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

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?

Commented bymr W last updated on 19/Jul/20

C(v)=((20)/v)+(v/(45))≥2(√(((20)/v)×(v/(45))))=(4/3)  minimum is at ((20)/v)=(v/(45)), i.e. v=(√(20×45))  =30  he should drive with speed 30.  don′t ask me 30 mph or 30 km/h,  because you didn′t give the unit of  v in C(v).

Commented bymathocean1 last updated on 19/Jul/20

Please sir can detail or explain this first line: C(v)=((20)/v)+(v/(45))≥2(√(((20)/v)×(v/(45))))=(4/3)

Commented bymathocean1 last updated on 19/Jul/20

my sorry is about this :  C(v)=((20)/v)+(v/(45))≥2(√(((20)/v)×(v/(45))))=(4/3)  i′m not understanding...

Commented bymr W last updated on 19/Jul/20

for any positive numbers a and b,  we have always  ((a+b)/2)≥(√(ab))  the equal sign is when a=b.

Commented bymr W last updated on 19/Jul/20

you can also do like this:  C(v)=((20)/v)+(v/(45))  such that C(v) is minimum,  (dC/dv)=−((20)/v^2 )+(1/(45))=0  ⇒v=(√(20×45))=30

Commented bymathocean1 last updated on 19/Jul/20

thank you very much sir!