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Question Number 96306 by bemath last updated on 31/May/20

$$\mathrm{Given}\mathrm{z}=\frac{\mathrm{xy}-{4\mathrm{y}}^2}{{\mathrm{x}}^2+{4\mathrm{y}}^2},\mathrm{x},\mathrm{y}\ne 0$$$$\mathrm{find}\mathrm{minimum}\mathrm{and}\mathrm{maximum}$$$$\mathrm{value}\mathrm{of}\mathrm{z}$$

Commented by john santu last updated on 31/May/20

$$\left(1\right)\frac{\partial z}{\partial x}=\frac{y(x^2+4y^2)-2x(xy-4y^2)}{{(x^2+4y^2)}^2}=0$$$$x^{2}y+4y^3-2x^{2}y+8{xy}^2=0$$$$4y^3+8{xy}^2-x^{2}y=0$$$$y(4y^2+8xy-x^2)=0\Rightarrow 4y^2+8xy-x^2=0$$$$\left(2\right)\frac{\partial z}{\partial \mathrm{y}}=\frac{(x-8y)(x^2+4y^2)-8y(xy-4y^2)}{{(x^2+4y^2)}^2}=0$$$$x^3+4{xy}^2-8x^{2}y-32y^3-8{xy}^2+32y^3=0$$$$x(x^2-4y^2-8y)=0\Rightarrow x^2-4y^2-8y=0$$$$adding\left(1\right)and\left(2\right)$$$$x^2-4y^2-8y=0$$$$-x^2+4y^2+8xy=0$$$$---------+$$$$8xy-8y=0\Rightarrow 8y(x-1)=0$$$$x=1\Rightarrow 4y^2+8y-1=0$$$$y^2+2y-\frac{1}{4}=0$$$${(y+1)}^2=\frac{5}{4};\mathrm{y}=-1\pm \frac{\sqrt{5}}{2}$$$$\mathrm{we}\mathrm{get}\mathrm{critical}\mathrm{point}\mathrm{are}$$$$(1,-1+\frac{\sqrt{5}}{2})\mathrm{\&}(1,-1-\frac{\sqrt{5}}{2})$$$$z(1,-1+\frac{\sqrt{5}}{2})=\frac{\frac{-2+\sqrt{5}}{2}-4(\frac{9}{4}-\sqrt{5})}{1+\frac{9}{4}-\sqrt{5}}$$$$=\frac{-4+2\sqrt{5}-36+16\sqrt{5}}{13-4\sqrt{5}}=\frac{18\sqrt{5}-40}{13-4\sqrt{5}}\approx 0.061$$$$z(1,-1-\frac{\sqrt{5}}{2})=\frac{\frac{-2-\sqrt{5}}{2}-4(\frac{9}{4}+\sqrt{5})}{1+\frac{9}{4}+\sqrt{5}}=\frac{-4-2\sqrt{5}-36-16\sqrt{5}}{13+4\sqrt{5}}$$$$=\frac{-18\sqrt{5}-40}{13+4\sqrt{5}}\approx -3.6569$$

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