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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
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?
$${Among}\:{all}\:{triangles}\:{in}\:{the}\:{first} \\ $$$${quadrant}\:{formed}\:{by}\:{the}\:{x}−{axis}, \\ $$$${the}\:{y}−{axis}\:{and}\:{tangent}\:{lines}\:{to} \\ $$$${the}\:{graph}\:{of}\:{y}\:=\:\mathrm{3}{x}−{x}^{\mathrm{2}} ,\:{what}\:{is} \\ $$$${the}\:{smallest}\:{possible}\:{area}? \\ $$
Answered by MJS_new last updated on 28/Nov/20
since the curve includes the origin, the smallest area is zero
$$\mathrm{since}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{includes}\:\mathrm{the}\:\mathrm{origin},\:\mathrm{the} \\ $$$$\mathrm{smallest}\:\mathrm{area}\:\mathrm{is}\:\mathrm{zero} \\ $$
Commented by benjo_mathlover last updated on 28/Nov/20
hahaha...no sir. but thank you
$${hahaha}…{no}\:\:{sir}.\:{but}\:{thank}\:{you} \\ $$
Commented by MJS_new last updated on 28/Nov/20
I know the other one was meant but the degenerated triangle A=B=C= ((0),(0) ) 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.
$$\mathrm{I}\:\mathrm{know}\:\mathrm{the}\:\mathrm{other}\:\mathrm{one}\:\mathrm{was}\:\mathrm{meant}\:\mathrm{but}\:\mathrm{the} \\ $$$$\mathrm{degenerated}\:\mathrm{triangle}\:{A}={B}={C}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\mathrm{is} \\ $$$$\mathrm{there}\:\mathrm{without}\:\mathrm{a}\:\mathrm{doubt}…\:\mathrm{but}\:\mathrm{of}\:\mathrm{course}\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{not}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangles}\:\mathrm{in}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \\ $$$$\mathrm{quadrant}\:\mathrm{but}\:\mathrm{in}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{one}. \\ $$
Answered by mr W last updated on 28/Nov/20
tangent point P(p,y_P ) y_P =3p−p^2 m=(dy/dx)=3−2x=3−2p eqn. of tangent: y=m(x−p)+y_P y=(3−2p)x+p^2 ⇒(x/(p^2 /(2p−3)))+(y/p^2 )=1 Area of triangle A=(p^4 /(2(2p−3))) ((d(2A))/dp)=((4p^3 )/(2p−3))−((2p^4 )/((2p−3)^2 ))=0 p=0 (rejected) or (p/(2p−3))=2 ⇒p=2 ⇒A_(min) =(2^4 /(2(2×2−3)))=8
$${tangent}\:{point}\:{P}\left({p},{y}_{{P}} \right) \\ $$$${y}_{{P}} =\mathrm{3}{p}−{p}^{\mathrm{2}} \\ $$$${m}=\frac{{dy}}{{dx}}=\mathrm{3}−\mathrm{2}{x}=\mathrm{3}−\mathrm{2}{p} \\ $$$${eqn}.\:{of}\:{tangent}: \\ $$$${y}={m}\left({x}−{p}\right)+{y}_{{P}} \\ $$$${y}=\left(\mathrm{3}−\mathrm{2}{p}\right){x}+{p}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{x}}{\frac{{p}^{\mathrm{2}} }{\mathrm{2}{p}−\mathrm{3}}}+\frac{{y}}{{p}^{\mathrm{2}} }=\mathrm{1} \\ $$$${Area}\:{of}\:{triangle} \\ $$$${A}=\frac{{p}^{\mathrm{4}} }{\mathrm{2}\left(\mathrm{2}{p}−\mathrm{3}\right)} \\ $$$$\frac{{d}\left(\mathrm{2}{A}\right)}{{dp}}=\frac{\mathrm{4}{p}^{\mathrm{3}} }{\mathrm{2}{p}−\mathrm{3}}−\frac{\mathrm{2}{p}^{\mathrm{4}} }{\left(\mathrm{2}{p}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$${p}=\mathrm{0}\:\left({rejected}\right)\:{or} \\ $$$$\frac{{p}}{\mathrm{2}{p}−\mathrm{3}}=\mathrm{2} \\ $$$$\Rightarrow{p}=\mathrm{2} \\ $$$$\Rightarrow{A}_{{min}} =\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{2}\left(\mathrm{2}×\mathrm{2}−\mathrm{3}\right)}=\mathrm{8} \\ $$
Commented by mr W last updated on 28/Nov/20
Commented by benjo_mathlover last updated on 28/Nov/20
yes..thank you
$${yes}..{thank}\:{you} \\ $$

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