Posted: 12/19/2006 6:37:12 AM EDT
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A carpenter is going to build two log cabins. He has two piles of logs. Each pile has 50 logs 80 foot in length and 12 inches in diameter. He determines that one house shall be 40'x40' square, so he cuts each 80' log in half to get 100 logs that are 40' in length. The other house shall be 60'x20', so he cuts each 80' log into a fifty 60' logs and gets fifty 20' logs. Each house shall be of equal height because each log is 12 inches in diameter. When he is complete, he used the exact amount of materials in each house. However, the 40x40 house has 400 more square feet of floor space than the other house. Where did the 400 square feet of floor space go? |
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Oh c'mon! That's basic math, not geometry. On edit: Taken to an extreme, the house could be 78' long by 1' wide, giving a floor space of 78 sq. ft. Maximize the interior area of a 4-sided shape by making it square. Maximize it further by making it round. 2*pi*r = 160 (circumference) -> r = 160 / (2*pi) =~ 25.46 ft. area = pi*r^2 =~2037.2 This is ~437 square feet more than the square building, ans still made with the same materials. Where did it come from? |
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Sure, you can easily calculate the difference between the two houses with different dimensions. 40x40=1600 60x20=1200 But explain exactly where the floorspace went, given the fact that you used exactly the same materials. Why do you lose exactly one square foot when you go from 40x40 to 39x41. Where does it go? The perimeter length is still exactly the same. Where does it go? ETA... I should have paid more attention in geometry. |
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I'm not sure if this is really a riddle, or just a question 'cause you're not sure. It'a a basic principle of geometry. Different shapes of equal perimter length will have different areas. The maximum being a circle, the minimum being a rectangle of infinitely small (or zero) width. I guess in a way you could think of it that the missing area is now on the outside of the shape. |
Where did it go? It went outside of the perimeter. If the floorspace is not inside, then it is outside the perimeter. The space does not actually "go" anywhere, its just that the perimeter of a square rectangle is able to enclose more area than the perimeter of a non-square rectangle (all squares are rectangles, not all rectangles are square). Taken to an extreme, the area enclosed by a constant-perimeter rectangle will approach ZERO as the lengths of the short sides approach ZERO. You could take the same logs, not cut anything, and build two walls that touch. There is no enclosed space, yet you have the same 160' outside perimeter. |
Bonus room |
CS major? Then you had to learn programming. Write a program to graph the interior area of a constant-perimeter rectangle as one side goes from 0 to 1/4th the perimeter. This will give you a visual representation of the relationship. |
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Think of it this way: start with a square, now change the square so that one quadrant comes to the center (makes an L shape). You can visualize the missing space that way. XXXXXXXXX X---------X X---------X X---------X X---------X X---------X X---------X X---------X X---------X XXXXXXXXX -----XXXXX -----X----X -----X----X -----X----X XXXXX----X X---------X X---------X X---------X XXXXXXXXX |
Well, yes, a polynomial can be used to describe the relationship between the length of one side and the area, but it doesn't really get to the concept. Also, it doesn't apply to non-rectangular shapes. And Y is not always zero, Y is the area. |
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BD, take a loop of string. the maximum area it can contain inside the loop is given by a circle. The minimum area it can contain in the loop is nothing, since you can grab it on either end and pull it tight, with no space inside the string. And anything in between... EDIT-- DOH! beat by a rubber band! that analogy is better than mine... |
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Try this one... One house is 80' long by 0' wide. So, the square footage is 0' Make the house 70' long x 10' wide. 700 sqare feet. 60' x 20' = 1200 sq feet 50' x 30' = 1500 sq feet 40' x 40' = 1600 sq feet For maximum floor area, the house should be round. So start bending those logs! |
Simple. Take the formula for the area of the rectangle. Since the perimeter is constant solve for the area as a function of one of the sides. Take the derivitive of the area as a function of one side and set it equal to zero. The answer will fall right out. (i.e. the maximum area occurs when the rectangle is a square) |