1. ## optimization boat problem

A man launches his boat from point A on a bank of a straight river, 1 km wide, and wants to reach point B, 1 km downstream on the opposite bank, as quickly as possible. He could row his boat directly across the river to point C and then run to B, or he could row directly to B, or he could row to some point D between B and C and then run to B. If he can row 6 km/h and run 8 km/h, where should he land to reach B as soon as possible? (We assume that the speed of the water is negligible compared to the speed at which the man rows.)

db= $1-x$
ad= $\sqrt (x^2 +1)$

$\frac{\sqrt (x^2 +1)}{6}$= $\frac{1-x}{8}$

derivative:
$\frac{(x)}{6\sqrt (x^2 +1)}$= $\frac{1}{8}$

$8x$= $6\sqrt (x^2 +1)$

$64x^2=36(x^2+1)$

$64x^2=36x^2+36$

$28x^2=36$

$x=\sqrt(36/28)$=1.1 km
t(0)=1.5 and t(8)=1.2..., so this should be the min, but it was marked wrong.

2. Originally Posted by hazecraze
db= $1-x$
ad= $\sqrt (x^2 +1)$

$\frac{\sqrt (x^2 +1)}{6}$= $\frac{1-x}{8}$
Why set them euqal? Add them, surely, then find the minimum of the this function.

3. In a similar example in class, we set them equal to each other.

hmmm...after simplifying (getting a common denominator and squaring both sides) I got $\frac{100x^2 +100x}{2304}=0$

derivative: $200x=-100$

$
x=-\frac{1}{2}$

?

5. You need to ask yourself - what is it I need to set to zero, and why? Surely not the function that tells you the time taken. Make sure you set your working out line by line and then if something isn't making sense just post this working and we'll point you forward.

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### a man launches his boat from point a on a bank of a straight river 4 km wide

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