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Question Number 194321 by CrispyXYZ last updated on 04/Jul/23
find min  −(((x−y)(((xy)/4)−4)^2 )/(xy))  s. t.  x>0>y
$$\mathrm{find}\:\mathrm{min}\:\:−\frac{\left({x}−{y}\right)\left(\frac{{xy}}{\mathrm{4}}−\mathrm{4}\right)^{\mathrm{2}} }{{xy}} \\ $$$$\mathrm{s}.\:\mathrm{t}.\:\:{x}>\mathrm{0}>{y} \\ $$
Answered by Frix last updated on 04/Jul/23
f(x,y)=−(((x−y)(((xy)/4)−4)^2 )/(xy))=−(((x−y)(xy−16)^2 )/(16xy))  (df/dy)=0  (((xy−16)(2y^2 −xy−16))/(16y^2 ))=0  x>0∧y<0 ⇒ y=((x−(√(x^2 +128)))/4)  ⇒  f(x,y)=−((x^4 −320x^2 −2048)/(128x))+(((x^2 +128)^(3/2) )/(138))  (df/dx)=0  −((3x^4 −320x^2 +2048)/(128x^2 ))+((3x(√(x^2 +128)))/(128))=0  x>0 ⇒ x=((4(√3))/3)  ⇒ y=−((4(√3))/3)  ⇒ minimum is ((128(√3))/9)
$${f}\left({x},{y}\right)=−\frac{\left({x}−{y}\right)\left(\frac{{xy}}{\mathrm{4}}−\mathrm{4}\right)^{\mathrm{2}} }{{xy}}=−\frac{\left({x}−{y}\right)\left({xy}−\mathrm{16}\right)^{\mathrm{2}} }{\mathrm{16}{xy}} \\ $$$$\frac{{df}}{{dy}}=\mathrm{0} \\ $$$$\frac{\left({xy}−\mathrm{16}\right)\left(\mathrm{2}{y}^{\mathrm{2}} −{xy}−\mathrm{16}\right)}{\mathrm{16}{y}^{\mathrm{2}} }=\mathrm{0} \\ $$$${x}>\mathrm{0}\wedge{y}<\mathrm{0}\:\Rightarrow\:{y}=\frac{{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{128}}}{\mathrm{4}} \\ $$$$\Rightarrow \\ $$$${f}\left({x},{y}\right)=−\frac{{x}^{\mathrm{4}} −\mathrm{320}{x}^{\mathrm{2}} −\mathrm{2048}}{\mathrm{128}{x}}+\frac{\left({x}^{\mathrm{2}} +\mathrm{128}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{138}} \\ $$$$\frac{{df}}{{dx}}=\mathrm{0} \\ $$$$−\frac{\mathrm{3}{x}^{\mathrm{4}} −\mathrm{320}{x}^{\mathrm{2}} +\mathrm{2048}}{\mathrm{128}{x}^{\mathrm{2}} }+\frac{\mathrm{3}{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{128}}}{\mathrm{128}}=\mathrm{0} \\ $$$${x}>\mathrm{0}\:\Rightarrow\:{x}=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\Rightarrow\:{y}=−\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{minimum}\:\mathrm{is}\:\frac{\mathrm{128}\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$

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