# Thread: Calculus Word Problem - No Solution! ? :O

1. ## Calculus Word Problem - No Solution! ? :O

The question was on our exam today, and several of us are dumbfounded after it. We knew how to solve this question, and we went about our business, and it turned out badly. I took it home and compared it to my method before, and I had the correct method, the numbers are correct, but the numbers work out badly when we get to a certain point, because it gives a negative instead of a positive like before.

There is an island 60 meters from land. 150 meters down shore is an ice cream store. A man on the island wants to get ice cream. He can swim at 3m/s and run on land at 5m/s . TO what point down shore should he swim to minimize the time to get to his beloved ice cream? (Must be Ben and Jerry's?)

Thanks guys... We called the distance down shore, x, and the distance he ran, 150-x . And the distance swam in the lake is (x^2 + 60^2)^(1/2)

Let me know if oyu are going to answer this :P Wet my tastebuds. get my excited for something cause we're down and out about this.

2. Originally Posted by mike_302
The question was on our exam today, and several of us are dumbfounded after it. We knew how to solve this question, and we went about our business, and it turned out badly. I took it home and compared it to my method before, and I had the correct method, the numbers are correct, but the numbers work out badly when we get to a certain point, because it gives a negative instead of a positive like before.

There is an island 60 meters from land. 150 meters down shore is an ice cream store. A man on the island wants to get ice cream. He can swim at 3m/s and run on land at 5m/s . TO what point down shore should he swim to minimize the time to get to his beloved ice cream? (Must be Ben and Jerry's?)

Thanks guys... We called the distance down shore, x, and the distance he ran, 150-x . And the distance swam in the lake is (x^2 + 60^2)^(1/2)

Let me know if oyu are going to answer this :P Wet my tastebuds. get my excited for something cause we're down and out about this.
time = distance/speed

$t = \frac{\sqrt{x^2+60^2}}{3} + \frac{150-x}{5}$

$\frac{dt}{dx} = \frac{x}{3\sqrt{x^2+60^2}} - \frac{1}{5}$

set $\frac{dt}{dx} = 0$ ...

$\frac{x}{3\sqrt{x^2+60^2}} = \frac{1}{5}$

$5x = 3\sqrt{x^2+60^2}$

$25x^2 = 9(x^2 + 60^2)$

$16x^2 = 9 \cdot 60^2$

$x = \frac{3 \cdot 60}{4} = 45$ m

min time will be 46 sec

3. Sorry if I'm getting this wrong, but do you not just differentiate and equate to zero? If so I don't see why it's working out badly.

$\frac{d}{dx} \left( \frac{\sqrt{x^2 + 60^2}}{3} - \frac{x}{5} \right) = -1/5 + \frac{x}{3 \sqrt{3600 + x^2}} = 0$

$\frac{x^2}{3600 + x^2} = \frac{3}{5}$

$x^2 = 2025$ $\Rightarrow$ $x = 45$

Hope this helped. Is this university level math?

-pomp

4. so we did it for cost... We multiplied instead of dividing... darn!

5. uni level intro calc. it was that we multiplied instead of dividing

6. Fair enough, it was an easy mistake to make and sometimes an easy mistake right at the get go can ruin any chance of getting the right answer!

7. Haha, it was THE ONLY question on our exam we messed up and had trouble with... oh well.

Thanks!