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-

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 \dots 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 \dots 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