More Calculus Fun

Ok here are some more for you guys. If you don't mind it would be cool if you could just give the answer and show how you got it instead of trying to "lead" me :D. Thanks again guys...

Optimization Problems (The prof and the book showed nowhere near enough examples of these problems)

1) If 1200 cm^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

I know the volume of an open box is (l-2x)(w-2x)(x). I just don't know what to do with the 1200 cm^2

2) Find the point on the line y = 4x + 7 that is closest to the origin.

3) Find the area of the largest rectangle that can be inscribed in the ellipse (x^2/a^2) + (y^2/b^2) = 1.

In this one I know you solve for either x or y and then substitute in the area of a rectangle formula which is length x width (or a = x*y). So after this you take the derivative and then what?

LEARN TO DO YOUR HOMEWORK

Optimization problems as far as I can remember are almost always solved by the fact that the maximum or minimum of a function is at a point where the derivative equals zero, i.e. where the slope is equal to zero. Therefore the strategy becomes.. what is the thing (e.g. volume, line distance, etc) we want to maximize/minimize, simplify it down to one variable and then we take the derivative with respect to that variable and set it equal to zero. Then we solve for that variable.

1) I'm guessing this is a square piece of material from which we are to cut a plus-shaped looking shape that can be folded up into an open top box. If not, then there is a different solution.

Solution #1: I like to keep everything written as a variable because it makes it much easier to check your solution after you are done. So, I define:

l = length of one side of what I assume is a square piece of material
w = width of the box's base
d = depth of box's base
h = height of the box

The thing we want to maximize is the volume. Volume in the variables we defined is w*d*h... the problem states that it has a square base, so w = d.. and we can write it as V = w^2*h. If we cut the plus pattern out of a square piece of material then l = 2h + w. Since l is a constant, we want an expression for volume that has only one variable... so we solve for w and plug that in to our volume expression.

w = l - 2h =>
V = (l - 2h)^2*h
V = (l^2 - 4*l*h - 4*h^2)*h
V = l^2*h - 4*l*h^2 - 4*h^3

So this is the expression for which we want to know at what value of h is it a maximum... so we take the derivative and set it equal to zero. (if you are uncomfortable taking a derivative with respect to h you could insert x for h.. but you should get comfy with taking the derivative with respect to any variable...)

dV/dh = l^2 - 8*l*h - 12*h^2

This equation now defines the slope of the function V as function h... we want to know when that slope equals zero, and then we solve for h... at this point I'll plug in the value for l, which I assume is the square root of 1200...

1200 - 8*sqrt(1200)*h - 12*h^2 = 0

Which when I solve gives me roots of 3.7 and -26.8... which has got to be an incorrect answer i think... hmm... shitty... Anybody see my mistake?... i'm fairly confident that my solution is appropriate.

The 2nd way to do it would be to figure out how one can make a box of maximum value with a surface area of exactly 1200 cm^2... something like 4*w*h+w^2 = 1200 for the area, were w^2*h is volume... solve the top equation:
(1200 - w^2)/4w = h;
plug into volume equation:
V = w^2*(1200 - w^2)/4w
V = 300w - (w^3)/4
and do the same thing as above....

dV/dw = 300 - 3/4*w^2 = 0
w = sqrt(-300*(-4/3))
w = 20 cm

Solve for h in the surface area equation:

h = (1200 - w^2)/4w
h = (1200 - (20^2)/(4*20)
h = 10 cm

check answer... 10cm * 20 cm = 200 cm^2*4 = 800 cm^2, and 20 cm*20cm = 400 cm^2... for a total surface area of 1200 cm^2.

The volume associated with that box is 400cm^2*10cm = 4000 cm^3.

Still bothered by my first solution... they should give you different answers.. but I still doubt that it is correct. Thoughts?

For the line question you are looking equation for the length of a line:

(x-x0)^2 + (y-yo)^2 = l^2 (generally speaking, where l is the length of the line)...

Your problem wants to know from the origin, so xo = 0 and yo = 0, So:

x^2 + y^2 = l^2

but you can't look at any point y and x, only those that are defined by the equation
y = 4x + 7... so

x^2 + (4x + 7)^2 = l^2 ->
l = [x^2 + (4x + 7)^2]^(1/2)

This is the equation which describes the length of a line from the origin to any point on the line defined by y = 4x + 7... so we need to find when this is at a minimum, i.e. the slope is equal to zero. The derivative is just a lot of product rule differentiation.

dl/dx = 1/2*[2x + 8(4x +7)]/[x^2 + (4x + 7)^2]^(1/2) = 0

Solve for x... then plug determine y from y = 4x + 7 and plug those into the line equation to get the length of the shortest line. Note.. the answer should be less than but close to 2. Good Luck.

