# World Problems

• Dec 5th 2007, 04:28 PM
FalconPUNCH!
World Problems
Consider the following Problem: A box with an open top is to be constructed from a square piece of cardboard, 3 ft. wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have.

(a) Draw several diagrams to illustrate the situation, some short boxes with large bases and some tall boxes with small bases. Find the volumes of several such boxes.

(b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with symbols.

(c) Write an expression for the volume

(d) Using the given info write an equation that relates the variables.

There are more parts after but I know how to do those parts. I'm not necessary good with word problems so I have a hard time understanding them. Please help as soon as possible.
• Dec 5th 2007, 05:17 PM
angel.white
• Dec 5th 2007, 05:20 PM
FalconPUNCH!
Quote:

Originally Posted by angel.white

I'll look into it when I come back from work. If anyone else wants to help me with this problem feel free to.
• Dec 6th 2007, 08:13 AM
FalconPUNCH!
Angel.White I read through the problem you gave me several times and I'm still having a hard time understanding it. :confused:
• Dec 6th 2007, 10:57 AM
angel.white
Quote:

Originally Posted by FalconPUNCH!
Angel.White I read through the problem you gave me several times and I'm still having a hard time understanding it. :confused:

Okay, the first thing I want you to do is find the formula for volume. I've drawn you a diagram to illustrate how the problem is set up. In the first illustration, you can see the dotted lines are the fold points, and in the second illustration they are starting to fold up to form a box. How would you determine the volume of this box?

Once you have that, consider what would happen if you increased the length of x, or if you decreased the length of x. How would that effect the box?

Once you've done that, you can do a)
And what I've done is basically b)
and finding the volume is c) so you'll have done that.

So that will be a,b,c (I don't understand how d is different from c, so we'll assume you've done d as well). Now you just have to find the maximum volume. So take the derivative of your equation for volume.

Once you've done all that, post your equation for the volume, and your equation for the derivative of the volume, and we'll see about finding the maximum volume.
• Dec 6th 2007, 12:25 PM
FalconPUNCH!
Thanks I have one more that I need help with

1. An Object with weight W is dragged along a horizontal plane by a force acting along a roope attached to the object. If the rope makes an angle T with a plane, then the magnitude of the force is:

$F = \frac{mW}{msin(T) + cos(T)}$

Where m is a constant called the coefficient of friction. For what value of T is F smallest?

Edit: I'll make a new thread for this since I just want to talk about the first problem here.
• Dec 6th 2007, 12:38 PM
FalconPUNCH!
The formula for the volume that I got for the first problem is

$V = (3-2x)^3$

Derivative I got is:

$V' = -6(3-2x)^2$
• Dec 6th 2007, 01:30 PM
angel.white
Quote:

Originally Posted by FalconPUNCH!
Thanks I have one more that I need help with

1. An Object with weight W is dragged along a horizontal plane by a force acting along a roope attached to the object. If the rope makes an angle T with a plane, then the magnitude of the force is:

$F = \frac{mW}{msin(T) + cos(T)}$

Edit: I just got done and realized this was looking for smallest. The same technique applies, read through the explanation, consider the examples, and then realize that you are looking for a minimum instead of a maximum. (valley instead of peak)

Where m is a constant called the coefficient of friction. For what value of T is F smallest?

This next question is very similar to the box one, you are simply trying to find where force is maximized. Now, you know that as t changes, F will change as well. So as t changes, force will either get higher or lower. The point where it is highest, it must be changing directions, meaning it was getting higher, but now it is getting lower (thus it is at a peak on the graph).

This means that going up to this point, it's slope will be increasing, and at this point, it's slope will be zero, and after this point, it's slope will be decreasing.

Now we know that the derivative of our function will give us the equation of the slope. So wherever the derivative is equal to zero, we know that the slope of our function must be equal to zero. So find the zeros of the derivative to find potential maximums of the function.

Now, we know that since we are dragging along a plane, t cannot be less than zero, or it would be going through the plane, and t cannot be greater than pi/2, or it would be dragging backwards. So our domain must be from 0 to pi/2.

List all of the zeros of the derivative that are within this domain, they are potential maximums. Then choose a test point before and after, and find whether the derivative is greater than zero or less than zero. If it is greater than zero, the slope is positive, so it must be increasing. If it is less than zero, the slope is negative, so it must be decreasing. This will allow you to identify whether your point is a maximum or minimum.

