Thread:

1. Applications of Quadratic Equations

A square piece of cardboard is formed into a box by cutting 10-centimeter squares from each of the four corners and then folding up the sides, as shown in the figure. If the volume of the box is to be 49,000cm^3, what size square piece of cardboard is needed?

I did the first two but am not sure how to go about doing this one.

2. Originally Posted by Mad
A square piece of cardboard is formed into a box by cutting 10-centimeter squares from each of the four corners and then folding up the sides, as shown in the figure. If the volume of the box is to be 49,000cm^3, what size square piece of cardboard is needed?

I did the first two but am not sure how to go about doing this one.
Let $x$ be the side length of the original cardboard square. if we cut 10 cm squares from the corners (DRAW A DIAGRAM!) and fold it into a box, then we would have:

length = width = $x - 20$

and height = $10$

use these in the volume equation to find $x$

3. Originally Posted by Mad
A square piece of cardboard is formed into a box by cutting 10-centimeter squares from each of the four corners and then folding up the sides, as shown in the figure. If the volume of the box is to be 49,000cm^3, what size square piece of cardboard is needed?

I did the first two but am not sure how to go about doing this one.

Lets call the length of the paper L (just for fun )

since we are cutting 10cm squares from each corner the length of base will be $L-20$ (we are cutting two corners from each side) so the volume will be

$V=(L-20)(L-20)10$ and we know that the volume is 49,000

so we need to solve the equation

$49,000=10(L-20)^2 \iff 4,900 = (L-20)^2$

taking the square root we get

$\pm 70 =L-20 \iff L=20 \pm 70$

Since we are talking about a length we will only use the positive solution so we get...

$L=90$

4. Ha Ha you beat me this time.

There are too many cooks in the kitchen

5. Thanks guys.

An 18-wheeler left a grain depot with a load of wheat and traveled 550 mi to deliver the wheat before returning to the depot. Because of the lighter load on the return trip, the average speed of the truck returning was 5 mph faster than its average speed going. Find the rate returning if the entire trip, not counting unloading time or rest stops, was 21 h.

I don't know why I'm so bad at these.

6. Originally Posted by TheEmptySet
Ha Ha you beat me this time.

There are too many cooks in the kitchen
hehe, yeah. you're right. new plan: i'll be the head chef and order everyone around

7. Originally Posted by Mad
Thanks guys.

An 18-wheeler left a grain depot with a load of wheat and traveled 550 mi to deliver the wheat before returning to the depot. Because of the lighter load on the return trip, the average speed of the truck returning was 5 mph faster than its average speed going. Find the rate returning if the entire trip, not counting unloading time or rest stops, was 21 h.

I don't know why I'm so bad at these.
Please post new questions in a new thread.

so we know that $d=rt \iff t=\frac{d}{r}$

we know that his rate was "r" on the way there and "r+5" on the return trip
and that the total travel time was 21 hours.

so the time there from the above equation is

$t=\frac{550}{r}$

and the time on the way back is

$t=\frac{550}{r+5}$

so if we add the above times they should equal the total time. so we get

$\frac{550}{r}+\frac{550}{r+5}=21$

P.S the answer should work out to be r =50.

Good luck

8. Oh, sorry, I was just trying to save space rather than start a new thread every time I need help. We go through quite a bit in a day and the professor never seems to cover the more complicated stuff, she always does the problems that I already know how to do so I end up here and didn't want to clog up the boards.

Thankfully, others have complained about this, as well, and so she's said that she'll try to do harder examples so, hopefully, I won't have to ask for help as often.

And thank you for your patience, I can never seem to figure out what type of problem I should be setting up.
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