Idaho Society of Professional Engineers

           PO Box 170239, Boise, ID 83717-0239  208-426-0636  Fax: 208-426-0639  E-Mail: ispe@idahospe.org

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Idaho Society of Professional Engineers
Friday Update – 11/19/10

What’s Inside:
• Upcoming Events

• Call for ISPE Award Nominations
• MATHCOUNTS Problem of the Week
 

WELCOME NEW ISPE MEMBERS!!

Robert Pharmer – Southwest Chapter

Teri Tyler – Southeast Chapter

In the News:
•  Why Are So Many Terrorists Engineers?
• Cloaked - Why is engineering still a mystery to students, and what does that mean for the next generation?

UPCOMING EVENTS:

IDABO EDUCATION INSTITUTE – January 24 – 28, 2011
Nampa Civic Center, 311 3rd Street, Nampa, Idaho
Idaho Association of Building Officials Presents….
2009 Code Training and More!!
 

CALL FOR ISPE AWARD NOMINATIONS
Each year ISPE selects outstanding Idahoans in recognition of their engineering accomplishments and contributions to the engineering profession. Awards will be presented on June 10, 2011 during the 2011 Annual Meeting in Coeur d'Alene. Nominations must be submitted no later than April 29, 2011. Award criteria and nomination forms, as well as lists of previous winners, can be obtained from the ISPE web site, or by contacting the ISPE office at 208-426-0636.

The awards for which we are looking for nominees include:

Idaho Engineering Hall of Fame: Given by ISPE to recognize Idahoans that have made engineering contributions beyond Idaho i.e. nationally or world wide.

Idaho Excellence in Engineering Award: To recognize an Idahoan who is distinguishing themselves in engineering.

Idaho Excellence in Engineering Educator Award: This award recognizes an Engineering Educator who has had a significant impact on the engineering profession in Idaho.

Young Engineer of the Year Award: To recognize an engineer that is making a contribution to their profession. Must be no more than 35 years old.

Self nominations are welcomed and encouraged.

MATHCOUNTS PROBLEM OF THE WEEK
Can you solve this MATHCOUNTS problem? The answer will appear in the next edition of the Friday Update!

Making the Grade
At Smart Middle School students are required to take 7 classes each grading period. At the end of each grading period a student is assigned a final grade (A, B, C, D or F) for each class for the grading period. A student’s overall grade point average for the grading period is determined using a 4-point scale, where a final class grade of A is worth 4.0 points, a B is worth 3.0 points, a C is worth 2.0 points, a D is worth 1.0 point and an F earns no points. Suppose Jared has learned his final grade for the first grading period for six of his classes. He earned a grade of A in two classes, a grade of B in three classes and a grade of C in one class. What is the positive difference between the highest possible overall grade point average and lowest possible overall grade point average that Jared can have for his seven classes? Express your answer as a decimal to the nearest hundredth.
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Given the previous information for Jared's first six grades, what is the lowest possible letter grade that Jared can earn in his seventh class to have an overall grade point average of at least 3.00 for the first grading period?
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On the last school day of the first grading period Michael learns his final grade for each class. If order is not important and we know he has earned at least 1 of each of the 5 letter grades, what is the total number of combinations of 7 final grades (one for each class) that Michael can possibly receive for the first grading period?
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Assume now that we do not know any of Michael's 7 final grades. If order is not important, what is the total number of combinations of 7 final grades (one for each class) that Michael can possibly receive for the first grading period?

Answer to last week’s problem:
Since the area Kelli wants to cover with stones is 2,510 square feet, we need to divide this by 4 because that’s the number of square feet that can be covered by each bag of stones. It will take 2510/4 = 627.5 = 628 bags of stones. Each bag holds 50 pounds of stones, so the total number of pounds that would need to be purchased to cover the shaded region is 628(50) = 31,400 lbs.
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From the previous question we know that Kelli will need 628 bags of the 50-lb bags of stones to cover the region shown shaded. If each bag costs $48.50, it will cost 628(48.50) = $30,458 to purchase the amount of stones needed from Garden Mart. From the previous question we know that Kelli would only need the amount of stones contained in 627.5 50-lb bags of stones to cover the desired region. Thus, the total number of pounds of stones required is (627.5)(50) = 31,375 lbs. A ton is 2000 lbs, so if purchased from Garden Wholesaler, the first two tons of stones would cost (4000)(1.50) = $6,000. Kelli would need an additional 31,375 – 4000 = 27,375 pounds of stones. The cost for these stones would be (27,375)(0.90) = $24,637.50. Therefore, the total cost to purchase the stones from Garden Wholesaler would be $6,000 +
$24,637.50 = $30,637.50. The positive difference between the cost of purchasing the stones at Garden Mart and the cost when the stones are purchased from Garden Wholesaler is $30,637.50 − $30,458 = $179.50.
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The region that will be covered with stones consists of the four rectangular paths and the inner octagon, less the area taken up by the fountain. First, to determine the area taken up by the fountain, we need to know its radius. We are told that the circumference of the fountain is 20π feet. Since C = dπ we know that the diameter is 20 feet, making the radius 20/2 = 10 feet. So for the fountain A = πr2 = π(10)2 = 100π square feet is the area.

Next we can determine the area of the inner octagon since we are told that its area is 4/9 the area of the outer octagon. Students may recall that the area of a regular octagon is 1/2 × apothem × perimeter. For those who are not familiar with this formula, the same calculation can be derived from the fact that the interior of a regular octagon contains 8 congruent isosceles triangles. So the area of the octagon is the sum of the areas of these triangles, or 8 times the area of one of the triangles. Since the entrance to the path, which is 10 feet wide, represents 1/3 of the side, the outer octagon has side length 30 feet. We are also told that the distance from the center of the garden to the entrance of a path (side of the outer octagon) is 36 feet. These two measures give you the base and altitude of an isosceles triangle with an area equal to 1/2bh = 1/2(30)(36) = 540 square feet. Thus, the total area of the outer octagon is 8(540) = 4320 ft2. The area of the inner octagon is then 4/9(4320) = 1920 square feet. Finally, we need to find the area of the four rectangular paths. We know the width is 10 feet. To determine the length we need to again use the fact that the distance from the center of the garden to the entrance of a path (side of the outer octagon) is 36 feet. If the inner octagon has an area that is 4/9 the area of the outer, it follows that the scale factor, or ratio of the lengths of corresponding parts, is √(4/9) = 2/3. Therefore, the distance from the center of the garden to the side of the inner octagon must be 2/3 the distance from the center to the side of the outer octagon. So the distance from the center to the side of the inner octagon is 2/3(36) = 24 feet. That means the distance between the side of the inner octagon and the side of the outer octagon, or the length of the rectangular path is 36 – 24 = 12 feet. Now we see that the area
of one of the rectangular paths is 10 × 12 = 120 square feet. Since there are four paths the total combined area of the paths is 4(120) = 480 square feet. Putting all this together we see that the area of the region shown shaded is 1920 ft2 + 480 ft2 – 100π ft2 = 2085.84 ≈ 2086 square feet.

If you want to see the problem again, click http://mathcounts.org/Page.aspx?pid=1573 and choose the problem for November 1, 2010.


Idaho Society of Professional Engineers
PO Box 170239
Boise, ID 83717-0239
208-426-0636
Fax: 208-426-0639
E-Mail: ispe@idahospe.org 
Web Site: www.idahospe.org 




 

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