Welcome to our community

Be a part of something great, join today!

Swag's question at Yahoo! Answers regarding maximizing a quadratic function

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Here is the question:

Problem Solving - Math?

The number of miles M that a certain automobile can travel on one gallon of gas at a speed of v mph is given by the following formula:
M=-(1/30)v^(2)+(5/2)v for 0 < v < 70

a. For the consumption of one gallon of gasoline, find the most economical speed.

b. Find the maximum number of miles that can be driven at the most economical speed.

I've always been bad at beginning these type of problems. I don't really need an answer unless it isn't too much trouble because I'd have something to check my work with. I'm just wondering how to go about beginning the processes of both of these problems.
Here is a link to the question:

Problem Solving - Math? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
Last edited:
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Swag's question at Yahoo! Ansers regarding maximizing a quadratic function

Hello Swag,

We are given a function $M(v)$ that tells us the number of miles that can be traveled at a speed of $v$, where $0<v<70$ (I assume $v$ is in mph.)

This function is:

$\displaystyle M(v)=-\frac{1}{30}v^2+\frac{5}{2}v$

a.) We may observe that this is a quadratic function, whose graph opens downward. The maximum value of such a quadratic occurs on the axis of symmetry, which will be midway between its roots. Factoring, we find:

$\displaystyle M(v)=\frac{1}{30}\left(v(75-v) \right)$

Now, it is easy to see the roots are:

$v=0,\,75$

and so the axis of symmetry is:

$\displaystyle v=\frac{75}{2}=37.5$

Thus, the most economical speed, that is the speed which maximized the distance that can be traveled, is 37.5 mph.

b.) To find this maximum number of miles, we simply need to evaluate:

$\displaystyle M(37.5)=\frac{1}{30}\left(37.5(75-37.5) \right)=\frac{75^2}{2^2\cdot30}=\frac{375}{8}=46.875$

Hence, the maximum number of miles that may be driven is 46.875.