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Question Number 109302 by peter frank last updated on 22/Aug/20

Answered by Dwaipayan Shikari last updated on 22/Aug/20

$$x=(10-y)$$$$x^2+y^2=y^2+y^2-20y+100=2{(y-5)}^2+50$$$$\text{Missing left or extra right}$$

Commented by peter frank last updated on 23/Aug/20

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

Answered by john santu last updated on 22/Aug/20

$$\frac{◻JS◻}{\mathrm{♡}\mathrm{♡}\mathrm{♡}}$$$$let{x}^2+y^2=k^2;since(x,y)satifiesthe$$$$linex+y=10sox+y=10istangent$$$$linetocircle.\Rightarrow k=\mid \frac{0+0-10}{\sqrt{2}}\mid$$$$\Rightarrow k=\frac{10}{\sqrt{2}}=5\sqrt{2}\Rightarrow x^2+y^2=k^2=50.$$

Answered by mr W last updated on 22/Aug/20

$$x^2+y^2=x^2+{(10-x)}^2$$$$=2(x^2-10x+25)+50$$$$=2{(x-5)}^2+50$$$$⩾50$$

Commented by peter frank last updated on 22/Aug/20

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

Answered by 1549442205PVT last updated on 23/Aug/20

$$$$$$2({\mathrm{x}}^2+{\mathrm{y}}^2)-{(\mathrm{x}+\mathrm{y})}^2=2({\mathrm{x}}^2+{\mathrm{y}}^2)-({\mathrm{x}}^2+2\mathrm{xy}+{\mathrm{y}}^2)$$$$=({\mathrm{x}}^2-2\mathrm{xy}+{\mathrm{y}}^2)={(\mathrm{x}-\mathrm{y})}^2⩾0$$$$\mathrm{Therefore},2({\mathrm{x}}^2+{\mathrm{y}}^2)-{10}^2⩾0$$$$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2⩾\frac{{10}^2}{2}=50.\mathrm{The}\mathrm{equality}\mathrm{ocurrs}$$$$\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}\{\begin{array}{l}\mathrm{x}-\mathrm{y}=0\\ \mathrm{x}+\mathrm{y}=10\end{array}\iff \mathrm{x}=\mathrm{y}=5$$$$\mathrm{Consequently},{\mathrm{x}}^2+{\mathrm{y}}^2\mathrm{has}\mathrm{smallest}\mathrm{value}$$$$\mathrm{equal}\mathrm{to}50\mathrm{when}(\mathrm{x},\mathrm{y})=(5,5)$$$$\mathbf{Other}\mathbf{way}:$$$$\mathrm{Put}\mathrm{x}=5+\mathrm{a},\mathrm{y}=5-\mathrm{a}.\mathrm{Then}\mathrm{we}\mathrm{have}$$$${\mathrm{x}}^2+{\mathrm{y}}^2={(5+\mathrm{a})}^2+{(5-\mathrm{a})}^2=25+10\mathrm{a}+{\mathrm{a}}^2$$$$+25-10\mathrm{a}+{\mathrm{a}}^2=50+{2\mathrm{a}}^2⩾50$$$${\mathrm{x}}^2+{\mathrm{y}}^2=50\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}\mathrm{a}=0\iff \mathrm{x}=\mathrm{y}=5$$$$\mathrm{Thus},{\mathrm{x}}^2+{\mathrm{y}}^2\mathrm{has}\mathrm{least}\mathrm{value}\mathrm{equal}\mathrm{to}$$$$50\mathrm{when}(\mathrm{x},\mathrm{y})=(5,5)$$$$\mathrm{why}\mathrm{do}\mathrm{my}\mathrm{post}\mathrm{has}\mathrm{space}\mathrm{below}?$$$$$$

Answered by $@y@m last updated on 22/Aug/20

$$Letz=x^2+y^2$$$$z=x^2+{(10-x)}^2$$$$\frac{dz}{dx}=2x-2(10-x)=4x-20....\left(1\right)$$$$Formaximaorminima,$$$$\frac{dz}{dx}=0$$$$\Rightarrow 4x-20=0$$$$\Rightarrow x=5$$$$Now,$$$$\frac{d^{2}y}{{dx}^2}=\frac{d}{dx}(4x-20)=4$$$$\because \frac{d^{2}y}{{dx}^2}>0$$$$\therefore zisminimumwhenx=5$$$$z=5^2+{(10-5)}^2=50$$

Commented by peter frank last updated on 23/Aug/20

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

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