Pre-Calc Max/Min problem: energy of fish swimming

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mk42

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A fish swims at a speed v relative to the water, against a current of 5 mi/h. Using a mathematical model of energy expenditure, it can be shown that the total energy E required to swim a distance of 10 mi is given by

E(v)=2.73v^3 (10/v-5)

Biologists believe that migrating fish try to minimize the total energy required to swim a fixed distance. Find the value of v that minimizes the energy required.

Please help--I don't know how to use the -b/2a answer on this, and the graph *has* no minimum.
 
... and the graph *has* no minimum.
Click to expand...

oh no?

If your energy function is \(\displaystyle \L E(v) = 2.73v^3 \left(\frac{10}{v-5}\right)\)

it has a minimum when v is real close to 7.5
 
actually i went back and messed with my calculator, and the minimum is nowhere near 7.5. It's actually at 4.999998, where the graph drops to about 50 thou.
why 7.5?
 
Using the function for E(v) in my previous post (you haven't confirmed or denied whether it is correct), E(v) is not defined at v = 5.

If a fish is swimming against a 5 mph current, doesn't it make sense that it would have to swim at a speed > 5 mph to get anywhere?

Once again I ask you, is this correct for E(v) ...

\(\displaystyle \L E(v) = 2.73v^3 \left(\frac{10}{v-5}\right)\)

:?:

If so, the minimum energy is at v approx 7.5 mph.
 
yes, that is the correct formula, however in the graph on my calculator there is no value at all for 7.5 The graph is simply not there.
 
what is your window setting for your y-values?

fyi, E(7.5) is approx 4607 ... units might be in calories.
 
my window is x: 0 to 8 and y:-67000 to 0.

On my graph, around 4 the curve takes an enormous dip down to just past 50,000, and then it rises up almost completely vertically to intersect with five, and then disappears.
 
You are having Domain problems and graphing problems.

First the graphing. The fact that you wondered about the value at v = 5 showing up around 50 indicates you are not aware of the asymptote at v = 5. Take a look at your function. There is an asymptote at v = 5. You should not have a value there. You may have to define more points - consider the granularity of the graph. It's unlikely to be continuous the way your calculator draws it.

Now the Domain. You have the right idea. A value greater than 5 is REQUIRED. So why are you looking at v < 5? Define your Domain only on v > 5 and see if you are any happier.
 
If this is an asymptotic graph, then is there really a minimum? Won't the curve continue to decrease infintesimally?
 
You forgot the Domain issues. If the function is defined for v > 5, there is no asymptote in the Domain. That would be v = 5. Trust me on this. Zoom on in at v = [7,8]. You will see it.
 
in that window, all I can see is a straight horizontal line above the x-axis.
 
Well obviously my calculator is posessed and is having fun with me. The graph was asymptotic with two curves that appproached the asymptotes, but now its parabolic on the other side of the asymptote. I don't know what changed. But thank you for your help.
 
You don't have to be condescending about it. My calculator is very old and sometimes the computer chip doesn't work.

I appreciate your help, but you really did make me feel bad.
 
I apologize.

so, go buy a new calculator. TI-83+ is cheap and versatile. Sells for about $65.

or ... seek out and download a graphing program.

Don't forget what tkh said about Domain.
 
mk42 said:
its parabolic on the other side of the asymptote. I don't know what changed. But thank you for your help.
Click to expand...
Whoops! Now you are trusting your eyes a little too much. You THINK it is "parabolic". Maybe it is and maybe it isn't. "Isn't" being far more likely in the grand scheme of things, but you may have a rational claim in this case. Maybe I'd be okay with "a kinda-parabolic-ish-looking sort of thing".
 
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