To do normally his commercial activities in some place situated at 150km from him, a driver use a car. the consumption of gazoil in liters for 10 km is defined by: C(v)=((20)/v)+(v/(45)) where v is the speed of car. How should be the speed of car to reduce minimally the consumption of gazoil?


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Question Number 104046 by mathocean1 last updated on 19/Jul/20

$To do normally his commercial$ $activities in some place situated at 150 km$ $from him, a driver use a car . the consumption$ $of gazoil in liters for 10 km is defined by :$ $C (v) = \frac{20}{v} + \frac{v}{45} where v is the speed of car .$ $How should be the speed of car to reduce$ $minimally the consumption of gazoil ?$
Commented by mr W last updated on 19/Jul/20

$C (v) = \frac{20}{v} + \frac{v}{45} ⩾ 2 \sqrt{\frac{20}{v} \times \frac{v}{45}} = \frac{4}{3}$ $m i n i m u m i s a t \frac{20}{v} = \frac{v}{45}, i . e . v = \sqrt{20 \times 45}$ $= 30$ $h e s h o u l d d r i v e w i t h s p e e d 30 .$ ${d o n}^{'} t a s k m e 30 m p h o r 30 k m / h,$ $b e c a u s e y o u {d i d n}^{'} t g i v e t h e u n i t o f$ $v i n C (v) .$
Commented by mathocean1 last updated on 19/Jul/20

$P l e a s e s i r c a n d e t a i l o r e x p l a i n t h i s f i r s t l i n e : C (v) = \frac{20}{v} + \frac{v}{45} ⩾ 2 \sqrt{\frac{20}{v} \times \frac{v}{45}} = \frac{4}{3}$
Commented by mathocean1 last updated on 19/Jul/20

$m y s o r r y i s a b o u t t h i s :$ $C (v) = \frac{20}{v} + \frac{v}{45} ⩾ 2 \sqrt{\frac{20}{v} \times \frac{v}{45}} = \frac{4}{3}$ $i^{'} m n o t u n d e r s t a n d i n g . . .$
Commented by mr W last updated on 19/Jul/20

$f o r a n y p o s i t i v e n u m b e r s a a n d b,$ $w e h a v e a l w a y s$ $\frac{a + b}{2} ⩾ \sqrt{a b}$ $t h e e q u a l s i g n i s w h e n a = b .$
Commented by mr W last updated on 19/Jul/20

$y o u c a n a l s o d o l i k e t h i s :$ $C (v) = \frac{20}{v} + \frac{v}{45}$ $s u c h t h a t C (v) i s m i n i m u m,$ $\frac{d C}{d v} = - \frac{20}{v^{2}} + \frac{1}{45} = 0$ $\Rightarrow v = \sqrt{20 \times 45} = 30$
Commented by mathocean1 last updated on 19/Jul/20

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