Let , , and be the side lengths of and . If the maximum area of is equal to , where and are integers with square-free, find the value of .
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For neatness, we'll maximise 1 6 times the square of the area first. Using Heron's formula, this is ( a + b + c ) ( a + b − c ) ( a − b + c ) ( − a + b + c )
We can now include the constraint using a Lagrangian multiplier; so define
F ( a , b , c , L ) = ( a + b + c ) ( a + b − c ) ( a − b + c ) ( − a + b + c ) + L ( 2 a + 7 b + 1 1 c − 1 2 0 )
Setting each of the partial derivatives to zero, 4 a ( − a 2 + b 2 + c 2 ) + 2 L 4 b ( a 2 − b 2 + c 2 ) + 7 L 4 c ( a 2 + b 2 − c 2 ) + 1 1 L 2 a + 7 b + 1 1 c − 1 2 0 = 0 = 0 = 0 = 0
Solving this (ie, getting Wolfram|Alpha to solve it) we find just one solution with positive lengths, a = 5 , b = 7 and c = 8 giving an area of 1 0 3 and an answer of 1 3 .