Question-212736

Question Number 212736 by RojaTaniya last updated on 22/Oct/24

Answered by efronzo1 last updated on 22/Oct/24

(x/4) = (9/(15)) ⇒x=((12)/5) (y/5)= (9/(15))⇒y= 3 (z/6)= (9/(15)) ⇒z=((18)/5)

$$\:\:\:\:\frac{\mathrm{x}}{\mathrm{4}}\:=\:\frac{\mathrm{9}}{\mathrm{15}}\:\Rightarrow\mathrm{x}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\:\:\:\:\frac{\mathrm{y}}{\mathrm{5}}=\:\frac{\mathrm{9}}{\mathrm{15}}\Rightarrow\mathrm{y}=\:\mathrm{3} \\ $$$$\:\:\:\:\:\frac{\mathrm{z}}{\mathrm{6}}=\:\frac{\mathrm{9}}{\mathrm{15}}\:\Rightarrow\mathrm{z}=\frac{\mathrm{18}}{\mathrm{5}} \\ $$

Commented by RojaTaniya last updated on 22/Oct/24

$$\:{Sir}\:{satisfying},\:\:{kindly}\: \\ $$$$\:{provide}\:{solutions}. \\ $$

Answered by mr W last updated on 22/Oct/24

f(x,y,z)=(√(16−x^2 ))+(√(25−y^2 ))+(√(36−z^2 )) F(x,y,z)=(√(16−x^2 ))+(√(25−y^2 ))+(√(36−z^2 ))+(1/λ)(x+y+z−9) (∂F/∂x)=((−x)/( (√(16−x^2 ))))+(1/λ)=0 ⇒(x/4)=(1/( (√(λ^2 +1)))) (∂F/∂y)=((−y)/( (√(25−y^2 ))))+(1/λ)=0 ⇒(y/5)=(1/( (√(λ^2 +1)))) (∂F/∂z)=((−z)/( (√(36−z^2 ))))+(1/λ)=0 ⇒(z/6)=(1/( (√(λ^2 +1)))) i.e. f_(max) (x,y,z) is at (x/4)=(y/5)=(z/6)=k or x=4k, y=5k, z=6k x+y+z=15k=9 ⇒k=(3/5) f_(max) (x,y,z)=(4+5+6)(√(1−k^2 ))=12 that means for f(x,y,z)=12, x=4×(3/5)=((12)/5), y=5×(3/5)=3, z=6×(3/5)=((18)/5)

$${f}\left({x},{y},{z}\right)=\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{25}−{y}^{\mathrm{2}} }+\sqrt{\mathrm{36}−{z}^{\mathrm{2}} } \\ $$$${F}\left({x},{y},{z}\right)=\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{25}−{y}^{\mathrm{2}} }+\sqrt{\mathrm{36}−{z}^{\mathrm{2}} }+\frac{\mathrm{1}}{\lambda}\left({x}+{y}+{z}−\mathrm{9}\right) \\ $$$$\frac{\partial{F}}{\partial{x}}=\frac{−{x}}{\:\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\lambda}=\mathrm{0}\:\Rightarrow\frac{{x}}{\mathrm{4}}=\frac{\mathrm{1}}{\:\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\frac{\partial{F}}{\partial{y}}=\frac{−{y}}{\:\sqrt{\mathrm{25}−{y}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\lambda}=\mathrm{0}\:\Rightarrow\frac{{y}}{\mathrm{5}}=\frac{\mathrm{1}}{\:\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\frac{\partial{F}}{\partial{z}}=\frac{−{z}}{\:\sqrt{\mathrm{36}−{z}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\lambda}=\mathrm{0}\:\Rightarrow\frac{{z}}{\mathrm{6}}=\frac{\mathrm{1}}{\:\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}} \\ $$$${i}.{e}.\:{f}_{{max}} \left({x},{y},{z}\right)\:{is}\:{at}\:\frac{{x}}{\mathrm{4}}=\frac{{y}}{\mathrm{5}}=\frac{{z}}{\mathrm{6}}={k} \\ $$$${or}\:{x}=\mathrm{4}{k},\:{y}=\mathrm{5}{k},\:{z}=\mathrm{6}{k} \\ $$$${x}+{y}+{z}=\mathrm{15}{k}=\mathrm{9}\:\Rightarrow{k}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${f}_{{max}} \left({x},{y},{z}\right)=\left(\mathrm{4}+\mathrm{5}+\mathrm{6}\right)\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }=\mathrm{12} \\ $$$${that}\:{means}\:{for}\:{f}\left({x},{y},{z}\right)=\mathrm{12}, \\ $$$${x}=\mathrm{4}×\frac{\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{12}}{\mathrm{5}},\:{y}=\mathrm{5}×\frac{\mathrm{3}}{\mathrm{5}}=\mathrm{3},\:{z}=\mathrm{6}×\frac{\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{18}}{\mathrm{5}} \\ $$

Answered by A5T last updated on 22/Oct/24

(√(4−x))(√(4+x))+(√(5−y))(√(5+y))+(√(6−z))(√(6+z)) ≤(√(4−x+5−y+6−z))×(√(4+x+5+y+6+z)) =(√(15^2 −(x+y+z)^2 ))=12 Equality holds when ((√(4−x)),(√(5−x)),(√(6−z)))=λ((√(4+x));(√(5+x));(√(6+z))) ⇒15−9=λ^2 (15+9)⇒λ=(1/2)⇒(x,y,z)=(((12)/5),3,((18)/5))

$$\sqrt{\mathrm{4}−{x}}\sqrt{\mathrm{4}+{x}}+\sqrt{\mathrm{5}−{y}}\sqrt{\mathrm{5}+{y}}+\sqrt{\mathrm{6}−{z}}\sqrt{\mathrm{6}+{z}} \\ $$$$\leqslant\sqrt{\mathrm{4}−{x}+\mathrm{5}−{y}+\mathrm{6}−{z}}×\sqrt{\mathrm{4}+{x}+\mathrm{5}+{y}+\mathrm{6}+{z}} \\ $$$$=\sqrt{\mathrm{15}^{\mathrm{2}} −\left({x}+{y}+{z}\right)^{\mathrm{2}} }=\mathrm{12} \\ $$$${Equality}\:{holds}\:{when}\: \\ $$$$\left(\sqrt{\mathrm{4}−{x}},\sqrt{\mathrm{5}−{x}},\sqrt{\mathrm{6}−{z}}\right)=\lambda\left(\sqrt{\mathrm{4}+{x}};\sqrt{\mathrm{5}+{x}};\sqrt{\mathrm{6}+{z}}\right) \\ $$$$\Rightarrow\mathrm{15}−\mathrm{9}=\lambda^{\mathrm{2}} \left(\mathrm{15}+\mathrm{9}\right)\Rightarrow\lambda=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\left({x},{y},{z}\right)=\left(\frac{\mathrm{12}}{\mathrm{5}},\mathrm{3},\frac{\mathrm{18}}{\mathrm{5}}\right) \\ $$

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