Foundation Degree so yeh Although we have'nt be taught 90% of this assignment so im having to teach myself alot of it
Yep Although my other teacher is the opposite he talked us through the assigment and let us go back to what we done crap to make it better Done alot of worked examples similiar to the assignment questions so we could understand properly what to do before hand
wow.. The optimization and vector stuff was all taught to me here in High School. I'm graduating in a couple of weeks and the assignments you have mentioned are all stuff I can be subjected to in our written exam. I should start uni in England ..
I assume this isn't for an actual maths degree though? Also foundation university work is designed to (if necessary) fill gaps between school and university, such as if you didn't do A-levels in all the subjects necessary for whatever you're reading. The type of questions for maths when doing a science subject at university (let alone a proper maths degree) can be really quite tricky, the papers I did last year are here if anyone's interested: http://www.maths.cam.ac.uk/undergrad/nst-pastpapers/2011/index.html Two of the less enjoyable 3 hours of my life...
I'm considering applying for Natural Sciences. I could do eight marks worth across both those papers. Bufo you don't mind if I ask you a few question do you? Particularly if your doing Physical science options?
The Foundation degree is like a slower and cheaper way of getting the degree in the subject it allows me to stay 2 years at the college i done my A-levels at and a 3rd year at the university of plymouth The maths is simply a module of it im doing a foundation degree which includes electronics and computing so maths involving complex numbers is useful in electronics but thats about it. Its difficult in terms of completely not understanding seeming not being taught anything, but some what fun in a way its challenging me mentally. @Picarro i've heard that alot tbh you probebly could do it i have a few friends in sweden and norway who do similiar stuff at a much younger age they seem to be on the ball with the curiculum there.
Guys.... I gotta ask. Aren't you over complicating the original question a bit? Or maybe I'm not getting the constrains of the problem 100% right... But the way I see it... To maximize the area of any rectangle (given its perimeter); you just make all four sides equal. I did that and a few minutes of very basic algebra/geometry gave me that the max area of the trapezoid is 200 cm^2. I did not need to use differential equations... Hell; all I used was the Perimeter formula, some Pythagoras and the Trapezoid area formula. Five minutes and done.
Sure, I'm in my second year now at Christ's doing chemical engineering (did first year physical natural sciences). PM me any questions you have.
They teach all that stufff in high school here but not everyone does the same modules. I'll have a look but I don't think it's that simple (although it might well be). I can vouch for Bufo, he was very hepful when I was applying to do Maths at Christs (I got an offer and then missed out by one grade on step which is generally fine but apparently they had a pretty busy year, and also I only barely scraped both grades, so I fell at the last hurdle). Speaking about tough maths exams, I don't think you get much tougher than STEP. My maths teacher told me I was the most talented student in the school (which I kinda doubted, although I didn't doubt his sincerity) and we sent off no less than 8 Mathematicians to Oxford and I still managed to muck up the STEP paper. Most of the Warwick people were getting 3s and Us.
It still only takes 5 mins if not less with differentiating, and to find the maximum "properly" (rather than just by inspection/trying values) you should be differentiating at some point. The perimeter is for the whole plate (including the triangles), you seem to have used it just for the rectangular section which as you said makes it simple.
Agree with you that doing differentiation would also be quick. Disagree with you that not using d/dx is not proper.... I did not (inspect/try) values; I simply applied a known property of rectangles... If you want to maximize the area of it given a perimeter; you simply divide the perimeter by 4 and that's the largest area rectangle you can make. No checking values there. After that I just used a bit of algebra and came up with an answer... My 200 square cm was wrong though... I typed the calculations into google as I did not want to go looking for my calculator and forgot a parenthesis somewhere. Either way... Here's how I did it.
I don't think that method takes into account the greater effect of x on the area than y, as x determines both one dimension of the rectangle and both sides of the triangle - this makes your answer slightly less than that calculated by differentiation (as shown by Picaro on the previous page, x=7.07 and the area is 70.7).
Think about it... If I make the inside rectangle a square and add/subtract to "x" whatever I took from "Y"; that eliminates that greater effect. - We know that the max area of any rectangle (given a perimeter) is to make it a square. - We know that the max area of any right triangle is to make b = h That tells me that to maximize the area of the trapezoid; I should make x = y so that I end up with a square and two right triangles where b = h. Doing that makes each triangle on the outside into a half of a square that has the same dimensions as the inner square and that gives me that the max area is the area of the "two squares" added... that's where the 2s² comes from... -> s² is the area of the inner square and the other s² is the sum of the areas of the two triangles.
You can't justify the side lengths by considering the rectangle and triangles separately, you have to consider the shape as a whole - by differentiating you get X=5sqrt(2) (or 7.07) and so Y=2.93, giving a total area of 70.7 which is higher than yours.
I guess I'm just thick... I see the logic of what picarro did and I see the logic of what you're saying. Oh well... I guess I should have used calculus from the get go instead of trying to simplyfy with geometry and algebra.... Still; the problem doesn't say that the ratio between x:y needs to be maintained (picarro's approach) anywhere so I started messing with the lengths to the sides to simplify it.
Its nice too see a little bit of debate going on with this! I was trying to refrain from asking for anymore help i've completed almost all (of 9 pages!! not bad for no clue when i started it right ? )except from 3 questions which i have been sitting for 2 days with no real understanding of how to do it, they sound simple enough but when you have'nt been properly taught then it can become impossible... I was considering just leaving them but i've got this far in my understanding why not take a few more steps On the sheet i seem to have written an answer (and only the answer) from around three weeks ago (when i first got the assignment) it was told to me by one of the class and i thought might be useful to keep a note of to check i get the same answer however it makes no sense now. Show the curve Y=(2x-1)^2(x-2)^2 Has the turning points x=1/2 and x=2 By examining the sign of the first derivative on both sides of each turning point or otherwise determine the nature of all three turning points The answer squiggled next to it is "X=0 ??????" i feel kinda silly because i have no idea what im supposed to be doing :/ its not something i've been ever told about or explained
Do you know how to differentiate? You find turning points by differentiating the equation to get another equation describing the gradient, and this =0 at turning points (as it goes from sloping one way to the other and so must be flat at one point). You can then solve this equation and should get x=1/2 and x=2 as solutions giving you the answer. Do you have any notes or other stuff to look at? Otherwise just search google for differentiation and there are various online guides. Also another useful website (especially for checking answers) is www.wolframalpha.com , you can type in functions and it will draw graphs, integrate, differentiate and find solutions. Link to wolfram showing answer: http://www.wolframalpha.com/input/?i=Y=((2x-1)^2)(x-2)^2 In this case you don't actually have to differentiate, as y=0 at both turning points and so you can just solve the equation for y=0.
Very minimal if im being completely honest most of differentiation i have been learning this week by google and youtube Its kinda frustrating really I had the same teacher for electronics two year back the entire class had no understanding and we all failed just because he would rush ahead even when we made it clear we had no clue what she was talking about Im hoping when he realizes we all have struggled to the point of not being able to even start it (one person atleast) he will grasp how screwed we potentially all are
