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Question Number 95924 by i jagooll last updated on 28/May/20

$$\mathrm{form}\mathrm{a}\mathrm{Lagrangian}\mathrm{to}\mathrm{maximize}$$$${\mathrm{x}}^2-{\mathrm{y}}^2\mathrm{subject}\mathrm{to}\mathrm{the}$$$$\mathrm{constraint}2\mathrm{x}+\mathrm{y}=3?$$

Commented by john santu last updated on 29/May/20

$$\mathrm{f}(\mathrm{x},\mathrm{y},\lambda )={\mathrm{x}}^2-{\mathrm{y}}^2+\lambda (2\mathrm{x}+\mathrm{y}-3)$$$$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=2\mathrm{x}+2\lambda =0\Rightarrow \lambda =-\mathrm{x}$$$$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=-2\mathrm{y}+\lambda =0\Rightarrow \lambda =2\mathrm{y}$$$$\frac{\partial \mathrm{f}}{\partial \lambda }=2\mathrm{x}+\mathrm{y}-3=0\Rightarrow -2\lambda +\frac{\lambda }{2}=3$$$$-3\lambda =6\Rightarrow \lambda =-2\Rightarrow \{\begin{array}{l}\mathrm{x}=2\\ \mathrm{y}=-1\end{array}$$$$\max \mathrm{f}(2,-1)=2^2-{(-1)}^2=3$$$$$$

Answered by mr W last updated on 28/May/20

$$f(x,y)=x^2-y^2$$$$F(x,y,\lambda )=x^2-y^2+\lambda (2x+y-3)$$$$\frac{\partial F}{\partial x}=2x+2\lambda =0\Rightarrow x=-\lambda$$$$\frac{\partial F}{\partial y}=-2y+\lambda =0\Rightarrow y=\frac{\lambda }{2}$$$$\frac{\partial F}{\partial \lambda }=2x+y-3=0\Rightarrow -2\lambda +\frac{\lambda }{2}-3=0\Rightarrow \lambda =-2$$$$\Rightarrow x=2,y=-1$$$$f_{max}=f(2,-1)=2^2-{(-1)}^2=3$$$$$$$$or$$$$f\left(x\right)=x^2-{(3-2x)}^2$$$$\frac{df}{dx}=2x-2(3-2x)(-2)=0$$$$\Rightarrow x=2$$$$f_{max}=f\left(2\right)=2^2-{(3-4)}^2=3$$

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