Im stuck on this really straight forward sounding question and i just can't seem to figure out how to get around it Im not asking for the answer just how the hell do i start it The summarised question is simply "A rectangular plate is joined onto two equal adjacent side triangular plates (like a trapezium) if the perimeter of the plate is (value in cm) find the dimesions of the plate when the area is maximum. Hence determine the maximum area" I have no clue where to start i have the perimeter but im trying to figure out with that how i find the area and dimensions of the trapezium shape the height of the rectangular section is labelled x same as the bottom of both of the triangles on either side Im suure there is a really easy way of doing this but im stumped... when i get it i'll be it will be a face palm moment
Looks like you'll be incorporating Pythagoras' theorem into this. I'm trying to figure out how though. What is the value of the perimeter of the rectangle?
The entire perimeter is 40cm thats the only value im given heres a rough paint its probebly so simple but all the other questions are so insane i figure there is some other difficult answer to it >.> Both X dimensions look the same on the diagram i have im it was just quick paint work
The approach (I think) you'll have to take is to work out the perimeter of the shape which is (if the triangles are isosceles): 2x+2y+2[pythagoras theory of hypotenuse]=40 You can then determine y in terms of x. Then you work out an equation for the area of the shape. xy + 2[area of triangles] and substitute in the values of y so you only have it in terms of x. Then you can find when this is at maximum using differentiation. You then have a value of x which you can use to find y because you determined the relationship earlier. Haven't done this on paper but this approach should work. Sorry if that's too step-by-step though.
If both of the sides to one of the triangles are the same (as shown by the Xs but not the diagram) then it's fairly straightforward, get equations for the perimeter and area, substitute into the area equation to get it in terms of X then differentiate to find the maximum - from there can then find the individual dimensions. If they're different then it's quite a bit trickier, and would probably have to iterate to find the solution. Edit: Actually looking at the problem again they must be the same, otherwise it's trivial as you just ignore the triangles.
Holy crap thanks for all your help guys! i will rep you when i finish this page! i got this question where im calculating the angle between two vectors using the formulae cos(theta) A.B/|A||B| All seems fine checked my calculations twice and i end up with cos(theta)=1.08 But i can't do angle(theta)=cos^-1(1.08) because its not between the range of (-1,1) so whats the angle... or is there no angle im nto sure if its a trick question or not
There's always an angle between two vectors - could you post the question so I could check your calculations?
Thats what im confused about :/ maybe i've written it down wrong Two lines are represented by the vectors a=4i-3j+2k b=2i-3j+6k Calculate the angle (theta) between the directions of these lines using the formula cos(theta) A.B/|A||B|
The equation is a.b=[a]cos(theta), where [a] is the magnitude, is that what you're using? Maybe check for a +/- sign error. You should get cos(theta)=29/(sqrt(29)sqrt(49)) if my calculations are right.
Yeh thats what i was using i ended it with cos(theta)=29/(sqrt(23)sqrt(31)) where the hell have i screwed up that much :S
You seem to be calculating the magnitudes of the vectors wrong, it should be sqrt(i^2+j^2+k^2) so for a is sqrt(4^2+3^2+2^2) which is sqrt(29)
As Bufo said - you should get cos(theta) = 29/(sqrt(29)sqrt(49)) It may sound silly but check that you're correctly entering (-3)^2 into your calculator. Many calculators will calculate -3^2 to be -9 not 9 (as they're effectively squaring 3 and then multiplying by -1)
Im actually smashing my face against the table right now.... I missed the brackets on the negative -3 for both j values all adds up now george ninjaed me i kinda feel ridiculous now Thanks fellas
Personally I just ignore minus signs when calculating magnitudes; it's the same value, just in the opposite direction, after all. (Just remember to take it back into consideration when doing |(a-b)| although that onl matters withing individual brackets. I'd be prepared to bet that 90% of all artithmetical errors I make are due to missing a minus sign somewhere, the rest are summing two and two and getting 5)
