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shall we start
follow this http://www.bbc.co.uk/scotland/learning/bitesize/higher/maths/calculus/
1.Find the maximum value of the product of 2 nos if their sum is 12
So, we're maximizing "xy" if "x + y = 12"... I think if we rewrite the second equation with an independent variable, then we can clean up the first expression as a one-variable expression, and then find the horizontal tangent of that x + y = 12 y = 12 - x xy <- y=12-x x(12 - x)
you can multiply it out and then take the derivative once you have the derivative, you set that equal to 0 to find the point of a horizontal tangent, since that will be the maximum (not really rigorous in showing that, but its an upside-down parabola so it has one horizontal tangent at the maximum point)
x(12 - x) = 12x - x^2 d/dx( 12x - x^2 ) = 12 - 2x 12 - 2x = 0 -2x = -12 x = 6 then we have to go back to that equation to find y's value y = 12 - x y = 12 - 6 y = 6 xy = 6(6) = 36
wow thanks next one
2 A printed page is required to contain k square units of printed matter .Side margins of width a and top bottom margins of width b are required .Find the length of the printed lines when the page is designed top use the least paper
is that all the information?
interesting... i'll have to think about that one a bit
k post it here when u get an idea iw ill wait
|dw:1333259293620:dw| Well, since there's no numerical information given, i think the final answer isn't going to be a number, probably an expression with the k. That's the picture... hmm
Well, I'll start with x := length of a printed line y := amount of printed matter (vertically) |dw:1333259779521:dw| "xy = k" is the equation for square units of printed matter on the page. The dimensions of the page itself are "2a + x" by "2b + y"
I'm not quite sure what the question is asking... anybody else know? :P
well u think abt it lets go to thenext one then
3.Prove that the fn f(x)=2sinx +tanx -3x is increasing in the interval (-90,90)
Is the paper a square?
dunno i hav only the info in the qn
acces any idea abt the 3rd qn?
i graphed it, it doesnt really look like that function is even increasing in that interval...
oh thn how?
pls try again i am sure the qn is crct
odd, the derivative is always positive on the interval (so the function should be increasing), but the actual function itself looks like its decreasing on the graph? o.O
wel u think ovr abt that one too next qn
Find the intervals of monotonicity of the function y=2x^2-log|x|,x!=0 .Hence find the points of maximum and minimum
you'd find the derivative and set it equal to 0... except im not sure on the absolute value there. 2x^2 - log|x| derivative of 2x^2 is 4x, the derivative of log(x) is 1/(x*ln(10))... but that absolute value gets me
y?? it is said in qn dont consider case of 0
oh, not only =0, also points where graph is undefined* the question only says x=/= 0
if i knew the derivative of log|x|, i could possibly give you a better explanation using an example but points where the derivative is undefined are also critical points when the function exists at that point
derivative of log mod x is 1/x
uhh, sorry, not sure i can help you with these questions. they're a bit outside of what i know. :)
20 m of wire is available to fence off a flower bed in the form of a circular sector.what must be the radius of the circle oif we wish to have a flower bed with greatest possible area?
May I help?
You may help. :) I have an idea of how to do that last one, unless that's what you know. :P
both of u bring in ur best n fast :P
As there are 20 m fence, 2r pi =20, r=10 A=r^2 pi A'=2 pi r A''= 2pi >0 therefore, max A occurs at r=10 A=(10)(10) pi =314
its not a full circle, but a sector of a circle
wt u think access?
tell ur method
So, the perimeter of the sector is: P = 2r + rt (t := angle between radii that make up the sector). P is also given to be 20. 20 = 2r + rt The area of the sector is A = r^2 t/2. If we use substitution to get a function of t with respect to r... 20 = 2r + rt 20 = r(2 + t) r=/=0 20/r = 2 + t 20/r - 2 = t A = r^2 (20/r - 2) / 2 = ( 20r^2 / r - 2r^2 ) / 2 = 10r - r^2 A' = 10 - 2r 10 - 2r = 0 -2r = -10 r = 5
yey thx a lot
do post here if u get better attempts on the qns we skipped
i actually found the solution online for the one concerning the page and print matter the answer just isnt intuitive for me, trying to think of a way to explain it. they basically used a function of area of the page to the x-value (substitution from k=xy) and minimized the area of the page
Any other question?
HMM?u knw integration and differentiation?
No, but the type of questions u posted, I think I can solve by any other method also.
did u get the book qn?
Which one? The "Page" Question????????
See bhai, @AccessDenied answer is right. But after that, how to proceed not exactly getting. What to find next????????????? Any more options and clues given?????????
In that case, @AccessDenied answer is rite.
can u help with that page qn?
ok. lemme try first. i didn't read the whole post was it resolved?
wt to do next?