Divide 10 into two parts so that twice the square of the first part plus thrice the square of the other part is the least.


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Question Number 6179 by 314159 last updated on 17/Jun/16

$D i v i d e 10 i n t o t w o p a r t s s o t h a t t w i c e t h e$ $s q u a r e o f t h e f i r s t p a r t p l u s t h r i c e t h e$ $s q u a r e o f t h e o t h e r p a r t i s t h e l e a s t .$
Commented by prakash jain last updated on 17/Jun/16
$one part x, second part (10−x) Minimize 2x^2 +3(10−x)^2 =2x^2 +3(100−2x+x^2 ) =5x^2 −6x+300 =5(x^2 −(6/5)x+60) =5(x^2 −(6/5)x+(9/(25))+60−(9/(25))) =5(x^2 −2∙((3/5))x+((3/5))^2 +((1491)/(25))) =5{(x−(3/5))^2 +((1491)/(25))} minimum value when x=(3/5) since (x−(3/5))^2 ≥0$
$o n e p a r t x, s e c o n d p a r t (10 - x)$ $Minimize$ $2 x^{2} + 3 {(10 - x)}^{2}$ $= 2 x^{2} + 3 (100 - 2 x + x^{2})$ $= 5 x^{2} - 6 x + 300$ $= 5 (x^{2} - \frac{6}{5} x + 60)$ $= 5 (x^{2} - \frac{6}{5} x + \frac{9}{25} + 60 - \frac{9}{25})$ $= 5 (x^{2} - 2 \cdot (\frac{3}{5}) x + {(\frac{3}{5})}^{2} + \frac{1491}{25})$ $= 5 {{(x - \frac{3}{5})}^{2} + \frac{1491}{25}}$ $m i n i m u m v a l u e w h e n x = \frac{3}{5}$ $s i n c e {(x - \frac{3}{5})}^{2} ⩾ 0$
Commented by Rasheed Soomro last updated on 18/Jun/16

$N i c e!!!$
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