Once you find a value for t which causes the derivative to go to zero, plug it into F(t) and find the y value. Then, just to be sure, check the y values of our endpoints for the domain, so try F(0) and F(pi/2) to make sure that they are not greater. (They may not be zeros, so could escape detection by the method we are using, so we need to check them to be sure)

Then, any maximums you found by finding F'(t)=0 and testing points before and after it should have their y values compared to the y values of our endpoints, and the highest will be the value which is our maximum.

-----
Sample Example 1:
Find the maximum from -2 to 3 of the function:
$f(x) = -x^2$

Our derivative
$f'(x) = -2x$

Find the zeros of the derivative:
0=-2x
0=x

Test a point before
test point x=-1
f'(-1)=-2(-1)=2
This is positive, so the slope is increasing up to the point (0,f(0)).

Test a point after
test point x=1
f'(1)=-2(1)=-2
This is negative, so the slope is decreasing after the point (0, f(0))

So we have a maximum, because it is at a peak.

Test endpoints:
$f(-2)=-(-2)^2=-4$
$f(3)=-(3)^2=-9$
Compare to our maximum:
$f(3)=-(0)^2=0$

So x=0, is our maximum, and our maximum point is (0,0)

Sample Example 2:
Find the maximum from -2 to 3 of the function:
$f(x) = x^2$

Our derivative:
f(x)=2x

Find the zeros of the derivative:
0=2x
0=x

Test a point before
test point x=-1
f'(-1)=2(-1)=-2
This is negative, so the slope is decreasing down to the point (0,f(0)).

Test a point after
test point x=1
f'(1)=2(1)=2
This is positive, so the slope is increasing after the point (0, f(0))

So this point is not at a peak, it is at a valley, you go down then you go up, that is a valley, so this point is actually at a minimum, and therefore not what we are looking for. But there are no other values where x=0, so what do we do? Well we test our endpoints.

Test endpoints:
$f(-2)=(-2)^2=4$
$f(3)=(3)^2=9$

So 9 is the highest value within our function, since it starts at (-2,4) goes down to (0,0), and goes back up to (3,9)

So x=3, is our maximum, and our maximum point is (3,9)

To illustrate why this is, I've attached a graph. Notice the three critical numbers of f(-2)=4, f(0)=0, f(3)=9

The derivative process will find us f(0)=0 but it needs to be tested, because it could be a minimum (or neither a max nor min as in f(0)=0 for the graph x^3). And we still need to consider the endpoints of our domain, as you can see in this case, they are not zeros and will not be found by the derivative, but they need to be checked anyway.
• Dec 6th 2007, 01:35 PM
angel.white
Quote:

Originally Posted by FalconPUNCH!
The formula for the volume that I got for the first problem is

$V = (3-2x)^3$

Close, you got the length and width right, but consider again, what will be the height of your box?
Quote:

Originally Posted by FalconPUNCH!
Derivative I got is:

$V' = -6(3-2x)^2$

Don't forget to apply the chain rule, in this case f(x)=x^3 and g(x)=3-2x

so V = f(g(x))
and the derivative of V is
V'=f'[g(x)]*g'(x)
• Dec 6th 2007, 09:22 PM
FalconPUNCH!
Quote:

Originally Posted by angel.white
Close, you got the length and width right, but consider again, what will be the height of your box?

Don't forget to apply the chain rule, in this case f(x)=x^3 and g(x)=3-2x

so V = f(g(x))
and the derivative of V is
V'=f'[g(x)]*g'(x)

Well I think I did apply the chain rule

$3(3-2x)^2(-2)$

Which I just simplified to

$-6(3-2x)^2$

Quote:

Originally Posted by angel.white
Close, you got the length and width right, but consider again, what will be the height of your box?

Not too sure about the height.

As for the second one I understand how to find the minimum and maximum but I'm having trouble with that problem because I'm not that good at trig. I try to take the derivative but it gets really complicating and I get some long equation, so I can't really find the zeros for it.
• Dec 7th 2007, 03:27 AM
angel.white
Quote:

Originally Posted by FalconPUNCH!
Well I think I did apply the chain rule

$3(3-2x)^2(-2)$

Which I just simplified to

$-6(3-2x)^2$

You're right, I'm sorry, I compared it to the derivative of my equation.
Quote:

Originally Posted by FalconPUNCH!
Not too sure about the height.

Okay, so the length and width of the paper are 3, and you're subtracting 2x from them. This you seem to have figured out. Then we fold the flaps up to create the sides of the box. Since the flaps are x units long, the box will be x units high. so the formula for volume will be $x(3-2x)^2$

Look at the pictures I drew you, do you see how the dotted lines are where the flaps fold up? And you can see that they are all x units long.

