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Question Number 79190 by mr W last updated on 23/Jan/20

$$if{x}^2+y^2=50,$$$$findtheminimumandmaximumof$$$${(x+y)}^2-8(x+y)+20$$

Commented by jagoll last updated on 23/Jan/20

$$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{xy}-8\mathrm{x}-8\mathrm{y}+20=$$$$2\mathrm{xy}-8\mathrm{x}-8\mathrm{y}+70$$$$\mathrm{let}\mathrm{f}(\mathrm{x},\mathrm{y},\lambda )=2\mathrm{xy}-8\mathrm{x}-8\mathrm{y}+70+\lambda ({\mathrm{x}}^2+{\mathrm{y}}^2-50)$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{we}\mathrm{use}\mathrm{differential}$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{right}\mathrm{sir}?$$

Commented by jagoll last updated on 23/Jan/20

$$\frac{\mathrm{df}}{\mathrm{dx}}=2\mathrm{y}-8+\lambda \left(2\mathrm{x}\right)=0$$$$\frac{\mathrm{df}}{\mathrm{dy}}=2\mathrm{x}-8+\lambda \left(2\mathrm{y}\right)=0$$$$\frac{\mathrm{df}}{\mathrm{d}\lambda }={\mathrm{x}}^2+{\mathrm{y}}^2-50=0$$

Commented by mind is power last updated on 23/Jan/20

$$niceButyourapproch{didn}^{\prime }tusethecontraint{x}^2+y^2=50$$$$trytoputg(x,y,\gamma )=2xy-8x-8y+70+\gamma (x^2+y^2-50)$$$$$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{oo}\mathrm{yes}.\mathrm{it}\mathrm{Langrange}\mathrm{method}$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{that}\mathrm{right}\mathrm{sir}?$$

Commented by jagoll last updated on 23/Jan/20

$$\frac{2\lambda \mathrm{x}}{2\lambda \mathrm{y}}=\frac{8-2\mathrm{y}}{8-2\mathrm{x}}\Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=\frac{4-\mathrm{y}}{4-\mathrm{x}}$$$$4\mathrm{x}-{\mathrm{x}}^2=4\mathrm{y}-{\mathrm{y}}^2$$$${(\mathrm{x}-2)}^2={(\mathrm{y}-2)}^2$$$$\mathrm{x}-2=\pm (\mathrm{y}-2)\Rightarrow \mathrm{x}=2\pm (\mathrm{y}-2)$$$${\mathrm{x}}_1=\mathrm{y}\vee {\mathrm{x}}_2=4-\mathrm{y}$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{case}\left(1\right){\mathrm{x}}^2+{\mathrm{y}}^2=50\Rightarrow \mathrm{x}=\pm 5=\mathrm{y}$$$$\mathrm{f}={\left(10\right)}^2-8\left(10\right)+20=40$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{case}\left(2\right){(4-\mathrm{y})}^2+{\mathrm{y}}^2-50=0$$$${2\mathrm{y}}^2-8\mathrm{y}-34=0$$$${\mathrm{y}}^2-4\mathrm{y}-17=0$$$${(\mathrm{y}-2)}^2-21=0\Rightarrow \mathrm{y}=2+\sqrt{21}$$

Commented by jagoll last updated on 23/Jan/20

$$\mathrm{mr}\mathrm{W}\mathrm{what}\mathrm{the}\mathrm{answer}?$$

Commented by mind is power last updated on 23/Jan/20

$$NiceSir$$

Answered by mind is power last updated on 23/Jan/20

$$x=\sqrt{50}cos\left(\theta \right)$$$$y=\sqrt{50}sin\left(\theta \right)$$$$x+y=\sqrt{50}.(cos\left(\theta \right)+sin\left(\theta \right))$$$$=10sin(\theta +\frac{\pi }{4})=10sin\left(\phi \right),\phi \in [0,2\pi [$$$$x+y\in [-10,10]$$$${(x+y)}^2-8(x+y)+20=f(x+y)$$$$f\left(t\right)=t^2-8t+20$$$$t\in [-10,10]$$$$f^{\prime }\left(t\right)=2t-8,minf=f\left(4\right)=16-32+20=4$$$$x+y=4=10sin\left(\phi \right),\phi ={\sin }^{-1}\left(\frac{2}{5}\right),\theta ={\sin }^{-1}\left(\frac{2}{5}\right)-\frac{\pi }{4}$$$$x=\sqrt{50}cos\left(\theta \right),y=\sqrt{50}sin\left(\theta \right)$$$$moxf(-10)=100+80+20=200$$$$-10=10sin\left(\phi \right)\Rightarrow \phi =-\frac{\pi }{2}\Rightarrow \theta =-\frac{3\pi }{4}$$$$x=\sqrt{50}.-\frac{\sqrt{2}}{2}=-5,y=-5$$$$$$

Answered by mr W last updated on 23/Jan/20

$$itistofindtheminimumand$$$$maximumoff={(x+y)}^2-8(x+y)+20$$$$undertheconditionthatx,ysatisfy$$$$x^2+y^2=50,i.e.(x,y)isonthecircle.$$$$letx=5\sqrt{2}\cos \theta ,y=5\sqrt{2}\sin \theta$$$$f={(x+y)}^2-8(x+y)+20$$$$={(x+y-4)}^2+4$$$$={(5\sqrt{2}\cos \theta +5\sqrt{2}\sin \theta -4)}^2+4$$$$={[10(\frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta )-4]}^2+4$$$$={[10\cos (\theta -\frac{\pi }{4})-4]}^2+4⩾4$$$$f_{min}iswhen\cos (\theta -\frac{\pi }{4})=\frac{4}{10},i.e.\theta =\frac{\pi }{4}+{\cos }^{-1}\frac{2}{5},$$$$f_{min}={(4-4)}^2+4=4$$$$f_{max}iswhen\cos (\theta -\frac{\pi }{4})=-1,i.e.\theta =\frac{5\pi }{4},$$$$f_{max}={(-10-4)}^2+4=200$$

Answered by key of knowledge last updated on 23/Jan/20

$$\mathrm{u}={(x+y)}^2-8(x+y)+20={((\mathrm{x}+\mathrm{y})-4)}^2+4$$$$\mathrm{if}((\mathrm{x}+\mathrm{y})-4)=0\Rightarrow \min \left(\mathrm{u}\right)=4$$$$\min (\mathrm{x}+\mathrm{y})=-10\mathrm{\&}\max (\mathrm{x}+\mathrm{y})=10$$$$\mathrm{ax}((\mathrm{x}+\mathrm{y})-4)=(-14)\Rightarrow \max \left(\mathrm{u}\right)=200$$

Answered by john santu last updated on 24/Jan/20

$$becausexandymeetthe$$$$constraintsofcircle{x}^2+y^2=50$$$$,letx+y=kistangentlinecircle,$$$$thendistanceofthecenter$$$$ofthecircletotangentisthe$$$$radius\Rightarrow r=\mid \frac{k}{\sqrt{2}}\mid ,\mid k\mid =10\Rightarrow k=\pm 10$$$$forx+y=-10\Rightarrow f={(-10)}^2-8(-10)+20=200$$$$\therefore f_{max}=200$$$$considerx+y=t$$$$f\left(t\right)=t^2-8t+20,aquadratic$$$$functionint.theminimum$$$$valuewhent=\frac{-b}{2a}=\frac{-(-8)}{2.1}=4$$$$the{f}_{min}=4^2-8\times 4+20=4$$$$$$

Commented by mr W last updated on 23/Jan/20

$$f_{min}\ne 40but=4.$$

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