SOLUTION: find all the possible rectangles whose sides are integers and the numerical value for the area is equal to the numerial value for the perimeter
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Question 333063: find all the possible rectangles whose sides are integers and the numerical value for the area is equal to the numerial value for the perimeter
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I did a little analysis using a spread sheet.
Your parameters are:
and
From the first equation we can derive:
From which we can restrict 
Consider the function
\ =\ 2l\ +\ 2w\ -\ lw)
Let 
and let 
take on the values 3, 4, 5, and 6. Then )
takes on values of 5, 6, 7, and 8 and will increase without bound as increases without bound. Clearly is not a possibility since we are looking for a result of \ =\ 0)
Let 
and let take on the values 3, 4, 5, and 6. Then )
has a constant value of 4 and will remain 4 as increases without bound. is not a possibility.
Let 
and let take on the values 3, 4, 5, and 6. Then )
takes on values of ... of ... wait for it... 3, 2, 1, and ZERO! We found one. continues to decrease without bound as increases without bound. Hence there will be no other values of such that \ =\ 0)
Let 
and let take on the values 3, 4, 5, and 6. Then )
takes on values of 2, 0, -2, -4, decreasing without bound as increases without bound. Another possibility found and another guarantee that it is the only possibility for .
Let 
and let take on the values 3, 4, 5, and 6. Then )
takes on values of 1, -2, -5, -8, decreasing without bound as increases without bound -- and skipping right over zero. No luck here.
Let 
and let take on the values 3, 4, 5, and 6. Then )
takes on values of 0, -4, -8, -12, decreasing without bound as increases without bound. Another possibility found and another guarantee that it is the only possibility for .
Notice that for , we found a possibility when was at its low limit of 3. What do you suppose is going to happen when we let 
?
Right. Let and let take on the values 3, 4, 5, and 6. Then )
takes on values of -1, -6, -11, -16, decreasing without bound as increases without bound. No zero here, or will there be one as far as you care to go.
Now, Let 
and let 
take on the values 8, 9, 10, and 11. Then Then )
takes on values of -2, -3, -4, -5, decreasing without bound as increases without bound. You should be convinced by now that we won't find any more instances of \ =\ 0)
Drop me a note and I'll share a graph of this.
John
My calculator said it, I believe it, that settles it
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