Another way of thinking about this might be that the left flap takes up x units, the base takes up 3-2x units, and the right flap takes up x units. So when you fold the flaps up, the base will be left at 3-2x, and the left height will be x, and the right height will be x. Anyway, look at the problem I drew again, and maybe try cutting some squares out of a piece of paper to see it.
Quote:

Originally Posted by FalconPUNCH!
As for the second one I understand how to find the minimum and maximum but I'm having trouble with that problem because I'm not that good at trig. I try to take the derivative but it gets really complicating and I get some long equation, so I can't really find the zeros for it.

For this we use the division rule.
If f(T)=mW
and g(T)=m*sin(T)+cos(T)

then $F=\frac{f(T)}{g(T)}$

So the division rule says to differentiate, this should become:
$F\prime=\frac{g(T)*f\prime(T) - f(T)*g\prime(T)}{\left(g(T)\right)^2}$

So we find f'(T), mW is a constant so it differentiates to zero. This is because m is told to you to be a constant, and while the weight can change, it won't change while it is being dragged. You could drag different weights, but the weight will not fluctuate according to the angle that it is being dragged. So the derivative of mW is 0
f'(T)=0

g(T) has it's terms added together, so they can be differentiated separately.
First we differentiate m*sin(T) and get m*cos(T)
Then we differentiate cos(T) and get -sin(T)
So the derivative:
g'(T) = m*cos(T)-sin(T)

So we said the division equation is:
$F\prime=\frac{g(x)*f\prime(x) - f(x)g\prime(x)}{\left(g(x)\right)^2}$

And when we substitute our values in:
$F\prime=\frac{\left(m*sin(T)+cos(T)\right)*0 - mW*\left(mcos(T)-sin(T)\right)}{\left(msin(T)+cos(T)\right)^2}$

Now the zero will cancel that term out
$F\prime=\frac{-mW\left(mcos(T)-sin(T)\right)}{\left(msin(T)+cos(T)\right)^2}$

-----
Okay, now we have the derivative. It's a little scary looking, but it's okay. Now we need to find the zeros. Notice that $\frac{0}{x}=0$ where x is any number. This means that if you divide zero by anything, you will get zero. So that means that if the numerator is equal to zero, the entire equation is equal to zero. (note also that there is nothing the denominator can equal that will cause the equation to become zero)

So look at our equation
$F\prime=\frac{-mW\left(mcos(T)-sin(T)\right)}{\left(msin(T)+cos(T)\right)^2}$

Set the numerator equal to zero
$0=-mW\left(mcos(T)-sin(T)\right)$

-mW is a constant, so it can never equal zero, so it is irrelevant and can be divided out
$0=mcos(T)-sin(T)$

$sin(T)=mcos(T)$

And divide cos(T)
$\frac{sin(T)}{cos(T)}=m$

And convert to sin/cos to tan
$tan(T)=m$

Take the arctangent
$T=arctan(m)$

------------

This had me confused for a bit, I figured your book didn't give you all the information you needed, but I think what they are looking for is this answer, that when T equals the arctangent of m, it will be maximized. But lets check all our values anyway:

Left domain restriction, t=0
$F(0) = \frac{mW}{mSin(0)+Cos(0)}= mW$

Right domain restriction, t=pi/2
$F\left(\frac{\pi}{2}\right) = \frac{mW}{mSin\left(\frac{\pi}{2}\right)+Cos\left( \frac{\pi}{2}\right)}= W$

And our value t=arctan(m)
$F\left(arctan(m)\right) = \frac{mW}{mSin\left(arctan(m)\right)+Cos\left(arct an(m)\right)}$

Now, this is a case where it will be helpful to draw a triangle. Remember arctan(m) is an angle so it is the angle who has a tangent of m. And we are taking the sine of this angle, so we can draw a triangle, we know tangent is opposite over adjacent, so the opposite side is m, and the adjacent side is 1, and by the Pythagorean theorem, the hypotenuse is $\sqrt{m^2+1}$. Now, the angle that gave us this triangle, we are taking the sine of that angle. So we are taking the sine on this same triangle. So our sine is $\frac{m}{\sqrt{m^2+1}}$ and our cosine is $\frac{1}{\sqrt{m^2+1}$ So now we can fill those values in

$F\left(arctan(m)\right) = \frac{mW}{m\frac{m}{\sqrt{m^2+1}} +\frac{1}{\sqrt{m^2+1}}}$

$F\left(arctan(m)\right) = \frac{mW}{\frac{m^2+1}{\sqrt{m^2+1}}}$

$F\left(arctan(m)\right) = \frac{mW}{(m^2+1)^{3/2}}$

And so, since m is positive (can't have negative friction) and m^2+1 is greater than 1, then $(m^2+1)^{3/2}$ must be greater than m, so this is the largest denominator, which results in the smallest actual value.

