# Quadratic function word problems

Printable View

• Nov 30th 2010, 04:11 PM
Jubbly
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.
• Nov 30th 2010, 04:55 PM
pickslides
Quote:

Originally Posted by Jubbly
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...
• Nov 30th 2010, 05:06 PM
Grep
Quote:

Originally Posted by Jubbly
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?
• Nov 30th 2010, 05:49 PM
Jubbly
Quote:

Originally Posted by pickslides
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?
• Nov 30th 2010, 05:57 PM
Grep
Quote:

Originally Posted by Jubbly
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.
• Nov 30th 2010, 06:06 PM
pickslides
Quote:

Originally Posted by Jubbly
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.
• Dec 1st 2010, 10:52 AM
RHandford
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:

$\\2x+2y=80\\
2y=80-2x\\
y=40-x$

Now solve the two together:

$\\40x-x^2=40-x
\\-x^2+41x-40=0$

Now find the vertex, I used $\\-\frac{b}{2a}
$
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!