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Question Number 111560 by dw last updated on 04/Sep/20

Commented by dw last updated on 04/Sep/20

Step by step solution

$${Step}\:{by}\:{step}\:{solution} \\ $$

Answered by Her_Majesty last updated on 04/Sep/20

(d/dz)[(√(x^2 +1))+(√((x−y)^2 +25))+(√((z−y)^2 +4))+(√((9−z)^2 +16))]= =((z−9)/( (√((9−z)^2 +16))))+((z−y)/( (√((z−y)^2 +4))))=0 (((z−9)^2 )/((9−z)^2 +16))=(((z−y)^2 )/((z−y)^2 +4)) (_(because of squaring!) ^(beware of false solutions) ) z^2 −(((8y)/3)−6)z+((4y^2 )/3)−27=0 z=2y−9 (false!) ∨ z=((2y)/3)+3 ⇒ (√(x^2 +1))+(√((x−y)^2 +25))+(√((y−9)^2 +36)) (d/dy)[(√(x^2 +1))+(√((x−y)^2 +25))+(√((y−9)^2 +36))]= =((y−x)/( (√((x−y)^2 +25))))+((y−9)/( (√((y−9)^2 +36))))=0 (((y−x)^2 )/((x−y)^2 +25))=(((y−9)^2 )/((y−9)^2 +36)) (_(because of squaring!) ^(beware of false solutions) ) y^2 −((72x−450)/(11))y+((36x^2 −2025)/(11))=0 y=6x−45 (false!) ∨ y=((6x+45)/(11)) ⇒ (√(x^2 +1))+(√((x−9)^2 +121)) (d/dx)[(√(x^2 +1))+(√((x−9)^2 +121))]= =(x/( (√(x^2 +1))))+((x−9)/( (√((x−9)^2 +121))))=0 (x^2 /(x^2 +1))=(((x−9)^2 )/((x−9)^2 +121)) (_(because of squaring!) ^(beware of false solutions) ) x^2 +(3/(30))x−((27)/(40))=0 x=−(9/(10)) (false!) ∨ x=(3/4) ⇒ y=(9/2)∧z=6 ⇒ answer is 15

$$\frac{{d}}{{dz}}\left[\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{25}}+\sqrt{\left({z}−{y}\right)^{\mathrm{2}} +\mathrm{4}}+\sqrt{\left(\mathrm{9}−{z}\right)^{\mathrm{2}} +\mathrm{16}}\right]= \\ $$$$=\frac{{z}−\mathrm{9}}{\:\sqrt{\left(\mathrm{9}−{z}\right)^{\mathrm{2}} +\mathrm{16}}}+\frac{{z}−{y}}{\:\sqrt{\left({z}−{y}\right)^{\mathrm{2}} +\mathrm{4}}}=\mathrm{0} \\ $$$$\frac{\left({z}−\mathrm{9}\right)^{\mathrm{2}} }{\left(\mathrm{9}−{z}\right)^{\mathrm{2}} +\mathrm{16}}=\frac{\left({z}−{y}\right)^{\mathrm{2}} }{\left({z}−{y}\right)^{\mathrm{2}} +\mathrm{4}}\:\left(_{{because}\:{of}\:{squaring}!} ^{{beware}\:{of}\:{false}\:{solutions}} \right) \\ $$$${z}^{\mathrm{2}} −\left(\frac{\mathrm{8}{y}}{\mathrm{3}}−\mathrm{6}\right){z}+\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{3}}−\mathrm{27}=\mathrm{0} \\ $$$${z}=\mathrm{2}{y}−\mathrm{9}\:\left({false}!\right)\:\vee\:{z}=\frac{\mathrm{2}{y}}{\mathrm{3}}+\mathrm{3} \\ $$$$\Rightarrow \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{25}}+\sqrt{\left({y}−\mathrm{9}\right)^{\mathrm{2}} +\mathrm{36}} \\ $$$$\frac{{d}}{{dy}}\left[\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{25}}+\sqrt{\left({y}−\mathrm{9}\right)^{\mathrm{2}} +\mathrm{36}}\right]= \\ $$$$=\frac{{y}−{x}}{\:\sqrt{\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{25}}}+\frac{{y}−\mathrm{9}}{\:\sqrt{\left({y}−\mathrm{9}\right)^{\mathrm{2}} +\mathrm{36}}}=\mathrm{0} \\ $$$$\frac{\left({y}−{x}\right)^{\mathrm{2}} }{\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{25}}=\frac{\left({y}−\mathrm{9}\right)^{\mathrm{2}} }{\left({y}−\mathrm{9}\right)^{\mathrm{2}} +\mathrm{36}}\:\left(_{{because}\:{of}\:{squaring}!} ^{{beware}\:{of}\:{false}\:{solutions}} \right) \\ $$$${y}^{\mathrm{2}} −\frac{\mathrm{72}{x}−\mathrm{450}}{\mathrm{11}}{y}+\frac{\mathrm{36}{x}^{\mathrm{2}} −\mathrm{2025}}{\mathrm{11}}=\mathrm{0} \\ $$$${y}=\mathrm{6}{x}−\mathrm{45}\:\left({false}!\right)\:\vee\:{y}=\frac{\mathrm{6}{x}+\mathrm{45}}{\mathrm{11}} \\ $$$$\Rightarrow \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{\left({x}−\mathrm{9}\right)^{\mathrm{2}} +\mathrm{121}} \\ $$$$\frac{{d}}{{dx}}\left[\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{\left({x}−\mathrm{9}\right)^{\mathrm{2}} +\mathrm{121}}\right]= \\ $$$$=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}+\frac{{x}−\mathrm{9}}{\:\sqrt{\left({x}−\mathrm{9}\right)^{\mathrm{2}} +\mathrm{121}}}=\mathrm{0} \\ $$$$\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}=\frac{\left({x}−\mathrm{9}\right)^{\mathrm{2}} }{\left({x}−\mathrm{9}\right)^{\mathrm{2}} +\mathrm{121}}\:\left(_{{because}\:{of}\:{squaring}!} ^{{beware}\:{of}\:{false}\:{solutions}} \right) \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{30}}{x}−\frac{\mathrm{27}}{\mathrm{40}}=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{9}}{\mathrm{10}}\:\left({false}!\right)\:\vee\:{x}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:{y}=\frac{\mathrm{9}}{\mathrm{2}}\wedge{z}=\mathrm{6} \\ $$$$\Rightarrow\:{answer}\:{is}\:\mathrm{15} \\ $$

