PREAMBLE
First, we work out some basic equations.
Given:
Cost of fuel: $p/gal
Consumption: g mpg
Driver hire: $w/hr.
Suppose the distance to be traveled is L miles. At constant v mph it will take t=L/v hours, so the cost of hiring the driver is $wt=$(Lw/v).
The fuel consumption for L miles is L/g gallons, and the cost of fuel is $(Lp/g).
Total cost=Lw/v+Lp/g dollars. If c=(total cost)/L is the overall cost per mile, then:
c=w/v+p/g.
(1) g(v)=v(85-v)/61.2 mpg.
g(0)=0 mpg. The fuel consumption cannot realistically be measured for a stationary vehicle.
g(42)=42×43/61.2=29.51 mpg. At 42mph, it would cost about $(p/29.51) per mile in fuel, or $0.034p per mile or about 3.4p¢ per mile.
The driver hire would be $(w/42), or about 2.38w¢, per mile.
g(61)=61×24/61.2=23.92 mpg. At 61mph, it would cost about $(p/23.92) per mile in fuel (about 4.2p¢ per mile), and $(w/61) per mile (about 1.64w¢) in driver hire.
(2) Fuel mileage is maximum when g'(v)=0=(85-2v)/61.2, v=42.5mph. g(42.5)=29.51 mpg. This is approximately the same value as in part (1) for g(42). We know this is a maximum because g(61) is lower than this. Also, g(42)=29.5098, whereas g(42.5)=29.5138, so g(42.5)>g(42).
(3) c=w/v+p/g dollars per mile. Therefore for L miles the cost is L(w/v+p/g). See preamble for the derivation of this formula. C(v)=L(w/v+61.2p/(v(85-v))).
(4) p=$3.96, w=$20.5, L=405 miles.
Although the maximum fuel mileage is at 42.5mph the much higher cost of driver hire offsets this economy. We are interested in the minimum speed taking into account the mileage and the driver hire.
c(v)=w/v+p/g dollars per mile.
dc/dv=-w/v²-p/g²(dg/dv); dg/dv=g'(v)=(85-2v)/61.2.
∴dc/dv=-w/v²-(p/g²)((85-2v)/61.2).
g²=v²(85-v)²/61.2², so p/g²=61.2²p/(v²(85-v)²).
dc/dv=0 at max/min points.
dc/dv=-w/v²-61.2²p/(v²(85-v)²)((85-2v)/61.2).
dc/dv=-w/v²-(85-2v)61.2p/(v²(85-v)²).
dc/dv=-(w(85-v)²+(85×61.2-2×61.2v)p)/(v²(85-v)²).
When dc/dv=0=(85-v)²+(85×61.2-2×61.2v)(p/w).
This gives us a quadratic in v:
85²-2×85v+v²+85×61.2p/w-2×61.2pv/w=0,
v²-2(85+61.2p/w)v=-85×61.2p/w-85².
Let r=p/w:
v²-2(85+61.2r)v=-85×61.2r-85².
Completing the square:
v²-2(85+2×61.2r)v+(85+61.2r)²=(85+61.2r)²-85×61.2r-85²,
(v-(85+61.2r))²=(85+61.2r)²-85×61.2r-85²,
(v-(85+61.2r))²=85²+2×85×61.2r+61.2²r²-85×61.2r-85²,
(v-(85+61.2r))²=85×61.2r+61.2²r²,
v=85+61.2r±√(61.2²r²+85×61.2r),
v=85+61.2r±61.2r√(1+(85×61.2r)/(61.2²r²)),
v=85+61.2r±61.2r√(1+85/(61.2r)).
v must be less than 85, so the negative square root applies:
v=85+61.2r-61.2r√(1+85/(61.2r)).
r=p/w=3.96/20.5=0.1932 approx.
v=85+11.82-11.82√(1+7.19)=62.99 mph.
So optimal speed is about 63mph.
c(63)=20.5/63+3.96×61.2/(63×22) dollars per mile.
c(63)=$0.5/mile approx.
Cost=405×0.5=$202.5 approx=€167.
(5) Cost = 500×0.5=$250=€206 approx.
(6) r=3.96/15=0.264.
Plug this value into v=85+61.2r-61.2r√(1+85/(61.2r)),
v=60.73mph.
C(60.73)=405(15/60.73+61.2×3.96/(60.73(85-60.73))).
Cost=$166.63 (€137.4).