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Question Number 96185 by bemath last updated on 30/May/20

what are critical points of this function z = xy+5xy^2 +10y

$$\mathrm{what}\:\mathrm{are}\:\mathrm{critical}\:\mathrm{points}\:\mathrm{of}\:\mathrm{this} \\ $$$$\mathrm{function}\:\mathrm{z}\:=\:\mathrm{xy}+\mathrm{5xy}^{\mathrm{2}} +\mathrm{10y} \\ $$

Answered by john santu last updated on 30/May/20

(1) (∂z/∂x) = y+5y^2 = 0 y(1+5y) = 0 { ((y = 0)),((y=−(1/5))) :} (2) (∂z/∂y) = x+10xy+10=0 ⇒x(1+10y)=−10. if y = 0 ⇒ x = −10 if y = −(1/5)⇒x = 10 critical point are (−10,0) ; (10,−(1/5)) .

$$\left(\mathrm{1}\right)\:\frac{\partial\mathrm{z}}{\partial{x}}\:=\:{y}+\mathrm{5}{y}^{\mathrm{2}} \:=\:\mathrm{0}\: \\ $$$${y}\left(\mathrm{1}+\mathrm{5}{y}\right)\:=\:\mathrm{0}\:\begin{cases}{{y}\:=\:\mathrm{0}}\\{{y}=−\frac{\mathrm{1}}{\mathrm{5}}}\end{cases} \\ $$$$\left(\mathrm{2}\right)\:\frac{\partial\mathrm{z}}{\partial\mathrm{y}}\:=\:{x}+\mathrm{10}{x}\mathrm{y}+\mathrm{10}=\mathrm{0} \\ $$$$\Rightarrow{x}\left(\mathrm{1}+\mathrm{10y}\right)=−\mathrm{10}. \\ $$$$\mathrm{if}\:\mathrm{y}\:=\:\mathrm{0}\:\Rightarrow\:{x}\:=\:−\mathrm{10} \\ $$$$\mathrm{if}\:\mathrm{y}\:=\:−\frac{\mathrm{1}}{\mathrm{5}}\Rightarrow{x}\:=\:\mathrm{10} \\ $$$${critical}\:{point}\:{are}\:\left(−\mathrm{10},\mathrm{0}\right)\:;\: \\ $$$$\left(\mathrm{10},−\frac{\mathrm{1}}{\mathrm{5}}\right)\:. \\ $$$$ \\ $$

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