odolwa99
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My answer to this question seems close to that of the book but I have only solved for +1/2 and not the -1/2. Can anyone help?
Many thanks.
Q. The area of a triangle is \frac{-4m^2 + 4m - 1}{m}. Find the value of m via differentiation.
Attempt: \frac{dA}{dm} = \frac{4m^2 - 4m - 1}{m^2} = 0 => 4m2 - 4m + 1 = 0 => (2m - 1)(2m - 1) => m = 1/2
Ans.: (From textbook): \frac{dA}{dm} = \frac{-4m^2 + 1}{m^2} = 0 => m = +/- 1/2
Many thanks.
Homework Statement
Q. The area of a triangle is \frac{-4m^2 + 4m - 1}{m}. Find the value of m via differentiation.
Homework Equations
The Attempt at a Solution
Attempt: \frac{dA}{dm} = \frac{4m^2 - 4m - 1}{m^2} = 0 => 4m2 - 4m + 1 = 0 => (2m - 1)(2m - 1) => m = 1/2
Ans.: (From textbook): \frac{dA}{dm} = \frac{-4m^2 + 1}{m^2} = 0 => m = +/- 1/2