x-y-12-minimum-value-of-x-2-4-y-2-9-

Question Number 218153 by Rojarani last updated on 31/Mar/25

x + y = 12

m i n i m u m v a l u e o f

\sqrt{x^{2} + 4} + \sqrt{y^{2} + 9} = ?

Answered by mr W last updated on 31/Mar/25

Commented by Ghisom last updated on 31/Mar/25

nice!

Commented by mr W last updated on 31/Mar/25

A C = \sqrt{x^{2} + 2^{2}} = \sqrt{x^{2} + 4}

B C = \sqrt{y^{2} + 3^{2}} = \sqrt{y^{2} + 9}

A B = \sqrt{12^{2} + {(2 + 3)}^{2}} = 13

A C + B C ⩾ A B

\Rightarrow \sqrt{x^{2} + 4} + \sqrt{y^{2} + 9} ⩾ 13

i . e . {(\sqrt{x^{2} + 4} + \sqrt{y^{2} + 9})}_{m i n} = 13

Commented by Rojarani last updated on 31/Mar/25

S i r, t h a n k s .

Commented by mehdee7396 last updated on 01/Apr/25

v e r y g o o d \underset{―}{\underset{⏟}{⋚}}

Commented by mr W last updated on 01/Apr/25

t h a n k s t o a l l!

Answered by efronzo1 last updated on 31/Mar/25

Answered by vnm last updated on 31/Mar/25

x = 6 + t, y = 6 - t

\frac{d}{d t} (\sqrt{{(6 + t)}^{2} + 4} + \sqrt{{(6 - t)}^{2} + 9}) =

\frac{6 + t}{\sqrt{{(6 + t)}^{2} + 4}} + \frac{6 - t}{\sqrt{{(6 - t)}^{2} + 9}} =

\frac{(6 + t) \sqrt{{(6 - t)}^{2} + 9} + (6 - t) \sqrt{{(6 + t)}^{2} + 4}}{\sqrt{{(6 + t)}^{2} + 4} \sqrt{{(6 - t)}^{2} + 9}} = 0

(t + 6) \sqrt{t^{2} - 12 t + 45} = (t - 6) \sqrt{t^{2} + 12 t + 40}

(t^{2} + 12 t + 36) (t^{2} - 12 t + 45) =

(t^{2} - 12 t + 36) (t^{2} + 12 t + 40)

(t^{2} + 12 t) 45 + 36 (- 12 t + 45) =

(t^{2} - 12 t) 40 + 36 (12 t + 40)

5 t^{2} + 12 t \cdot 85 - 36 \cdot 24 t + 36 \cdot 5 = 0

5 t^{2} + 156 t + 180 = 0

t = \frac{- 78 \pm 72}{5}

t_{1} = - 30 i n t h i s p o i n t \frac{d}{d t} (x^{2} + 4) > 0 a n d \frac{d}{d t} (y^{2} + 9) > 0,

t h e r e f o r t h i s p o i n t {i s n}^{'} t t h e p o i n t o f m i n i m u m

t_{2} = - \frac{6}{5}

\sqrt{{(6 - \frac{6}{5})}^{2} + 4} + \sqrt{{(6 + \frac{6}{5})}^{2} + 9} = 13

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