Asked by chev
                the hypotenuse  of a right triangle  is  12 cm. find one the legs  if the area is maximum?
            
            
        Answers
                    Answered by
            Reiny
            
    Intuitively, one should conclude that the sides would be equal, let's prove it
let one side by x, then the other is √(144 - x^2)
area = (1/2)x√(144-x^2) = (1/2)x(144-x^2)^(1/2)
d(area)/dx
= (1/2)[ x(1/2)(144-x^2)^(-1/2) (-2x) + (144-x^2)^(1/2) ]
= 0 for a max of area
-x^2/√(144-x^2) + √(144-x^2) = 0
x^2/√(144-x^2) = √(144-x^2)
x^2 = 144-x^2
2x^2 = 144
x^2 = 72
x = √72 = 6√2
one side is 6√2 , and the other side is
√(144 - 72) = √72 = 6√2
as anticipated.
check:
(√72)^2 + (√72)^2
= 144
    
let one side by x, then the other is √(144 - x^2)
area = (1/2)x√(144-x^2) = (1/2)x(144-x^2)^(1/2)
d(area)/dx
= (1/2)[ x(1/2)(144-x^2)^(-1/2) (-2x) + (144-x^2)^(1/2) ]
= 0 for a max of area
-x^2/√(144-x^2) + √(144-x^2) = 0
x^2/√(144-x^2) = √(144-x^2)
x^2 = 144-x^2
2x^2 = 144
x^2 = 72
x = √72 = 6√2
one side is 6√2 , and the other side is
√(144 - 72) = √72 = 6√2
as anticipated.
check:
(√72)^2 + (√72)^2
= 144
                    Answered by
            chev
            
    tnx 
    
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