What speeds will result in gas mileage of at least 45 for a compact car?

[Bitter]

Here is the problem I am having trouble with right now.

The number of miles M that a certain compact car can travel on 1 gallon of gasoline is related to its speed v (in mi/hr) by

M=-1/30v^2+5/2v for 0<v<70

For what sppeds will M be at least 45?

I start working the problem like this:

45=-1/30v^2 +5/2v

I then put the problem into a polynomial

-1/30v^2 + 5/2v -45

I'm fairly certain you can factor that but if not I use the quadratic.

-5/2 +- Sqr of .25 all over -1/15

That comes out to be 75/2 plus or .5

Which is equal to 38 and 37

Now three things are possible. a) I am heading in the wrong direction or b)I did something wrong or c)all the above

 

[0rthodontist]

It is not +/- .5--you divide everything by -1/15, not just the -5/2.

Anyway, once you've fixed that you'll get the same kind of result (two solutions). It means that the car's mileage will be exactly equal to 45 mpg (miles per gallon) at those two speeds. Call the speeds A and B. Now, you have divided up the speeds into three parts:
1. Those speeds less than A
2. Those speeds greater than A but smaller than B
3. Those speeds greater than B

Now, if the car has a mpg higher than 45 at some point, since the mpg function is continuous, the only way its mpg could drop below 45 by switching speeds is if at some point it exactly equalled 45. So if your car has mpg greater than 45 at one speed in 2., it has mpg greater than 45 at every point in 2. If your car has mpg less than 45 at one speed in 1., it has mpg less than 45 at every point in 1. So all you have to do is test each of 1., 2., and 3. at a single point to see if they have better or worse mileage than 45 mpg. (before you do that, based on common sense, which speed interval would you think has better mileage?)

 

[Bitter]

Thank you very much for catching my error.

 

[Bitter]

So basically, the values are numbers between 30 and 45?

 

[0rthodontist]

Yes, that's the answer.