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Originally Posted by martindcx1e

Ok here are some more for you guys. If you don't mind it would be cool if you could just give the answer and show how you got it instead of trying to "lead" me :D.

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Learn it yourself you cheeky fucker

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If you don't mind it would be cool if you could just give the answer and show how you got it instead of trying to "lead" me

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People like working to help people. People don't like doing work for other people.

kingnat is friggin smart. Calculus with real world application, I love it.

These are pretty easy man. Sounds like you're trying to memorize ways to do things instead of understanding wtf you're doing.

No offense <3

My bad, I was trying to help you learn the stuff. If my help's not appreciated, lots of luck.

http://verbatim.blogs.com/verbatim/images/keitel.jpg

Why don't you just pay someone $50 an hour to do your homework for you instead of begging people on FTR to do it.

wtf responses? like a month ago i asked if anyone here would be able to help out with calculus problems throughout the semester and then when i post them ppl flip out and tell me to do it myself. guys, your help is appreciated. i'm not just posting my hw without trying to do it. these are the problems i don't know how to do based on what i've heard in class and read in my book. if i see the whole problem done it is way more helpful to me than taking like 3 or 4 responses with somebody to figure it out.

i know these problems are not hard for you guys, and i'm sorry i can't post something more challenging. i am just learning this stuff. like i said my prof and the book show minimal examples, and there are lots of different problems that can be derived from the concepts. all i see is the basic concept and then like 1 or 2 examples. thanks to those who have helped me so far.

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If you don't mind it would be cool if you could just give the answer and show how you got it instead of trying to "lead" me.

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Originally Posted by spoonitnow

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If you don't mind it would be cool if you could just give the answer and show how you got it instead of trying to "lead" me.

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Originally Posted by martindcx1e

if i see the whole problem done it is way more helpful to me than taking like 3 or 4 responses with somebody to figure it out.

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I don't want a 2nd teacher here. I just want someone to be straight forward with me and show me how to do a friggin problem since my prof sucks and my book is cryptic.

Also, thank you very much kingnat for your help in this thread. It was much appreciated. Thank you so much for just posting how you did the problems. You rock :D

EDIT

go bucks

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If you don't mind it would be cool if you could just give the answer and show how you got it instead of trying to "lead" me.

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This mentality is what's probably bothering people.

But, in your defense, you probably have an auditory or kinesthetic learning type so you need a different approach to really learn this stuff. It's basically what leads to the feeling that you know a little bit of what you need to do as far as writing it down but then you stare at it and don't really know what you're doing with it, like here:

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I know the volume of an open box is (l-2x)(w-2x)(x). I just don't know what to do with the 1200 cm^2.

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Let's start you off on the right path. Let's say y = 3 - 4x - x^2 and we want to maximize y. Well how do we find maximums and minimums with Calculus? Derivatives. So y' = -4 - 2x, let y'=0 to find the maximum x value, so 0 = -4 - 2x and x = -2. Then our maximum y is 3 - 4(-2) - (-2)^2 = 7. Graph y = 3 - 4x - x^2 to verify and maybe that will help your understanding as well.

Edit: Waiting on my girl to get out of the shower so graphed it for you...

http://img138.imageshack.us/img138/8666/graphzv8.jpg

Note that the maximum y-value is 7, just like we figured out. (/edit)

Here's another for you to think about: Suppose we're going to fence in a rectangular area with 100 feet of fencing but we're fencing beside of a river so we only need to do three sides, and we want to maximize the area we can fence in. Here's a picture to make sure you get the idea:

http://img233.imageshack.us/img233/3805/fenceoj2.jpg

Well we need an equation for area first and foremost. The area of a rectangle is length times width, so let's define those. If we let the left side of the fence be length x, then the opposite side is also length x, and the top side is 100 - 2x (since we have 100 feet total of fencing). Now we have an equation for area: A = width * length = x(100-2x).

We want to maximize area here, so how would we do that?

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Originally Posted by martindcx1e

if i see the whole problem done it is way more helpful to me than taking like 3 or 4 responses with somebody to figure it out.

