calculus
posted by Jim .
Amy wants to reach, as quickly as possible, a destination of 5km downstream and on the opposite side of the river that is 1.5km wide. If she can run at 16kmp and can swim at 5kmp, how far should she run down the shore before turning into the water to swim? Consider the river current to be negligible.

draw the figure.
she runs d1 down stream. that leaves 5d1 left. The distance from that point to the end point is then
sqrt((5d1)^2+1.5^2)= sqrt(xxxx)
time = d1/16 p + sqrt(xxxx )/5 p I don't know what the p is in kmp.
dtime/d1=0=1/16+1/2 ( xxxx )^1/2 *2(5d1)(1)
1/16=(5d1)/sqrt(xxxx)
multiply both sides by sqrt(xxxx)
then square both sides, you should get a quadratic and can solve d1.
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