Josie is planning out a new community park in a suburban neighborhood. The community agrees that the park (which must be rectangular) should be as big as possible, but there are several homeowners who refuse to surrender their properties to the new development.
If the following grid shows (in black) the homeowners who are not willing to sell, what is the largest possible size for the park (in lots)?
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Probably, next step in these set of problems is this: "If you are allowed to forcefully evict one person from the grid so that you get the maximum area, who will that be?"
The rectangle is at the bottom and has a width of 2 8 and a height of 3 . 2 8 × 3 = 8 4
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Actually, on my first try, I answered 8 1 because I counted 2 7 for the width, and on my second try, I answered 8 7 because I counted 2 9 for the width :/
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On my first try, I answered 2 8 ⋅ 3 = 8 4 because I can count. :-/ (Hmm Arman, I guessed you never learned to count since the Algebra II midterm where you got a B because you counted wrong.)
Finn, we will be making this feature available to more people once we are satisfied with it.
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Hello! That's the first time you've ever replied to me! :D
Image files are extremely hard to parse,hence the lack of CS solutions..It would have been great if the grid had been provided in a text file or something.
Thought a lot about this problem. Then, eventually I used my old friend MS Paint and got it correct in second shot ^_^
Bottom of the grid. In Y axis value=3 In X axis value=28 Area of the rectangle=(28*3)=84
28*3 = 84 .. see the last portion of the diagram
I dont know why this question of counting number of blocks here. I solved it to increase my rating only.
In last three row make biggest rectangular park 28*3=54
oh my first try can solve the problem by mistake write 54 instead 84
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Okay... I just solved this but I forgot which rectangle I found... But pretty much you look for large gaps and then do some counting... Or you could write an immensely complex computer program... But I like my solution. But how is it that as soon as he had posted this problem, it was assigned a rating? How come I can't do that with my problems? Wahh... How do I do the same?