Question 121999
MOST TEACHERS DON'T HAVE THE OPPORTUNITY NOR THE INCENTIVE TO FIND ADDITIONAL SIMPLE SOLUTIONS TO MANY MATH PROBLEMS. tHEY ARE RESTRICTED TO A DEFINED TEXT BOOK WIT A SINGLE METHOD. WE VOLUNTEERS (EX. ENGINERES, STATISTICIANS, COMPUTER SPECIALISTS, CARPENTERS, ETC.) HAVE MANY DIFFERENT SPECIALITIES THAT FORCE US TO FIND THE SIMPLEIST & MOST EFFICIENT SOLUTION TO MANY REAL PROBLEMS NOT JUST CLASS ROOM EXERCISES. ALSO, TOO MANY MATH TEACHERS ARE NOT QUALIFIED NOR CERTIFIED TO TEACH MATH & CAN THEREFORE ONLY RELY ON WHATEVER THEY CAN 'LEARN FROM THE TEXT BOOKS'.
MOST OF THESE VOLUNTEERS I'D GUESS ARE FORMER PROFESSIONAL PEOPLE AS OPPOSED TO TEACHERS. THEREFORE YOU'LL GET MANY DIFFERENT METHODS TO SOLVE A PROBLEM. MANY MANY MATH PROBLEMS HAVE AT LEAST 3 DIFFERENT MATHODS TO SOLVE THEM. MY BROTHER, A RETIRED MATH TEACHER, TAUGHT ME ALL THE METHODS HE LEARNED IN 35 YEARS OF MIDDLE & HIGH SCHOOL TEACHING. 
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HAVING SOLVED MANY OF THESE 3 SIDED RECTANGLES FOR MAXIMUN AREA I'VE FOUND THE PROPORTION TO BE 1-2-1=4 
60/4=15. THUS YOU HAVE 15-30-15.
15*30=450M^2 IS THE MAX AREA.
PROOF:
REDUCE THE SHORT SIDE BY 1 M. 
14-32-14
14*32=448  
REDUCE THE LONG SIDE BY 1 M.
15.5-29-15.5
15.5*29=449.5