Determine-the-smallest-value-of-the-form-f-u-v-5v-2-5u-2-1-2u-v-where-u-v-R-

Question Number 514 by 112358 last updated on 25/Jan/15

D e t e r m i n e t h e s m a l l e s t v a l u e o f

t h e f o r m

f (u, v) = \frac{5 v^{2} + 5 u^{2} + 1}{2 u + v}

w h e r e u, v \in R^{+} .

Answered by prakash jain last updated on 23/Jan/15

\frac{\partial f}{\partial u} = 0 \Rightarrow (2 u + v) (10 u) - (5 u^{2} + 5 v^{2} + 1) \cdot 2 = 0

\Rightarrow 20 u^{2} + 10 u v - 10 u^{2} - 10 v^{2} - 2 = 0

\Rightarrow 5 u^{2} + 5 u v - 5 v^{2} - 1 = 0 \dots . (i)

\frac{\partial f}{\partial v} = 0 \Rightarrow (2 u + v) (10 v) - (5 u^{2} + 5 v^{2} + 1) = 0

\Rightarrow - 5 u^{2} + 20 u v + 5 v^{2} - 1 = 0 \dots (ii)

Adding (i) and (ii)

25 u v = 2

5 u^{2} + \frac{10}{25} - 5 {(\frac{2}{25 u})}^{2} - 1 = 0

5 u^{2} - \frac{4}{125 u^{2}} - \frac{3}{5} = 0

625 u^{4} - 75 u^{2} - 4 = 0

u^{2} = \frac{75 \pm \sqrt{{(75)}^{2} + 16 (625)}}{2 \times 625} = \frac{3 \pm \sqrt{3^{2} + 16}}{2 \times 25} = \frac{8}{50} = \frac{4}{25}

u = \frac{2}{5}, v = \frac{1}{5}

f (u, v) = 2

S o m e a d d i t i o n a l t e s t a r e n e e d e d w i t h

d o u b l e d e r i v a t i v e s t o c o n f i r m t h a t (\frac{2}{5}, \frac{1}{5})

i s a m i n i m a .