12
put x=\omega
S=22010
Not dbts, but action here has slowed up a bit, so trying to bring them up....
1) Reals x,y satisfy x^2+y^2+xy=1.
Find the minimum value of x^3y+xy^3.
2) This is a bit common sum.
(3+x^{2008}+x^{2009})^{2010} =\sum_{i=0}^{i=n}{a_ix^i}.
Find S=a_0-\frac{1}{2}a_1-\frac{1}{2}a_2+a_3-\frac{1}{2}a_4-\frac{1}{2}a_5+a_6+.....
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16 Answers
1((x^2+y^2)+xy)24≥xy(x^2+y^2)
x3y+xy3≤14
so got the maxima
but u asked for minima ???
39fir the furst question
add xy in both sides of the given equation
u have (x+y)2=1+xy
for this to be valid
minimum value xy can take is -1
when xy takes -1 , x2+y2=2
so minimum of x3y+xy3=-2
341The full solution will tell you more I guess.
basically you are minimizing f(t) = t-t2 where t = xy. From AM-GM you see that xy \le \frac{1}{3}.
Again as b555 has indicated xy≥-1.
So you have to check the behaviour of the quadratic between those values of t. You do not really need calculus for this.(though i did, it was faster to reckon :D)
11H'mmm.....
Alternately, x=rcos\alpha and y=rsin\alpha.
That gives r^2=\frac{2}{2+sin2\alpha }.
That also gives \frac{2}{3}\le r^2 \le 2.
Put r2=t so now let p=(1-t)t
p is min. when t is max, i.e t=2. that gives p=-2.