Optimization: maximum area of a triangle in a parabola
Inside a curve ($x^2-25$ – Parabola) a triangle is drawn with A as the vertex at the origin and the line joining points B and C lie on the parabola such that BC is parallel to x-axis and is under it. How do I get the equation for finding the maximum area of triangle ABC?
Is it something like $x(x^2-25)$?
Best Answer
Yes, you have to maximize $ (x_0) (x_0^2-25)$ just as at first indicated by you.
Easy way to differentiate a product of two terms using quotient rule with a negative sign:
If $ u\cdot v = $ constant, then by simplification of Product Rule in the format setting in numerator and denominator format
$$ \dfrac{u}{v}= -\dfrac{u^\prime}{v^\prime}$$
$$ \dfrac {(x_0^2-25)}{(x_0)} = -\dfrac {(2 x_0)}{1} \tag{1} $$
EDIT1:
Continuing from last year, simplify (1) to get
$$ x_{0\,for\, maximum \,area}= \frac{5}{\sqrt 3} $$