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Given-z-xy-4y-2-x-2-4y-2-x-y-0-find-minimum-and-maximum-value-of-z-

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Question Number 96306 by bemath last updated on 31/May/20
Given z = ((xy−4y^2 )/(x^2 +4y^2 )) , x,y≠0 find minimum and maximum value of z
$$\mathrm{Given}\:\mathrm{z}\:=\:\frac{\mathrm{xy}−\mathrm{4y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{4y}^{\mathrm{2}} }\:,\:\mathrm{x},\mathrm{y}\neq\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{z}\: \\ $$
Commented by john santu last updated on 31/May/20
(1) (∂z/∂x) = ((y(x^2 +4y^2 )−2x(xy−4y^2 ))/((x^2 +4y^2 )^2 )) =0 x^2 y+4y^3 −2x^2 y+8xy^2 =0 4y^3 +8xy^2 −x^2 y=0 y(4y^2 +8xy−x^2 )=0⇒4y^2 +8xy−x^2 =0 (2) (∂z/∂y) = (((x−8y)(x^2 +4y^2 )−8y(xy−4y^2 ))/((x^2 +4y^2 )^2 )) =0 x^3 +4xy^2 −8x^2 y−32y^3 −8xy^2 +32y^3 =0 x(x^2 −4y^2 −8y) =0⇒x^2 −4y^2 −8y=0 adding (1) and (2) x^2 −4y^2 −8y = 0 −x^2 +4y^2 +8xy =0 −−−−−−−−−+ 8xy−8y = 0 ⇒8y(x−1) = 0 x = 1 ⇒4y^2 +8y−1=0 y^2 +2y−(1/4) =0 (y+1)^2 = (5/4) ; y = −1± ((√5)/2) we get critical point are (1, −1+ ((√5)/2)) & (1,−1−((√5)/2)) z(1,−1+((√5)/2)) = ((((−2+(√5))/2)−4((9/4)−(√5)))/(1+(9/4)−(√5))) = ((−4+2(√5)−36+16(√5))/(13−4(√5))) = ((18(√5)−40)/(13−4(√5))) ≈ 0.061 z(1,−1−((√5)/2))=((((−2−(√5))/2)−4((9/4)+(√5)))/(1+(9/4)+(√5))) = ((−4−2(√5)−36−16(√5))/(13+4(√5))) =((−18(√5)−40)/(13+4(√5) )) ≈ −3.6569
$$\left(\mathrm{1}\right)\:\frac{\partial{z}}{\partial{x}}\:=\:\frac{{y}\left({x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} \right)−\mathrm{2}{x}\left({xy}−\mathrm{4}{y}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\mathrm{0} \\ $$$${x}^{\mathrm{2}} {y}+\mathrm{4}{y}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} {y}+\mathrm{8}{xy}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{4}{y}^{\mathrm{3}} +\mathrm{8}{xy}^{\mathrm{2}} −{x}^{\mathrm{2}} {y}=\mathrm{0} \\ $$$${y}\left(\mathrm{4}{y}^{\mathrm{2}} +\mathrm{8}{xy}−{x}^{\mathrm{2}} \right)=\mathrm{0}\Rightarrow\mathrm{4}{y}^{\mathrm{2}} +\mathrm{8}{xy}−{x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\frac{\partial{z}}{\partial\mathrm{y}}\:=\:\frac{\left({x}−\mathrm{8}{y}\right)\left({x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} \right)−\mathrm{8}{y}\left({xy}−\mathrm{4}{y}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\mathrm{0} \\ $$$${x}^{\mathrm{3}} +\mathrm{4}{xy}^{\mathrm{2}} −\mathrm{8}{x}^{\mathrm{2}} {y}−\mathrm{32}{y}^{\mathrm{3}} −\mathrm{8}{xy}^{\mathrm{2}} +\mathrm{32}{y}^{\mathrm{3}} =\mathrm{0} \\ $$$${x}\left({x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} −\mathrm{8}{y}\right)\:=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} −\mathrm{8}{y}=\mathrm{0} \\ $$$${adding}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right) \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} −\mathrm{8}{y}\:=\:\mathrm{0} \\ $$$$−{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +\mathrm{8}{xy}\:=\mathrm{0} \\ $$$$−−−−−−−−−+ \\ $$$$\mathrm{8}{xy}−\mathrm{8}{y}\:=\:\mathrm{0}\:\Rightarrow\mathrm{8}{y}\left({x}−\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$${x}\:=\:\mathrm{1}\:\Rightarrow\mathrm{4}{y}^{\mathrm{2}} +\mathrm{8}{y}−\mathrm{1}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} +\mathrm{2}{y}−\frac{\mathrm{1}}{\mathrm{4}}\:=\mathrm{0}\: \\ $$$$\left({y}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{5}}{\mathrm{4}}\:;\:\mathrm{y}\:=\:−\mathrm{1}\pm\:\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{critical}\:\mathrm{point}\:\mathrm{are} \\ $$$$\left(\mathrm{1},\:−\mathrm{1}+\:\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:\&\:\left(\mathrm{1},−\mathrm{1}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$${z}\left(\mathrm{1},−\mathrm{1}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:=\:\frac{\frac{−\mathrm{2}+\sqrt{\mathrm{5}}}{\mathrm{2}}−\mathrm{4}\left(\frac{\mathrm{9}}{\mathrm{4}}−\sqrt{\mathrm{5}}\right)}{\mathrm{1}+\frac{\mathrm{9}}{\mathrm{4}}−\sqrt{\mathrm{5}}} \\ $$$$=\:\frac{−\mathrm{4}+\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{36}+\mathrm{16}\sqrt{\mathrm{5}}}{\mathrm{13}−\mathrm{4}\sqrt{\mathrm{5}}}\:=\:\frac{\mathrm{18}\sqrt{\mathrm{5}}−\mathrm{40}}{\mathrm{13}−\mathrm{4}\sqrt{\mathrm{5}}}\:\approx\:\mathrm{0}.\mathrm{061} \\ $$$${z}\left(\mathrm{1},−\mathrm{1}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)=\frac{\frac{−\mathrm{2}−\sqrt{\mathrm{5}}}{\mathrm{2}}−\mathrm{4}\left(\frac{\mathrm{9}}{\mathrm{4}}+\sqrt{\mathrm{5}}\right)}{\mathrm{1}+\frac{\mathrm{9}}{\mathrm{4}}+\sqrt{\mathrm{5}}}\:=\:\frac{−\mathrm{4}−\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{36}−\mathrm{16}\sqrt{\mathrm{5}}}{\mathrm{13}+\mathrm{4}\sqrt{\mathrm{5}}} \\ $$$$=\frac{−\mathrm{18}\sqrt{\mathrm{5}}−\mathrm{40}}{\mathrm{13}+\mathrm{4}\sqrt{\mathrm{5}}\:}\:\approx\:−\mathrm{3}.\mathrm{6569} \\ $$

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