<PRE><B>Question 30623</B>
For a right triangle, if the length of the hypotenuse is 6 and one side is x. How large could x be to maximize the area.
USING PYTOGARUS THEOREM 
OTHER LEG =SQUARE ROOT OF (6^2-X^2)=SQRT.(36-X^2)
AREA = (1/2)*(X)*SQRT.(36-X^2)
SQUARING AREA WILL BE MAXIMUM WHEN
 Y= X^2(36-X^2) IS MAXIMUM...PUT P=X^2........
Y=P(36-P)=36P-P^2 SHALL BE MAXIMUM
=-{P^2-2*P*18+18^2-18^2)=18^2-(P-18)^2...SHALL BE MAXIMUM..THAT HAPPENS WHEN 
(P-18)=0...OR....P=18=X^2...OR X=SQRT(18)=3*SQRT.(2)=3*1.41=4.23
</PRE>