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Yes and no. Read my response above and I'm gonna go take a quick shower. If you've replied by then I'll talk some more but if not I'll be back later tonight.

P.S. I've tutored all sorts of math (mostly Calculus II-IV) for the past 5-6 years on a very regular basis, so trust me. :wink:

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Originally Posted by spoonitnow

Well we need an equation for area first and foremost. The area of a rectangle is length times width, so let's define those. If we let the left side of the fence be length x, then the opposite side is also length x, and the top side is 100 - 2x (since we have 100 feet total of fencing). Now we have an equation for area: A = width * length = x(100-2x).

We want to maximize area here, so how would we do that?

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thank you spoon. i guess my original statement comes off as lazy. sorry about that everyone. i assure you that's not the case.

ok so to find the max area of your example you would distribute the x, differentiate, set it equal to 0, solve for x then plug that into 100 - 2x to find the long side. so x = 25 and 100 - 2x = 50 and the max area is 1250. right?

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Originally Posted by martindcx1e

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Originally Posted by spoonitnow

Well we need an equation for area first and foremost. The area of a rectangle is length times width, so let's define those. If we let the left side of the fence be length x, then the opposite side is also length x, and the top side is 100 - 2x (since we have 100 feet total of fencing). Now we have an equation for area: A = width * length = x(100-2x).

We want to maximize area here, so how would we do that?

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thank you spoon. i guess my original statement comes off as lazy. sorry about that everyone. i assure you that's not the case.

ok so to find the max area of your example you would distribute the x, differentiate, set it equal to 0, solve for x then plug that into 100 - 2x to find the long side. so x = 25 and 100 - 2x = 50 and the max area is 1250. right?

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That's right. Now look at this one:

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1) If 1200 cm^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

I know the volume of an open box is (l-2x)(w-2x)(x). I just don't know what to do with the 1200 cm^2

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Square base, open top. Draw a picture of what the box looks like laid out flat and you'll probably get somewhere on this one pretty quickly. Here, I'll get you started:

http://img504.imageshack.us/img504/6162/boxua9.jpg

Letting x be the length of the base and h be the height.

Once you have the correct equation these problems are cake. So, we want to maximize volume which will be V = length * width * height, or V = x * x * h. We need everything to be in terms of x so we can take the derivative, so we need a way to write h in terms of x (kind of like when we wrote the length in terms of x in the fence problem and came up with the length being 100 - 2x).

So see if you can find a way to write h in terms of x and finish the problem. (Hint: "I just don't know what to do with the 1200 cm^2.")

I'm going to go see Saw 4 later so if you post before I leave I'll get back at you sooner :P

so x^2 + 4xh = 1200.
h = (1200 - x^2)/(4x)

Volume = (x^2) * [(1200 - x^2)/(4x)]
Which = 300x - (1/4)x^3

Deriv of that = 300 - (3/4)x^2
x = 20

(20)^2 + (4)(20)(h) = 1200
h = 10

Ok so you find the equation for what you want to minimize or maximize. Then you find another equation to let you isolate one variable. Then substitute into the original equation, take the derivative, set it to 0, and solve for that variable. Then plug that answer back into your 2nd equation to find the other variable. Is that pretty much it?

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Originally Posted by martindcx1e

so x^2 + 4xh = 1200.
h = (1200 - x^2)/(4x)

Volume = (x^2) * [(1200 - x^2)/(4x)]
Which = 300x - (1/4)x^3

Deriv of that = 300 - (3/4)x^2
x = 20

(20)^2 + (4)(20)(h) = 1200
h = 10

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You catch on quick, just make sure you answer the question by stating what exactly the maximum volume is. :wink: (Not being nit-picky, it's just an easy thing to forget).

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Originally Posted by martindcx1e

Ok so you find the equation for what you want to minimize or maximize. Then you find another equation to let you isolate one variable. Then substitute into the original equation, take the derivative, set it to 0, and solve for that variable. Then plug that answer back into your 2nd equation to find the other variable. Is that pretty much it?

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You're very close, but try not to think about it as a linear process of this many steps because every problem is different. The process of what you're doing is really like this:

1. Find the equation for you want to minimize or maximize.
2. Take the derivative and get information that lets you answer the question.

It's just that each step will sometimes be straight-forward and sometimes it won't be. In the box question you had to manipulate an extra variable h, but sometimes you'll be manipulating two or more extra variables, and sometimes you won't be manipulating any variables at all (think of the first example I gave of y = 3 - 4x - x^2 or whatever it was).

