2x+3y=5 (x^2 +y^2 )


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Question Number 98914 by shaxzod last updated on 17/Jun/20

$2 x + 3 y = 5$ $Missing \left or extra \right$
Commented by john santu last updated on 17/Jun/20
$y = ((5−2x)/3) ⇔y^2 = ((25−20x+4x^2 )/9) let f(x)= x^2 +((25−20x+4x^2 )/9) f ′(x) = 2x+((8x−20)/9) = 0 26x−20=0 ⇒x = ((10)/(13)) ⇔y=((5−((20)/(13)))/3) y = ((45)/(3.13)) = ((15)/(13)) min {x^2 +y^2 } = ((325)/(169))$
$y = \frac{5 - 2 x}{3} \Leftrightarrow y^{2} = \frac{25 - 20 x + {4 x}^{2}}{9}$ $let f (x) = x^{2} + \frac{25 - 20 x + {4 x}^{2}}{9}$ $f^{'} (x) = 2 x + \frac{8 x - 20}{9} = 0$ $26 x - 20 = 0 \Rightarrow x = \frac{10}{13} \Leftrightarrow y = \frac{5 - \frac{20}{13}}{3}$ $y = \frac{45}{3.13} = \frac{15}{13}$ $\min {x^{2} + y^{2}} = \frac{325}{169}$
Commented by mr W last updated on 17/Jun/20

$\frac{325}{169} = \frac{25}{13} = c o r r e c t$
	Answered by Farruxjano last updated on 17/Jun/20

	${L e t}^{'} s u s e C a u c h y - S h v a r z :$ $5 = 2 \times x + 3 \times y ⩽ \sqrt{(2^{2} + 3^{2}) (x^{2} + y^{2})}$ $\sqrt{13 (x^{2} + y^{2})} ⩾ 5 \Rightarrow x^{2} + y^{2} ⩾ \frac{25}{13}$ $A n s w e r : {(x^{2} + y^{2})}_{m i n} = \frac{25}{13}$
	Commented by john santu last updated on 17/Jun/20
	$with Lagrange multiplier f(x,y,λ) = x^2 +y^2 +λ(2x+3y−5) (∂f/∂x) = 2x+2λ = 0 , x=−λ (∂f/∂y) = 2y+3λ = 0 , y=−(3/2)λ we get 2x+3y=5 −2λ−(9/2)λ = 5 , λ = −((10)/(13)) { ((x = ((10)/(13)))),((y=−(3/2)×−((10)/(13)) = ((15)/(13)))) :} min { x^2 +y^2 } = ((325)/(169))$
	$with Lagrange multiplier$ $f (x, y, λ) = x^{2} + y^{2} + λ (2 x + 3 y - 5)$ $\frac{\partial f}{\partial x} = 2 x + 2 λ = 0, x = - λ$ $\frac{\partial f}{\partial y} = 2 y + 3 λ = 0, y = - \frac{3}{2} λ$ $we get 2 x + 3 y = 5$ $- 2 λ - \frac{9}{2} λ = 5, λ = - \frac{10}{13}$ ${\begin{cases} x = \frac{10}{13} \\ y = - \frac{3}{2} \times - \frac{10}{13} = \frac{15}{13} \end{cases}$ $\min {x^{2} + y^{2}} = \frac{325}{169}$
	Answered by maths mind last updated on 17/Jun/20
	$2x+3y≤(√(4+9)).(√(x^2 +y^2 )) ⇔(x^2 +y^2 )^(1/2) ≥(5/(√(13)))⇒x^2 +y^2 ≥((25)/(13))=((25.13)/(13.13))=((325)/(169)) {x^2 +y^2 }_(min) =((25)/(13))$
	$2 x + 3 y ⩽ \sqrt{4 + 9} . \sqrt{x^{2} + y^{2}}$ $\Leftrightarrow {(x^{2} + y^{2})}^{\frac{1}{2}} ⩾ \frac{5}{\sqrt{13}} \Rightarrow x^{2} + y^{2} ⩾ \frac{25}{13} = \frac{25.13}{13.13} = \frac{325}{169}$ ${x^{2} + y^{2}}_{m i n} = \frac{25}{13}$
	Answered by mr W last updated on 17/Jun/20

	$Missing \left or extra \right$
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