Posted on 03/10/2015 5:48:37 PM PDT by MNDude
My daughter has this problem-solving question for her homework. I'm feeling kind of dumb on this one. What do you think is the correct answer?
Mrs. Feltner wants to put a border on a baby blanket. The area of the blanket is 12 square units. Which shows how many units of materials she needs for the border?
A 12 units B 14 units C 15 units D 21 units
One of my profs did this: He gave some silly math problem and wrote to answers on the board. One was 3.6 and one was 3.6454265 (or something like that). Neither answer was correct, but the entire class believed the second was correct. He was teaching us how to BS.
However, using a 2-sig-dig constraint to restate 12,450 to 12,000 makes quite a bit of real world difference if the units are "ounces of gold."
At that point no calculus is needed, just algebra:
You want to solve L*W = 12 and 2L + 2W = P (for P any of the numbers, or for that matter any other number greater than 4 times the square root of 12 that you’d like for a perimeter and L, W positive, since they must represent lengths).
Thus L = P/2 - W, and we can substitute into the other equation get
(P/2 - W)*W = 12, which is equivalent to
0 = W^2 - (P/2)W + 12
Using the quadratic formula gives
W = [P/2 +/- sqrt ( (P/2)^2 - 4*1*12 )]/2*1
So as long as (1/4)P^2 >= 48 this has real solutions, which are easily seen to be positive.
So if you’d like P = 21, you get W = [21/2 +/- sqrt(441/4 - 48)]/2.
Pick the - for and W the + for L (If you go back to the equation we used to get rid of L and get an equation in W only you can see this is right.) Giving (approximately)
W = 1.30507 and L = 9.19493 (Ihe actual values are irrational numbers — it happens that the round-off errors in those approximations exactly cancel when computing the perimeter to give 21 on the nose, and multiplying gives 12 correct to six significant digits, just like the approximations I gave).
The problem would have been well-posed with the answer B if it had been specified that the blanket was rectangular with sides of lengths given by whole numbers of units, or even with sides of rational length.
(And it didn’t even say the blanket was rectangular. I’ve seen baby blankets with scalloped edges, or rounded-corners.)
to=two. Indefensible typo for a math thread. But I’m at a gas station.
Even A is correct. It does not specify width of border or whether corners are needed. It does not specify a uniform border. There certainly is a border solution whose total area is 12 square units.
Math is above my pay grade.
10 units form the borders on all four sides.
Well, having laid quite a few floors, the limits of significance rears its ugly head with ceramic tiles as well. If you think you can measure a ceramic tile to better than 4 binary significant digits in inches, you are kidding yourself. And If you think you can cut it closer than 4 significant digits, seek help: you’ve got delusions of grandeur. (4 significant binary digits of inches is sixteenths of an inch.)
Now you’re just trying to exhaust me into conceding. I want L*W=12. And the sky is blue. And 2+2=4, not 3.999989877577899
Throw away the homework. Go out and play!
Only if you can in fact measure the difference. If you can’t, they may very well be the same number, or it may even be possible that the “12,000” ounces of gold are more than the “12,450” ounces of gold.
Mom doesn’t let me out after dark on a school night!
But I still maintain that, depending on the value/merit of what is being measured, significant digit limitations can be costly and impractical.
Okay
W = [21/2 - sqrt(441/4 - 48)]/2
L = [21/2 + sqrt(441/4 - 48)]/2
When you add them and double the result, they give exactly 21. When you multiply them they give exactly 12.
I specifically said the problem became well-posed with answer B if you added the assumption that the lenght and width were rational numbers. Lengths usually aren’t: remember the Pythagorean theorem? A right triangle with legs of length 1 has a hypotenuse of length sqrt(2), which can’t be written as a fraction, and can’t be written in finitely many decimal places. If pressed I could give a compass and straight edge construction of the lengths of the sides needed to get the perimeter of exactly 21 and area exactly 12, but no one can give finite decimal approximations of them.
Take mom with you! Get out! And stay out! ;D
The baby has just thrown up on the blanket.
Start over.
I multiplied and got 11.99952750039849
Ok. You’ve probably given the best justification for 15. I like your use of “flaps” to help dummies like me visualize. However I still don’t see this as a big indictment of the educational system.
They want the child, for example, to draw this on graph paper. Count -2 boxes inside the quilt and get 12. She may draw up to three quilts if she is really motivated. Then she will count the left side of the border all the way around until she finds a “quilt” that matches one of the given answers.
I just don’t think this is some controversial anti-public education example here.
Well then you didn’t multiply the numbers I gave you, only decimal approximations of them in your calculator.
Here we’ll do it the old school way, no decimal approximations, no calculators:
{[21/2 - sqrt(441/4 - 48)]/2}*{[21/2 + sqrt(441/4 - 48)]/2}
= [21/2 - sqrt(441/4 - 48)]*[21/2 + sqrt(441/4 - 48)]/4 (since ratios multiply by multiplying the numerators to give the new numerator, and the denominators to give the new denominator)
= [(21/2)^2 - (441/4 -48)]/4
(since as you should remember from HS algebra (a - b)*(a + b) = a^2 - b^2 and the square of a square root is whatever you took the square root of in the first place)
= (441/4 - 441/4 + 48)/4 (since squaring a ration squares the numerator and squares the denominator)
= 48/4 = 12.
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