Question Number 78425 by arkanmath7@gmail.com last updated on 17/Jan/20
$${how}\:{to}\:{solve} \\ $$$${find}\:{inf}\:{and}\:{sup}\:{of}\:{A} \\ $$$$ \\ $$$$\mathrm{1}.\:{A}=\left\{\frac{{m}}{{n}}+\frac{\mathrm{4}{n}}{{m}}\:\:\:{m},{n}\:\in{N}\right\} \\ $$$$\mathrm{2}.\:{A}=\left\{\frac{{mn}}{\mathrm{4}{m}^{\mathrm{2}} \:+\:{n}^{\mathrm{2}} }\:\:{m}\in{Z},{n}\:\in{N}\right\} \\ $$$$\mathrm{3}.\:{A}=\left\{\frac{{m}}{\mid{m}\mid\:+\:{n}}\:\:{m}\in{Z},{n}\:\in{N}\right\} \\ $$$$ \\ $$$$ \\ $$
Answered by MJS last updated on 17/Jan/20
![(d/dm)[(m/n)+((4n)/m)]=(1/n)−((4n)/m^2 )=0 ⇒ m=±2n (d^2 /dm^2 )[(m/n)+((4n)/m)]=((8n)/m^3 )= { ((−(1/n^2 ); m=−2n)),(((1/n^2 ); m=+2n)) :} local max at m=−2n but m, n ∈N ⇒ ⇒ A_(max) =+∞ [m=1∧n→+∞ or n=1∧m→+∞ ⇒ A→+∞] local min at m=2n∧m, n ∈N ⇒ A_(min) =4](https://www.tinkutara.com/question/Q78431.png)
$$\frac{{d}}{{dm}}\left[\frac{{m}}{{n}}+\frac{\mathrm{4}{n}}{{m}}\right]=\frac{\mathrm{1}}{{n}}−\frac{\mathrm{4}{n}}{{m}^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:{m}=\pm\mathrm{2}{n} \\ $$$$\frac{{d}^{\mathrm{2}} }{{dm}^{\mathrm{2}} }\left[\frac{{m}}{{n}}+\frac{\mathrm{4}{n}}{{m}}\right]=\frac{\mathrm{8}{n}}{{m}^{\mathrm{3}} }=\begin{cases}{−\frac{\mathrm{1}}{{n}^{\mathrm{2}} };\:{m}=−\mathrm{2}{n}}\\{\frac{\mathrm{1}}{{n}^{\mathrm{2}} };\:{m}=+\mathrm{2}{n}}\end{cases} \\ $$$$\mathrm{local}\:\mathrm{max}\:\mathrm{at}\:{m}=−\mathrm{2}{n}\:\mathrm{but}\:{m},\:{n}\:\in\mathbb{N}\:\Rightarrow\: \\ $$$$\:\:\:\:\:\Rightarrow\:{A}_{\mathrm{max}} =+\infty\:\left[{m}=\mathrm{1}\wedge{n}\rightarrow+\infty\:\mathrm{or}\:{n}=\mathrm{1}\wedge{m}\rightarrow+\infty\:\Rightarrow\:{A}\rightarrow+\infty\right] \\ $$$$\mathrm{local}\:\mathrm{min}\:\mathrm{at}\:{m}=\mathrm{2}{n}\wedge{m},\:{n}\:\in\mathbb{N}\:\Rightarrow\:{A}_{\mathrm{min}} =\mathrm{4} \\ $$
Commented by MJS last updated on 17/Jan/20
$$\mathrm{because}\:\mathrm{at}\:\mathrm{a}\:\mathrm{max}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{derivate}\:\mathrm{is}\:\mathrm{smaller} \\ $$$$\mathrm{than}\:\mathrm{0} \\ $$
Commented by arkanmath7@gmail.com last updated on 17/Jan/20
$${why}\:{did}\:{you}\:{take}\:{local}\:{max}\:{at} \\ $$$${m}=−\mathrm{2}{n}\:{not}\:{at}\:{m}=\mathrm{2}{n}?? \\ $$
Commented by arkanmath7@gmail.com last updated on 17/Jan/20
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$
Answered by MJS last updated on 17/Jan/20
![(d/dm)[((mn)/(4m^2 +n^2 ))]=−((n(4m^2 −n^2 ))/((4m^2 +n^2 )^2 ))=0 ⇒ m=±(n/2) (d^2 /dm^2 )[((mn)/(4m^2 +n^2 ))]=((8mn(4m^2 −3n^2 ))/((4m^2 +n^2 )^3 ))= { (((1/n^2 ); m=−(n/2))),((−(1/n^2 ); m=(n/2))) :} min at m=−(n/2); A=−(1/4) max at m=(n/2); A=(1/4) try ((mn)/(4m^2 +n^2 ))<−(1/4) 4mn<−4m^2 −n^2 4m^2 +4mn+n^2 <0 (2m+n)^2 <0 false ((mn)/(4m^2 +n^2 ))>(1/4) leads to (2m−n)^2 <0 ⇒ false](https://www.tinkutara.com/question/Q78432.png)
$$\frac{{d}}{{dm}}\left[\frac{{mn}}{\mathrm{4}{m}^{\mathrm{2}} +{n}^{\mathrm{2}} }\right]=−\frac{{n}\left(\mathrm{4}{m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)}{\left(\mathrm{4}{m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:{m}=\pm\frac{{n}}{\mathrm{2}} \\ $$$$\frac{{d}^{\mathrm{2}} }{{dm}^{\mathrm{2}} }\left[\frac{{mn}}{\mathrm{4}{m}^{\mathrm{2}} +{n}^{\mathrm{2}} }\right]=\frac{\mathrm{8}{mn}\left(\mathrm{4}{m}^{\mathrm{2}} −\mathrm{3}{n}^{\mathrm{2}} \right)}{\left(\mathrm{4}{m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)^{\mathrm{3}} }=\begin{cases}{\frac{\mathrm{1}}{{n}^{\mathrm{2}} };\:{m}=−\frac{{n}}{\mathrm{2}}}\\{−\frac{\mathrm{1}}{{n}^{\mathrm{2}} };\:{m}=\frac{{n}}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{min}\:\mathrm{at}\:{m}=−\frac{{n}}{\mathrm{2}};\:{A}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{max}\:\mathrm{at}\:{m}=\frac{{n}}{\mathrm{2}};\:{A}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{try} \\ $$$$\frac{{mn}}{\mathrm{4}{m}^{\mathrm{2}} +{n}^{\mathrm{2}} }<−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{4}{mn}<−\mathrm{4}{m}^{\mathrm{2}} −{n}^{\mathrm{2}} \\ $$$$\mathrm{4}{m}^{\mathrm{2}} +\mathrm{4}{mn}+{n}^{\mathrm{2}} <\mathrm{0} \\ $$$$\left(\mathrm{2}{m}+{n}\right)^{\mathrm{2}} <\mathrm{0}\:\mathrm{false} \\ $$$$\frac{{mn}}{\mathrm{4}{m}^{\mathrm{2}} +{n}^{\mathrm{2}} }>\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{leads}\:\mathrm{to}\:\left(\mathrm{2}{m}−{n}\right)^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\:\mathrm{false} \\ $$
Answered by MJS last updated on 17/Jan/20
$$−\mathrm{1}<\frac{{m}}{\mid{m}\mid+{n}}<+\mathrm{1} \\ $$