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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 $x(x^2-25)$?

parabola triangle sketch

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  • $\begingroup$ Your question is far from being clear. $\endgroup$ – rlartiga Jan 12 '15 at 20:44
  • $\begingroup$ It is not clear what you are asking. Either improve the verbal description or give a sketch. $\endgroup$ – Narasimham Jan 12 '15 at 20:45
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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

$$ x_{0\,for\, maximum \,area}= \frac{5}{\sqrt 3} $$

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Take $y_0$ the coordinate of $B$ and $C$ therefore the area is equal to $y_0$ (height) multiplied by the base ($x_0-(-x_0)=2x_0$) therefore the area is equal to:

$$A(x_0)=\frac{(x_0^2-25)\cdot (2x_0)}{2}$$

Of course you are changing from $x^2+25$ to $x^2-25$ from the formula to the image

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