The key is that before you can take the derivative of the equation you find, it has to be all of the same variable. Like when we had volume = x * x * h, we can't differentiate that equation for volume yet because it's not all of the same variable. If h was a constant we could do it and it would be V' = 2hx but h is a function of x so we can't.

So instead of thinking of it as a linear process, maybe think of it as a series of goals or objectives. How you achieve those objectives might be different going from problem to problem; note the differences in the fence problem from the box program for an example of this. So, onward:

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2) Find the point on the line y = 4x + 7 that is closest to the origin.

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You're minimizing distance, so your first objective is to get an equation for distance in terms of one variable. :cool:

Here's an example I just remembered giving a student one time like a million years ago that uses the distance equation.

Find the point on x = 5 that is closest to (1,2).

Now no shit sherlock it's going to be (5,2), but I'm going to use this as a less complicated example of what's going on for you.

This is basically an exercise where we want to minimize distance, so first we want to get an equation for distance using the information we're given. The distance from any given point to (1,2) is:

Distance = [(x-1)^2 + (y-2)^2]^(1/2)

Then we think: Oh snap! We have two variables instead of one! How can we fix that? We look back to the information we were given:

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Find the point on x = 5 that is closest to (1,2).

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Well that x = 5 bit might be useful...

Distance = [(5-1)^2 + (y-2)^2]^(1/2)
Distance = [16 + (y-2)^2]^(1/2)

Oh boy, we're left with one variable so we can take the derivative!

Distance' = (1/2) (2*(y-2))[16 + (y-2)^2]^(-1/2)

So we set it equal to zero and solve for y...

0 = (1/2) (2*(y-2))[16 + (y-2)^2]^(-1/2)

You quickly realize that you have just (y-2) on the top, so y = 2.

So x = 5, y = 2, the point is (5,2). GG.

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Originally Posted by martindcx1e

so x^2 + 4xh = 1200.
h = (1200 - x^2)/(4x)

Volume = (x^2) * [(1200 - x^2)/(4x)]
Which = 300x - (1/4)x^3

Deriv of that = 300 - (3/4)x^2
x = 20

(20)^2 + (4)(20)(h) = 1200
h = 10

Ok so you find the equation for what you want to minimize or maximize. Then you find another equation to let you isolate one variable. Then substitute into the original equation, take the derivative, set it to 0, and solve for that variable. Then plug that answer back into your 2nd equation to find the other variable. Is that pretty much it?

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For full credit you probably need to show that it is a maximum (as opposed to a minimum or an infliction point)

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Originally Posted by miracleriver

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Originally Posted by martindcx1e

so x^2 + 4xh = 1200.
h = (1200 - x^2)/(4x)

Volume = (x^2) * [(1200 - x^2)/(4x)]
Which = 300x - (1/4)x^3

Deriv of that = 300 - (3/4)x^2
x = 20

(20)^2 + (4)(20)(h) = 1200
h = 10

Ok so you find the equation for what you want to minimize or maximize. Then you find another equation to let you isolate one variable. Then substitute into the original equation, take the derivative, set it to 0, and solve for that variable. Then plug that answer back into your 2nd equation to find the other variable. Is that pretty much it?

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For full credit you probably need to show that it is a maximum (as opposed to a minimum or an infliction point)

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Depends on the instructor. You're right obv but in my experience most instructors don't require it for entry-level Calculus students.

Three years ago I could have done this standing on my head.
Now I cant.
I feel old.

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Originally Posted by Taxi

Three years ago I could have done this standing on my head.
Now I cant.
I feel old.

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I had Calculus I six years ago. I know how you feel.

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Originally Posted by martindcx1e

wtf responses? like a month ago i asked if anyone here would be able to help out with calculus problems throughout the semester and then when i post them ppl flip out and tell me to do it myself. guys, your help is appreciated. i'm not just posting my hw without trying to do it. these are the problems i don't know how to do based on what i've heard in class and read in my book. if i see the whole problem done it is way more helpful to me than taking like 3 or 4 responses with somebody to figure it out.

i know these problems are not hard for you guys, and i'm sorry i can't post something more challenging. i am just learning this stuff. like i said my prof and the book show minimal examples, and there are lots of different problems that can be derived from the concepts. all i see is the basic concept and then like 1 or 2 examples. thanks to those who have helped me so far.

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I feel yah. I just misinterpreted your tone.