Question 319256
The quickest time the car can travel is going the fastest speed.
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You didn't mention it but are you constrained to only use 1 tank of fuel?
Rate*Time=Distance
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40 kph Range=400 km
80 kph Range=200 km
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First equation, distance traveled at times T1 and T2
{{{40*T1+80*T2=240}}}
{{{T1+2T2=6}}}
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Now let's look at efficiency.
We need an equation that relates the fuel usage (liters/hr) to speed (km/hr) and the efficiency (km/liter).
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Let's say the tank holds Z liters.
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Let's define the fuel rate as Q,
(400 km/Z liter)*Q1(liters/hr)=40 km/hr
{{{Q1=(40Z)/400=0.4*Z}}} liters/hr
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(100 km/Z liters)*Q2(liters/hr)=80 km/hr
{{{Q2=(80Z)/100=0.8*Z}}} liters/hr
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{{{Q1*T1+Q2*T2=Z}}}
{{{0.1*ZT1+0.8*ZT2=Z}}}
{{{0.1T1+0.8T2=1}}}
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So now you have a linear programming problem.
You have two constraint equations.
The solution lies on the line,
{{{T1+8*T2=6}}}
{{{drawing(300,300,-2,8,-2,4,grid(1),graph(300,300,-2,8,-2,4,(240-40x)/80))}}}

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The feasible region vertices to minimize the function, {{{T1+T2}}} occur at 
{{{T1=6}}}, {{{T2=0}}}
{{{T1=0}}}, {{{T2=3}}}
and the intersection point of the two lines.
{{{0.1T1+0.8T2=1}}}
1.{{{T1+8T2=8}}}
2.{{{T1+2T2=6}}}
From eq. 1,
{{{T1=8-8T2}}}
Substitute into eq. 2,
{{{8-8T2+2T2=6}}}
{{{-6T2=-2}}}
{{{highlight(T2=1/3)}}}
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{{{T1=8-8/3}}}
{{{highlight(T1=16/3)}}}
{{{drawing(300,300,-2,8,-2,4,grid(1),circle(16/3,1/3,.2),
circle(0,3,.2),circle(6,0,.2),graph(300,300,-2,8,-2,4,(240-40x)/80,(8-x)/8))}}}
Check the function at each point:


{{{T1=6}}}, {{{T2=0}}}:{{{T1+T2=6}}}
{{{T1=0}}}, {{{T2=3}}}:{{{T1+T2=3}}}
{{{T1=16/3}}},{{{T2=1/3}}}:{{{T1+T2=17/3}}}
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The fastest time occurs when {{{T1=0}}} and {{{T2=3}}}.