Question
the hypotenuse of a right triangle is 12 cm. find one the legs if the area is maximum?
Answers
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
chev
tnx