, and Time:

Time = Distance / Speed

Let Point 1 (x1, y1) represent Bangkok (-115, 170) and Point 2 represents Phnom Penh (65, 70).  Hence, Distance =√( (-115 – 65)^2 + (170 – 70)^2 ) , and Speed = 400. 

Then:

Time = √( (-115 – 65)^2 + (170 – 70)^2 ) / 400

Time = √( 180^2 + 100^2 ) / 400

Time ≈ 0.5

The plane will take about a half an hour to travel from Bangkok to Phnom Penh.

Making It Rain

Problem:  

A rain gutter (as shown above) is made from aluminum sheet 20 inches wide.  Edges that are x inches wide are turned up on each side to form the gutter.  Find x to allow a cross sectional area of 13 square inches.  Round off the answer to the nearest tenth of an inch.  [pg. 764, number 81 – Blitzer]

Discussion:

If you fold the edges up, you get a rectangle with sides x in. and 20 – 2*x in.  We want to find x such that the area is 13 in^2.  Then:

x * (20 – 2 * x) = 13

20 * x – 2 *x^2 = 13

Bringing all terms to one side:

0 = 2*x^2 – 20 *x + 13

We have a quadratic equation.  This time, we’ll use the quadratic formula:

x = ( 20 ± √(20^2 – 4 * 2 * 13) ) / (2 * 2)

With a calculator the results are (results rounded to tenths):

x ≈ 9.3 in and x ≈ 0.7 in.

There is no specification which measure is preferable, so both answers are acceptable. 

Plot Enclosure

Problem:  

You have 600 feet of fencing to enclose a rectangular plot (see figure).  The side facing the river will not be enclosed.  What is the largest area that can be enclosed?  [pg. 782, number 65 – Blitzer]

The area of the enclosure is:

A = x * (600 – 2*x)

A = 600 * x – 2 * x^2  (I)

The equation is in the form of a quadratic equation y = a*x^2 + b*x + c.  In general, if a < 0, the quadratic equation will have a maximum at x = -a / (2*b).

From (I), a = -2, b = 600, and c = 0.  Since a < 0, we will have a maximum at x = -600 / (2 * -2) = 150.

The sides will have 150 feet by 300 feet.   The maximum area is 150 ft * 300 ft = 45,000 ft^2.

Eddie

This blog is property of Edward Shore, 2017.

Fun with the Fractional Part function and Integers

Fun with the Fractional Part function and Integers

Let n be a integer and frac(n) be the fractional part function.  For example, frac(28.38) = 0.38.

Alternating Ones and Zeros – Ones Assigned to Odd Numbers

f(n) = 2 * frac(n/2) = n mod 2

n	1	2	3	4	5	6	7	8	9	10
f(n)	1	0	1	0	1	0	1	0	1	0

Alternating Ones and Zeros – Ones Assigned to Even Numbers

f(n) = 2 * frac((n + 1)/2)

n	1	2	3	4	5	6	7	8	9	10
f(n)	0	1	0	1	0	1	0	1	0	1

Alternating Integers – Integers Assigned to Odd Numbers, Zeros to Even

f(n) = (2*n) * frac(n/2)

n	1	2	3	4	5	6	7	8	9	10
f(n)	1	0	3	0	5	0	7	0	9	0

Alternating Integers – Integers Assigned to Even Numbers, Zeroes to Odd

f(n) = (2*n) * frac((n+1)/2)

n	1	2	3	4	5	6	7	8	9	10
f(n)	0	2	0	4	0	6	0	8	0	10

Alternate +1 and -1.  Use of superposition of functions

f(n) = (2) * frac((n+1)/2) + (-2) * frac(n/2)

n	1	2	3	4	5	6	7	8	9	10
f(n)	-1	1	-1	1	-1	1	-1	1	-1	1

The Number of Petals of a Rose Function

Odd + Even = (2*n)*frac(n/2) + (4*n)*frac((n+1)/2)

This determines the number of pedals a rose gets.  The rose is represented by the polar equation r = cos(n*θ).  If n is odd, the rose has n petals.  If n is even, the rose has 2*n petals.

n	1	2	3	4	5	6	7	8	9	10
f(n)	1	4	3	8	5	12	7	16	9	20

Other Examples

f(n) = 3 * frac(n/3) = n mod 3

n	1	2	3	4	5	6	7	8	9	10
f(n)	1	2	0	1	2	0	1	2	0	1

f(n) = 3 * n * frac(n/3)

n	1	2	3	4	5	6	7	8	9	10
f(n)	1	2	0	4	10	0	7	14	0	10

f(n) = 3 * n * frac((n+1)/3)

n	1	2	3	4	5	6	7	8	9	10
f(n)	2	0	3	8	0	6	14	0	9	20

f(n) = 4 * frac(n/4) = n mod 4

n	1	2	3	4	5	6	7	8	9	10
f(n)	1	2	3	0	1	2	3	0	1	2

f(n) = 4 * frac(2*n/4) = 4 * frac(n/2)

n	1	2	3	4	5	6	7	8	9	10
f(n)	2	0	6	0	10	0	14	0	18	0

f(n) = 4 * frac(3*n/4) (sort of “reversing” n mod 4)

n	1	2	3	4	5	6	7	8	9	10
f(n)	3	2	1	0	3	2	1	0	3	2

Generalizing:

Let m and n be positive integers.  Hence, m * frac(n/m)  = n mod m

To reverse the sequence, use m * frac((m-1)*n/n)

Don’t forget to play sometimes.  Math can be really fun when you let go. 

Eddie

This blog is property of Edward Shore, 2017.