Among all triangles in the first quadrant formed by the x−axis, the y−axis and tangent lines to the graph of y = 3x−x^2 , what is the smallest possible area?


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Question Number 123817 by benjo_mathlover last updated on 28/Nov/20

$A m o n g a l l t r i a n g l e s i n t h e f i r s t$ $q u a d r a n t f o r m e d b y t h e x - a x i s,$ $t h e y - a x i s a n d t a n g e n t l i n e s t o$ $t h e g r a p h o f y = 3 x - x^{2}, w h a t i s$ $t h e s m a l l e s t p o s s i b l e a r e a ?$
	Answered by MJS_new last updated on 28/Nov/20

	$since the curve includes the origin, the$ $smallest area is zero$
	Commented by benjo_mathlover last updated on 28/Nov/20

	$h a h a h a . . . n o s i r . b u t t h a n k y o u$
	Commented by MJS_new last updated on 28/Nov/20

	$I know the other one was meant but the$ $degenerated triangle A = B = C = (\begin{matrix} 0 \\ 0 \end{matrix}) is$ $there without a doubt . . . but of course {it}^{'} s$ $not the limit of the triangles in the 1^{st}$ $quadrant but in the 2^{nd} one .$
	Answered by mr W last updated on 28/Nov/20

	$t a n g e n t p o i n t P (p, y_{P})$ $y_{P} = 3 p - p^{2}$ $m = \frac{d y}{d x} = 3 - 2 x = 3 - 2 p$ $e q n . o f t a n g e n t :$ $y = m (x - p) + y_{P}$ $y = (3 - 2 p) x + p^{2}$ $\Rightarrow \frac{x}{\frac{p^{2}}{2 p - 3}} + \frac{y}{p^{2}} = 1$ $A r e a o f t r i a n g l e$ $A = \frac{p^{4}}{2 (2 p - 3)}$ $\frac{d (2 A)}{d p} = \frac{4 p^{3}}{2 p - 3} - \frac{2 p^{4}}{{(2 p - 3)}^{2}} = 0$ $p = 0 (r e j e c t e d) o r$ $\frac{p}{2 p - 3} = 2$ $\Rightarrow p = 2$ $\Rightarrow A_{m i n} = \frac{2^{4}}{2 (2 \times 2 - 3)} = 8$
	Commented by mr W last updated on 28/Nov/20

	Commented by benjo_mathlover last updated on 28/Nov/20

	$y e s . . t h a n k y o u$
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