Question 998202
Let's say Anne runs at 5 mph down the road,
until she is {{{x}}} miles before the point directly in front of the helicopter.
Then, she runs across the field in a straight line towards the helicopter.
The road, field, helicopter, and Anne's path look like this: {{{drawing(400,300,-0.1,2.1,-1.2,0.25,
line(-0.2,0.01,2.5,0.01),triangle(2,-1,1.4,0,2,0),
green(line(-0.2,-0.01,2.5,-0.01)),
red(line(0,0,1.4,0)),red(line(2,-1,1.4,0)),
red(circle(0,0,0.02)),red(circle(2,-1,0.02)),
locate(0.4,0.2,road),locate(0.6,-0.6,green(field)),
locate(-0.08,-0.03,red(start)),
locate(1.7,-1.05,red(helicopter)),
locate(1.7,0,x),locate(2.01,-0.45,1),
locate(0.7,-0.01,red(2-x)),
locate(1.4,-0.45,red(sqrt(x^2+1)))
)}}}
So Anne runs on the road for {{{2-x}}} miles at {{{5}}} mph in {{{(2-x)/5}}} hours.
Then, she runs across the field for {{{sqrt(x^2+1)}}} miles at {{{3}}} mph in {{{sqrt(x^2+1)/3}}} hours.
Anne's total time as a function of {{{x}}} is
{{{t(x)=(2-x)/5+sqrt(x^2+1)/3=(6-3x+5sqrt(x^2+1))/15}}} hours.


If Anne wants to reach the helicopter in exactly 42 minutes (0.7 hour), {{{x}}} must be such that
{{{(6-3x+5sqrt(x^2+1))/15=0.7}}}
{{{6-3x+5sqrt(x^2+1)=15*0.7}}}
{{{6-3x+5sqrt(x^2+1)=10.5}}}
{{{5sqrt(x^2+1)=10.5-6+3x}}}
{{{5sqrt(x^2+1)=4.5+3x}}}
{{{25(x^2+1)=(4.5+3x)^2}}}
{{{25x^2+25=20.25+27x+9x^2}}}
{{{25x^2-9x^2-27x+25-20.25=0}}}
{{{16x^2-27x+4.75=0}}}
{{{x = (-(-27) +- sqrt((-27)^2-4*16*4.75 ))/(2*16) }}}
{{{x = (27 +- sqrt(729-304))/32}}}
{{{x = (27 +- sqrt(425))/32}}}
The two solutions are:
{{{x[1]=about0.20}}} (rounded), meaning Anne runs
{{{2-0.2=highlight(1.8)}}} miles down the road, and then leaves the road, and
{{{x[2]=about1.49}}} (rounded), meaning Anne runs
{{{2-1.49=highlight(0.51)}}} miles down the road, and then leaves the road.




If Anne wants to reach the helicopter as soon as possible, we must chose {{{x}}} to make {{{t(x)}}} minimum.
I cannot think of another way other than resorting to calculus,
unless you are expected to use a graphing calculator (or a spreadsheet) to get an approximate answer.
The derivative of {{{t(x)=(2-x)/5+sqrt(x^2+1)/3=(1/5)(2-x)+(1/3)(x^2+1)^("1/2")}}} is
{{{dt/dx=(1/5)(-1)+(1/3)(1/2)(2x)(x^2+1)^("-1/2")=-1/5+(1/3)x(x^2+1)^("-1/2")}}}
{{{dt/dx=-1/5+x/3sqrt(x^2+1)=(-3sqrt(x^2+1)+5x)/15sqrt(x^2+1)}}}
When {{{t(x)}}} is minimum, {{{dt/dx)=0}}} , meaning that
{{{-3sqrt(x^2+1)+5x=0}}}-->{{{5x=3sqrt(x^2+1)}}}-->{{{25x^2=9(x^2+1)}}}-->{{{25x^2=9x^2+9}}}-->{{{25x^2-9x^2=9}}}-->{{{16x^2=9}}}-->{{{x^2=9/16}}}-->{{{x=3/4=0.75}}}
So, if Anne wants to reach the helicopter as soon as possible,
she should leave the road after running (((2-0.75=highlight(1.25)}}} miles.