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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 (x^2 +y^2 )_(min) =50

$${x}=\left(\mathrm{10}−{y}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={y}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{20}{y}+\mathrm{100}=\mathrm{2}\left({y}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{50} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \underset{{min}} {\right)}=\mathrm{50} \\ $$

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

thank you

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

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

((□JS□)/(♥♥♥)) let x^2 +y^2 = k^2 ; since (x,y) satifies the line x+y = 10 so x+y = 10 is tangent line to circle. ⇒ k = ∣((0+0−10)/( (√2)))∣ ⇒k = ((10)/( (√2))) = 5(√2) ⇒ x^2 +y^2 = k^2 = 50.

$$\:\:\:\frac{\Box{JS}\Box}{\heartsuit\heartsuit\heartsuit} \\ $$$${let}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:{k}^{\mathrm{2}} \:;\:{since}\:\left({x},{y}\right)\:{satifies}\:{the}\: \\ $$$${line}\:{x}+{y}\:=\:\mathrm{10}\:{so}\:{x}+{y}\:=\:\mathrm{10}\:{is}\:{tangent} \\ $$$${line}\:{to}\:{circle}.\:\Rightarrow\:{k}\:=\:\mid\frac{\mathrm{0}+\mathrm{0}−\mathrm{10}}{\:\sqrt{\mathrm{2}}}\mid\: \\ $$$$\Rightarrow{k}\:=\:\frac{\mathrm{10}}{\:\sqrt{\mathrm{2}}}\:=\:\mathrm{5}\sqrt{\mathrm{2}}\:\Rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:{k}^{\mathrm{2}} \:=\:\mathrm{50}. \\ $$

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

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={x}^{\mathrm{2}} +\left(\mathrm{10}−{x}\right)^{\mathrm{2}} \\ $$$$=\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{25}\right)+\mathrm{50} \\ $$$$=\mathrm{2}\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{50} \\ $$$$\geqslant\mathrm{50} \\ $$

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

thank you

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

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

2(x^2 +y^2 )−(x+y)^2 =2(x^2 +y^2 )−(x^2 +2xy+y^2 ) =(x^2 −2xy+y^2 )=(x−y)^2 ≥0 Therefore,2(x^2 +y^2 )−10^2 ≥0 ⇒x^2 +y^2 ≥((10^2 )/2)=50.The equality ocurrs if and only if { ((x−y=0)),((x+y=10)) :}⇔x=y=5 Consequently,x^2 +y^2 has smallest value equal to 50 when (x,y)=(5,5) Other way : Put x=5+a,y=5−a.Then we have x^2 +y^2 =(5+a)^2 +(5−a)^2 =25+10a+a^2 +25−10a+a^2 =50+2a^2 ≥50 x^2 +y^2 =50 if and only if a=0⇔x=y=5 Thus,x^2 +y^2 has least value equal to 50 when (x,y)=(5,5) why do my post has space below?

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

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

Let z=x^2 +y^2 z=x^2 +(10−x)^2 (dz/dx)=2x−2(10−x)=4x−20 ....(1) For maxima or minima, (dz/dx)=0 ⇒4x−20=0 ⇒x=5 Now, (d^2 y/dx^2 ) =(d/dx)(4x−20)=4 ∵ (d^2 y/dx^2 ) >0 ∴ z is minimum when x=5 z=5^2 +(10−5)^2 =50

$${Let}\:{z}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${z}={x}^{\mathrm{2}} +\left(\mathrm{10}−{x}\right)^{\mathrm{2}} \\ $$$$\frac{{dz}}{{dx}}=\mathrm{2}{x}−\mathrm{2}\left(\mathrm{10}−{x}\right)=\mathrm{4}{x}−\mathrm{20}\:....\left(\mathrm{1}\right) \\ $$$${For}\:{maxima}\:{or}\:{minima}, \\ $$$$\frac{{dz}}{{dx}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}{x}−\mathrm{20}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{5} \\ $$$${Now}, \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\frac{{d}}{{dx}}\left(\mathrm{4}{x}−\mathrm{20}\right)=\mathrm{4} \\ $$$$\because\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:>\mathrm{0}\: \\ $$$$\therefore\:{z}\:{is}\:{minimum}\:{when}\:{x}=\mathrm{5} \\ $$$${z}=\mathrm{5}^{\mathrm{2}} +\left(\mathrm{10}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{50} \\ $$

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

thank you

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

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