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Question Number 140367 by aliibrahim1 last updated on 06/May/21
Answered by meetbhavsar25 last updated on 06/May/21
Answer is (9/(25)). x^2 has minimum value 0. y^2 has minimum value 0. (z−1)^2 has minimum value 0...but if we take so then equation 2x−4y+5z=2 is not satisfying. so we have to take x=0 y=0 z=(2/5) and we get x^2 +y^2 +(z−1)^2 = (9/(25)).
$${Answer}\:{is}\:\frac{\mathrm{9}}{\mathrm{25}}. \\ $$$${x}^{\mathrm{2}} \:{has}\:{minimum}\:{value}\:\mathrm{0}. \\ $$$${y}^{\mathrm{2}} \:{has}\:{minimum}\:{value}\:\mathrm{0}. \\ $$$$\left({z}−\mathrm{1}\right)^{\mathrm{2}} \:{has}\:{minimum}\:{value}\:\mathrm{0}…{but}\:{if}\:{we}\:{take}\:{so}\:{then}\:{equation}\:\mathrm{2}{x}−\mathrm{4}{y}+\mathrm{5}{z}=\mathrm{2}\:{is}\:{not}\:{satisfying}. \\ $$$${so}\:{we}\:{have}\:{to}\:{take}\: \\ $$$${x}=\mathrm{0} \\ $$$${y}=\mathrm{0} \\ $$$${z}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${and}\:{we}\:{get}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left({z}−\mathrm{1}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{9}}{\mathrm{25}}. \\ $$
Commented by aliibrahim1 last updated on 06/May/21
thx dir appreciate it
$${thx}\:{dir}\:{appreciate}\:{it} \\ $$
Commented by aliibrahim1 last updated on 06/May/21
your answer is incorrect sir
$${your}\:{answer}\:{is}\:{incorrect}\:{sir} \\ $$
Commented by meetbhavsar25 last updated on 07/May/21
What is the answer?
$${What}\:{is}\:{the}\:{answer}? \\ $$
Commented by meetbhavsar25 last updated on 07/May/21
Okay....i got the answer (1/5)
$${Okay}….{i}\:{got}\:{the}\:{answer}\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$
Commented by aliibrahim1 last updated on 07/May/21
yes correct
$${yes}\:{correct} \\ $$
Answered by mr W last updated on 06/May/21
z=((2−2x+4y)/5) F=x^2 +y^2 +(z−1)^2 =x^2 +y^2 +(((2−2x+4y)/5)−1)^2 =(1/(25))(29x^2 +41y^2 −16xy+12x−24y+9) (∂F/∂x)=0 ⇒29x−8y=−6 (∂F/∂y)=0 ⇒−8x+41y=12 29×41x−8×41y=−6×41 −8×8x+41×8y=12×8 x=((12×8−6×41)/(29×41−8×8))=−(2/(15)) y=((12×29−6×8)/(41×29−8×8))=(4/(15)) F_(min) =(−(2/(15)))^2 +((4/(15)))^2 +(((−2×(2/(15))−4×(4/(15))+3)/5))^2 =(1/5)
$${z}=\frac{\mathrm{2}−\mathrm{2}{x}+\mathrm{4}{y}}{\mathrm{5}} \\ $$$${F}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left({z}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left(\frac{\mathrm{2}−\mathrm{2}{x}+\mathrm{4}{y}}{\mathrm{5}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}\left(\mathrm{29}{x}^{\mathrm{2}} +\mathrm{41}{y}^{\mathrm{2}} −\mathrm{16}{xy}+\mathrm{12}{x}−\mathrm{24}{y}+\mathrm{9}\right) \\ $$$$\frac{\partial{F}}{\partial{x}}=\mathrm{0}\:\Rightarrow\mathrm{29}{x}−\mathrm{8}{y}=−\mathrm{6} \\ $$$$\frac{\partial{F}}{\partial{y}}=\mathrm{0}\:\Rightarrow−\mathrm{8}{x}+\mathrm{41}{y}=\mathrm{12} \\ $$$$\mathrm{29}×\mathrm{41}{x}−\mathrm{8}×\mathrm{41}{y}=−\mathrm{6}×\mathrm{41} \\ $$$$−\mathrm{8}×\mathrm{8}{x}+\mathrm{41}×\mathrm{8}{y}=\mathrm{12}×\mathrm{8} \\ $$$${x}=\frac{\mathrm{12}×\mathrm{8}−\mathrm{6}×\mathrm{41}}{\mathrm{29}×\mathrm{41}−\mathrm{8}×\mathrm{8}}=−\frac{\mathrm{2}}{\mathrm{15}} \\ $$$${y}=\frac{\mathrm{12}×\mathrm{29}−\mathrm{6}×\mathrm{8}}{\mathrm{41}×\mathrm{29}−\mathrm{8}×\mathrm{8}}=\frac{\mathrm{4}}{\mathrm{15}} \\ $$$${F}_{{min}} =\left(−\frac{\mathrm{2}}{\mathrm{15}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{15}}\right)^{\mathrm{2}} +\left(\frac{−\mathrm{2}×\frac{\mathrm{2}}{\mathrm{15}}−\mathrm{4}×\frac{\mathrm{4}}{\mathrm{15}}+\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}} \\ $$
Commented by mr W last updated on 06/May/21
an other way without calculus: F=(1/(25))(29x^2 +41y^2 −16xy+12x−24y+9) say 29x^2 +41y^2 −16xy+12x−24y+9=k 29x^2 +(12−16y)x+41y^2 −24y+9−k=0 Δ=(12−16y)^2 −4×29×(41y^2 −24y+9−k)≥0 (6−8y)^2 −29×(41y^2 −24y+9−k)≥0 1125y^2 −600y+225−29k≤0 Δ=600^2 −4×1125×(225−29k)≥0 300^2 −1125×225+1125×29k≥0 k≥5 F≥(1/(25))×5=(1/5) i.e. F_(min) =(1/5)
$${an}\:{other}\:{way}\:{without}\:{calculus}: \\ $$$${F}=\frac{\mathrm{1}}{\mathrm{25}}\left(\mathrm{29}{x}^{\mathrm{2}} +\mathrm{41}{y}^{\mathrm{2}} −\mathrm{16}{xy}+\mathrm{12}{x}−\mathrm{24}{y}+\mathrm{9}\right) \\ $$$${say}\:\mathrm{29}{x}^{\mathrm{2}} +\mathrm{41}{y}^{\mathrm{2}} −\mathrm{16}{xy}+\mathrm{12}{x}−\mathrm{24}{y}+\mathrm{9}={k} \\ $$$$\mathrm{29}{x}^{\mathrm{2}} +\left(\mathrm{12}−\mathrm{16}{y}\right){x}+\mathrm{41}{y}^{\mathrm{2}} −\mathrm{24}{y}+\mathrm{9}−{k}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{12}−\mathrm{16}{y}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{29}×\left(\mathrm{41}{y}^{\mathrm{2}} −\mathrm{24}{y}+\mathrm{9}−{k}\right)\geqslant\mathrm{0} \\ $$$$\left(\mathrm{6}−\mathrm{8}{y}\right)^{\mathrm{2}} −\mathrm{29}×\left(\mathrm{41}{y}^{\mathrm{2}} −\mathrm{24}{y}+\mathrm{9}−{k}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{1125}{y}^{\mathrm{2}} −\mathrm{600}{y}+\mathrm{225}−\mathrm{29}{k}\leqslant\mathrm{0} \\ $$$$\Delta=\mathrm{600}^{\mathrm{2}} −\mathrm{4}×\mathrm{1125}×\left(\mathrm{225}−\mathrm{29}{k}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{300}^{\mathrm{2}} −\mathrm{1125}×\mathrm{225}+\mathrm{1125}×\mathrm{29}{k}\geqslant\mathrm{0} \\ $$$${k}\geqslant\mathrm{5} \\ $$$${F}\geqslant\frac{\mathrm{1}}{\mathrm{25}}×\mathrm{5}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${i}.{e}.\:{F}_{{min}} =\frac{\mathrm{1}}{\mathrm{5}} \\ $$
Commented by aliibrahim1 last updated on 07/May/21
thx sir really appreciate it
$${thx}\:{sir}\:{really}\:{appreciate}\:{it} \\ $$

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