Area of the triangle = 1/2 AC x BC = y
AC = 10-BC
BC = x
AC = 10 - x
y= 1/2 x (x)(10 - x)
y = 5x -(x^2/2)
Then differentiate 😊
AC = 10-BC
BC = x
AC = 10 - x
y= 1/2 x (x)(10 - x)
y = 5x -(x^2/2)
Then differentiate 😊
(edited 8 years ago)
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