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Question Number 85241 by Power last updated on 20/Mar/20

Answered by jagoll last updated on 20/Mar/20

minimum ((4s^2 +t^2 )/(5st)) = ((4s^2 )/(5st)) + (t^2 /(5st)) = (4/5)((s/t)) + (1/5)((t/s)). let (s/t) = u ⇒f(u) = (4/5)u+(1/5)u^(−1) f ′(u)= (4/5)−(1/(5u^2 )) = 0⇒ u = (1/2) minimum f((1/2)) = (4/5)×(1/2)+ (1/5)×2 =(2/5)+(2/5)= (4/5)

$$\mathrm{minimum}\:\frac{\mathrm{4s}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }{\mathrm{5st}}\:=\:\frac{\mathrm{4s}^{\mathrm{2}} }{\mathrm{5st}}\:+\:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{5st}} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{5}}\left(\frac{\mathrm{s}}{\mathrm{t}}\right)\:+\:\frac{\mathrm{1}}{\mathrm{5}}\left(\frac{\mathrm{t}}{\mathrm{s}}\right). \\ $$$$\mathrm{let}\:\frac{\mathrm{s}}{\mathrm{t}}\:=\:\mathrm{u}\:\Rightarrow\mathrm{f}\left(\mathrm{u}\right)\:=\:\frac{\mathrm{4}}{\mathrm{5}}\mathrm{u}+\frac{\mathrm{1}}{\mathrm{5}}\mathrm{u}^{−\mathrm{1}} \\ $$$$\mathrm{f}\:'\left(\mathrm{u}\right)=\:\frac{\mathrm{4}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{5u}^{\mathrm{2}} }\:=\:\mathrm{0}\Rightarrow\:\mathrm{u}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{minimum}\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{2}}+\:\frac{\mathrm{1}}{\mathrm{5}}×\mathrm{2} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{5}}=\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$

Commented by mr W last updated on 20/Mar/20

we can get this result without calculus. we knowfor positive A and B: A^2 +B^2 −2AB=(A−B)^2 ≥0 ⇒A^2 +B^2 ≥2AB now with A=2s, B=t ((4s^2 +t^2 )/(5st))=(((2s)^2 +t^2 )/(5st))≥((2(2s)t)/(5st))=(4/5) i.e. the minimum of ((4s^2 +t^2 )/(5st)) is (4/5).

$${we}\:{can}\:{get}\:{this}\:{result}\:{without}\:{calculus}. \\ $$$${we}\:{knowfor}\:{positive}\:{A}\:{and}\:{B}: \\ $$$${A}^{\mathrm{2}} +{B}^{\mathrm{2}} −\mathrm{2}{AB}=\left({A}−{B}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\Rightarrow{A}^{\mathrm{2}} +{B}^{\mathrm{2}} \geqslant\mathrm{2}{AB} \\ $$$${now}\:{with}\:{A}=\mathrm{2}{s},\:{B}={t} \\ $$$$\frac{\mathrm{4s}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }{\mathrm{5st}}=\frac{\left(\mathrm{2}{s}\right)^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }{\mathrm{5st}}\geqslant\frac{\mathrm{2}\left(\mathrm{2}{s}\right){t}}{\mathrm{5}{st}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${i}.{e}.\:{the}\:{minimum}\:{of}\:\frac{\mathrm{4s}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }{\mathrm{5st}}\:{is}\:\frac{\mathrm{4}}{\mathrm{5}}. \\ $$

Commented by jagoll last updated on 20/Mar/20

good improvment

$$\mathrm{good}\:\mathrm{improvment} \\ $$

Commented by Power last updated on 20/Mar/20

thanks

$$\mathrm{thanks} \\ $$

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