Commented by dw last updated on 05/Sep/20

Thank yiu Sir

$${Thank}\:{yiu}\:{Sir} \\ $$

Answered by 1549442205PVT last updated on 04/Sep/20

Applying the inequality : (√(a_1 ^2 +b_1 ^2 ))+(√(a_2 ^2 +b_2 ^2 ))+...+(√(a_n ^2 +b_n ^2 )) ≥(√((a_1 +a_2 +...+a_n )^2 +(b_1 +b_2 +...+b_n )^2 )) we have: P=(√(x^2 +1))+(√((x−y)^2 +25))+(√((z−y)^2 +4))+(√((9−z)^2 +16)) =(√(x^2 +1))+(√((y−x)^2 +25))+(√((z−y)^2 +4))+(√((9−z)^2 +16)) ≥(√((x+y−x+z−y+9−z)^2 +(1+5+2+4)^2 )) =(√(9^2 +12^2 ))=15 The equality ocurrs if and only if (x/1)=((y−x)/5)=((z−y)/2)=((9−z)/4)=((x+y−x+z−y+9−z)/(1+5+2+4))=(9/(12))=(3/4) ⇒x=0.75;y=4.5; z=6 Thus,P has smallest value equal to 15 when (x,y,z)=((3/4);(9/2);6)

$$\mathrm{Applying}\:\mathrm{the}\:\mathrm{inequality}\:: \\ $$$$\sqrt{\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{b}_{\mathrm{1}} ^{\mathrm{2}} }+\sqrt{\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{b}_{\mathrm{2}} ^{\mathrm{2}} }+...+\sqrt{\mathrm{a}_{\mathrm{n}} ^{\mathrm{2}} +\mathrm{b}_{\mathrm{n}} ^{\mathrm{2}} } \\ $$$$\geqslant\sqrt{\left(\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} +...+\mathrm{a}_{\mathrm{n}} \right)^{\mathrm{2}} +\left(\mathrm{b}_{\mathrm{1}} +\mathrm{b}_{\mathrm{2}} +...+\mathrm{b}_{\mathrm{n}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{we}\:\mathrm{have}: \\ $$$$\mathrm{P}=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} +\mathrm{25}}+\sqrt{\left(\mathrm{z}−\mathrm{y}\right)^{\mathrm{2}} +\mathrm{4}}+\sqrt{\left(\mathrm{9}−\mathrm{z}\right)^{\mathrm{2}} +\mathrm{16}} \\ $$$$=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{\left(\mathrm{y}−\mathrm{x}\right)^{\mathrm{2}} +\mathrm{25}}+\sqrt{\left(\mathrm{z}−\mathrm{y}\right)^{\mathrm{2}} +\mathrm{4}}+\sqrt{\left(\mathrm{9}−\mathrm{z}\right)^{\mathrm{2}} +\mathrm{16}} \\ $$$$\geqslant\sqrt{\left(\mathrm{x}+\mathrm{y}−\mathrm{x}+\mathrm{z}−\mathrm{y}+\mathrm{9}−\mathrm{z}\right)^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{5}+\mathrm{2}+\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{9}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\mathrm{15} \\ $$$$\mathrm{The}\:\mathrm{equality}\:\mathrm{ocurrs}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\frac{\mathrm{x}}{\mathrm{1}}=\frac{\mathrm{y}−\mathrm{x}}{\mathrm{5}}=\frac{\mathrm{z}−\mathrm{y}}{\mathrm{2}}=\frac{\mathrm{9}−\mathrm{z}}{\mathrm{4}}=\frac{\mathrm{x}+\mathrm{y}−\mathrm{x}+\mathrm{z}−\mathrm{y}+\mathrm{9}−\mathrm{z}}{\mathrm{1}+\mathrm{5}+\mathrm{2}+\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{12}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{0}.\mathrm{75};\mathrm{y}=\mathrm{4}.\mathrm{5};\:\mathrm{z}=\mathrm{6} \\ $$$$\mathrm{Thus},\mathrm{P}\:\mathrm{has}\:\mathrm{smallest}\:\mathrm{value}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{15}\:\boldsymbol{\mathrm{when}}\:\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}\right)=\left(\frac{\mathrm{3}}{\mathrm{4}};\frac{\mathrm{9}}{\mathrm{2}};\mathrm{6}\right) \\ $$

Commented by dw last updated on 05/Sep/20

thank you Sir

$${thank}\:{you}\:{Sir} \\ $$

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