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-

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

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!!!