If-x-amp-y-satisfy-the-equation-x-2-y-2-4x-6y-1-0-find-minimum-value-of-x-y-

Question Number 97266 by bobhans last updated on 07/Jun/20

If x & y satisfy the equation x^{2} + y^{2} - 4 x - 6 y - 1 = 0

find minimum value of x + y ?

Commented by bobhans last updated on 07/Jun/20

f (x, y, λ) = x + y + λ (x^{2} + y^{2} - 4 x - 6 y - 1)

\frac{\partial f}{\partial x} = 1 + λ (2 x - 4) = 0 \Rightarrow λ = \frac{1}{4 - 2 x}

\frac{\partial f}{\partial y} = 1 + λ (2 y - 6) = 0 \Rightarrow λ = \frac{1}{6 - 2 y}

so 4 - 2 x = 6 - 2 y; y = 1 + x

substitute to constraint

x^{2} + {(x + 1)}^{2} - 4 x - 6 (1 + x) - 1 = 0

{2 x}^{2} - 8 x - 6 = 0; x^{2} - 4 x - 3 = 0

x = \frac{4 \pm \sqrt{28}}{2} = 2 \pm \sqrt{7} \Rightarrow {\begin{cases} x = 2 + \sqrt{7} \land y = 3 + \sqrt{7} \\ x = 2 - \sqrt{7} \land y = 3 - \sqrt{7} \end{cases}

so minimum value x + y = 5 - 2 \sqrt{7} or 5 - \sqrt{28}

Answered by john santu last updated on 07/Jun/20

let x + y = k . because (x, y)

satisfy the circle x^{2} + y^{2} - 4 x - 6 y - 1 = 0

then r = \frac{∣ 2 + 3 - k ∣}{\sqrt{2}} = \sqrt{14}

\Leftrightarrow∣ k - 5 ∣ = \sqrt{28}

{\begin{cases} k_{\max} = 5 + \sqrt{28} \\ k_{\min} = 5 - \sqrt{28} \end{cases}

Commented by bobhans last updated on 07/Jun/20

great sir your short cut

Answered by Farruxjano last updated on 07/Jun/20

x^{2} + y^{2} - 4 x - 6 y - 1 = 0, m i n {x + y} = ?

x^{2} + y^{2} - 4 x - 6 y - 1 = 0 \Rightarrow {(x - 2)}^{2} + {(y - 3)}^{2} = 14

{(\frac{x - 2}{\sqrt{14}})}^{2} + {(\frac{y - 3}{\sqrt{14}})}^{2} = 1 \Rightarrow {\begin{cases} \frac{x - 2}{\sqrt{14}} = s i n α \\ \frac{y - 3}{\sqrt{14}} = c o s α \end{cases} \Rightarrow

\Rightarrow {\begin{cases} x = \sqrt{14} s i n α + 2 \\ y = \sqrt{14} c o s α + 3 \end{cases} \Rightarrow x^{2} + y^{2} = 14 +

+ 4 \sqrt{14} s i n α + 6 \sqrt{14} c o s α + 4 + 9 =

= [u s i n g t h i s i n e q u a l i t y : a s i n α + b c o s α ⩾

- \sqrt{a^{2} + b^{2}}] = 27 + [4 \sqrt{14} s i n α + 6 \sqrt{14} c o s α] ⩾

⩾ 27 + \sqrt{16 \cdot 14 + 36 \cdot 14} = 27 - \sqrt{52 \cdot 14} = 27 - \sqrt{728}

Commented by bobhans last updated on 07/Jun/20

sorry sir . it wrong

Commented by bobhans last updated on 07/Jun/20

it should be {\begin{cases} x = 2 + \sqrt{14} \cos α \\ y = 3 + \sqrt{14} \sin α \end{cases}

then x + y = 5 + \sqrt{28} \cos (α - 45^{o});

so \min {x + y} = 5 - \sqrt{28} .

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