Results 1 to 7 of 7

Math Help - Quadratic function word problems

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    71

    Quadratic function word problems

    Hey guys I need can someone me help get started on these two word problems.

    An archway over a road is modeled by the quadratic function y < -.1x^2+12.
    Can a car 6 feet wide and 7 feet high fit under the arch without crossing the median line?

    You have 80 feet of fence to enclose a rectangular garden, The function a = 40x-x^2 gives you the area of the garden in sqft where x is the width. What width gives you the maximum gardening area? what is the maximum area?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
    Quote Originally Posted by Jubbly View Post
    You have 80 feet of fence to enclose a rectangular garden, The function a = 40x-x^2 gives you the area of the garden in sqft where x is the width. What width gives you the maximum gardening area? what is the maximum area?

    Thanks.
    This is a parabola, find it's turning point (half way between the x-intercepts) and you will have the max width.

    Post your workings...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    58
    Quote Originally Posted by Jubbly View Post
    Hey guys I need can someone me help get started on these two word problems.

    An archway over a road is modeled by the quadratic function y < -.1x^2+12.
    Can a car 6 feet wide and 7 feet high fit under the arch without crossing the median line?
    Did you mean y = and not y < perhaps? Hard to say for sure otherwise.

    Ok, if it can't cross the median, then the road must be 12 feet wide at a height of 7 feet or more. Or put another way, 6 feet from the centerline (y axis), the curve must be at a height of y > 7 feet. This is easy by simply graphing it, but not much harder to work out without graphing it.

    So, how high is the curve at x = 6?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2008
    Posts
    71
    Quote Originally Posted by pickslides View Post
    This is a parabola, find it's turning point (half way between the x-intercepts) and you will have the max width.

    Post your workings...
    Uhm, I don't think I'm doing it right, but i'm doing the completing the square method but I'm getting x = 0. By turning point do you mean the vertex?

    Nevermind. I think I got it. Would x = 20?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2009
    Posts
    58
    Quote Originally Posted by Jubbly View Post
    Uhm, I don't think I'm doing it right, but i'm doing the completing the square method but I'm getting x = 0. By turning point do you mean the vertex?
    That's indeed what he meant. The vertex is right between the two intercepts. Completing the square is overkill in the extreme. What's the GCF (greatest common factor) of that equation? x=0 is indeed one solution, but there's another important one.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
    Quote Originally Posted by Jubbly View Post
    Uhm, I don't think I'm doing it right, but i'm doing the completing the square method but I'm getting x = 0. By turning point do you mean the vertex?

    Nevermind. I think I got it. Would x = 20?
    Yep.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member RHandford's Avatar
    Joined
    Sep 2010
    From
    N.E. Lincolnshire
    Posts
    70
    Hi Jubbly

    Let me see if I can help you with the second one:

    We have area= 40x-x^2 and we can also deduce as we have 80 feet of fence, so 2x+2y=80. Now we need to turn this into one equation so:

    \\2x+2y=80\\<br />
2y=80-2x\\<br />
y=40-x

    Now solve the two together:

    \\40x-x^2=40-x<br />
\\-x^2+41x-40=0

    Now find the vertex, I used \\-\frac{b}{2a}<br />
which gives 20.5. Now substitute 20.5 into the quadratic and solve for y.

    I get the maximum area to be 380.25 feet^2.

    I hope this helps and that I got it correct - I am sure someone will correct me if I am wrong!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quadratic function word problems
    Posted in the Algebra Forum
    Replies: 5
    Last Post: December 5th 2010, 03:26 AM
  2. help with quadratic word problems?
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: December 9th 2009, 06:01 PM
  3. Quadratic Word Problems
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 29th 2009, 03:27 PM
  4. Quadratic Word problems ?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 29th 2008, 07:40 PM
  5. Quadratic word problems
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 23rd 2007, 11:09 AM

Search Tags


/mathhelpforum @mathhelpforum