So when T=arctan(m) we have three test cases, our two domain boundaries, and the arctan(m). The arctan(m) results in the lowest actual value of F, so it is the minimum.

***The minimum value of F comes when T = arctan(m)

*whew* that got hairy for a while :/
Note, in retrospect, it would have been slightly simpler to take the maximum of the denominator instead of the minimum of the whole thing, that would have saved you from having to apply the division rule. But then it would have taken the same form as everything else we did.

Recap:
We determined the domain of our variable
We found the derivative of our function
We found the zero of our derivative
We plugged the zero and domain values into our function
We compared results to find the smallest
• Dec 7th 2007, 07:17 AM
FalconPUNCH!
Wow that's pretty confusing when you get detailed. I don't think I would have ever gotten to the end without your help. Well what was holding me back was that W I didn't know if it were supposed to turn into a 0 when you took the derivative because they don't say it's a constant. Thank you for your help. I'm going to spend time to digest this and understand it.

Quote:

Okay, so the length and width of the paper are 3, and you're subtracting 2x from them. This you seem to have figured out. Then we fold the flaps up to create the sides of the box. Since the flaps are x units long, the box will be x units high. so the formula for volume will be http://www.mathhelpforum.com/math-he...8e048408-1.gif

Look at the pictures I drew you, do you see how the dotted lines are where the flaps fold up? And you can see that they are all x units long.

Another way of thinking about this might be that the left flap takes up x units, the base takes up 3-2x units, and the right flap takes up x units. So when you fold the flaps up, the base will be left at 3-2x, and the left height will be x, and the right height will be x. Anyway, look at the problem I drew again, and maybe try cutting some squares out of a piece of paper to see it.
Yeah I looked over it last night and thought the height was x but I wasn't really sure about it. Now that I looked at your picture it makes sense.

Quote:

Now, this is a case where it will be helpful to draw a triangle. Remember arctan(m) is an angle so it is the angle who has a tangent of m. And we are taking the sine of this angle, so we can draw a triangle, we know tangent is opposite over adjacent, so the opposite side is m, and the adjacent side is 1, and by the Pythagorean theorem, the hypotenuse is . Now, the angle that gave us this triangle, we are taking the sine of that angle. So we are taking the sine on this same triangle. So our sine is and our cosine is So now we can fill those values in
So you get $\sqrt{m^2+1}$ by using Pythagoreans theorem and then just replace COS and SIN each with their respective inputs? I don't think I would have thought about that because, like you said, the book doesn't really give enough details on how to do this.
• Dec 7th 2007, 08:08 AM
angel.white
Quote:

Originally Posted by FalconPUNCH!
So you get $\sqrt{m^2+1}$ by using Pythagoreans theorem and then just replace COS and SIN each with their respective inputs? I don't think I would have thought about that because, like you said, the book doesn't really give enough details on how to do this.

Right, I always draw a triangle when doing these, label the angle "t" then say "the tangent of t is m" so label the opposite side m, and the adjacent side 1, use Pythagorean theorem to find the hypotenuse. Then, say "the sin of t is..." and see that your angle is labeled t, so the triangle you take the sine from is the same as the triangle you just drew.

So any trigonometric function of an inverse trigonometric function will follow this same principle.

And yeah, I know what you mean about books, mine blows too (Sadsmile) I spent probably 40 hours one weekend learning how to differentiate. Not because it's hard, but because when I would go to my book for help I would just get more confused.
• Dec 7th 2007, 08:11 AM
FalconPUNCH!
Quote:

Originally Posted by angel.white
Right, I always draw a triangle when doing these, label the angle "t" then say "the tangent of t is m" so label the opposite side m, and the adjacent side 1, use Pythagorean theorem to find the hypotenuse. Then, say "the sin of t is..." and see that your angle is labeled t, so the triangle you take the sine from is the same as the triangle you just drew.

So any trigonometric function of an inverse trigonometric function will follow this same principle.

And yeah, I know what you mean about books, mine blows too (Sadsmile) I spent probably 40 hours one weekend learning how to differentiate. Not because it's hard, but because when I would go to my book for help I would just get more confused.

Oh ok I understand it now. Yeah some of the books aren't that great that's why I come here, to get help with the problems that the book can't help me with :P well thank you for all your help. :)
• Dec 7th 2007, 12:50 PM
angel.white
What did you come up with for max volume of your box?
(Also, it will probably be important for you to determine your domain for x. If you have difficulty with that, look at the pictures I drew you when you do this, and say "how big can x be" and "how small can x be")