Word problems (open-top box and radius of a pipe)

[foahchon]

Hi, I'm having some problems with word problems (again). The first one:

1) Imogene wants to make an open-top box for packing baked goods by cutting equal squares from each corner of an 11-in by 14-in piece of cardboard as shown in the diagram. She figures that for versatility the area of the bottom must be 80in squared. What size square should she cut from each corner?

I realize this is a pretty common question, with a twist, but I can't get it to work out. Here's what I've got so far:

\(\displaystyle x(14 - 2x)(11 - 2x) = 0\)
\(\displaystyle 4x^3 - 50x^2 - 154x = 0\)
\(\displaystyle 4x^2 - 50x - 154 = 0\)

x should equal 1.72 but I can't see how. I know the 80 is supposed to go in there somewhere, but I don't know where.

The second one is this:

2) A small pipe is placed against a wall, but no block is used to keep it in place. There is a point on the edge of this pipe that is both 5in and 10in from the wall. Find two possibilities for the radius of the pipe.

I have no idea where to even start with this one. Any help would be appreciated, thanks.

 

[arthur ohlsten]

The bottom is not squared, or you can't get the desired solution
the bottom is [11-2x][14-2x]=80 where x is edge of cut out
154-50x+4x^2=80
2x^2-25x+37=0
x=25+/-[625-296]^1/2 all over 4
x=6.25+/-[329/16]^1/2
x=6.25-4.53
x=1.72 answer

if the bottom is a square then s^2=80
s=sqrt80
x= 1/2[11-s] and
x=1/2 [14-s]

x=1 from 1st equation
x=2.5 from 2nd equation

the box would have unequal sides

Arthur==============================================

I don't understand the 2nd problem. Is there other data?

 

[galactus]

I am taking a leap here, but maybe this is something like what they're getting at.

Some point on the pipe that is 5 and 10 inches from the wall.

 

[foahchon]

Hi, thanks a lot for the help, I have the first answer for the first (open-top box) question. I knew I was close. As for the second problem, though, I've asked a tutor about it, and he also had problems with it. My teacher basically gave us the equation (x - r)^2 + (y - r)^2 = r^2 in class, which produces a quadratic equation. I should also mention that my teacher wanted us to treat the ground as the x-axis, and the wall as the y-axis. Here is the quadratic equation I got:

(10 - r)(10 - r) + (5 - r)(5 - r) = r^2
100 - 10r -10 r + r^2 + 25 - 5r - 5r + r^2 = r^2
r^2 - 30r + 125 = 0

The possible answers are 25 and 5.

 

[Denis]

NO comments...but I'd sure like to YELL at your teacher...