Positive reals a and b are such that a + b = s . The maximum value of s a + b is c . Find c .
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@Alak Bhattacharya , for square root you can enter \sqrt 2 2 , \sqrt {2\pi} 2 π , \sqrt[3]8 = 2 3 8 = 2 . Put all formulas a + b = s , all variables x , y , z and constants a , b , c in LaTex.
Smart substitution 2:
Given (√a+√b)/√(a+b)=√c
Write a=s(1+x)/2 b=s(1−x)/2
So √c=[√(1+x)+√(1−x)]/√2
Then c=1+1√(1-x²)
Whence √c=√[1+1√(1-x²)]
Thus √c≤√2
Max.c=2
Here's a somewhat inelegant solution.
Without loss of generality, we can set s = 1 (this just scales a and b by the same factor and doesn't affect the final result). So we have b = 1 − a , and we want to maximise f ( a ) = a + 1 − a , where 0 ≤ a ≤ 1 .
We can use calculus for this, but it's neater to observe that both a and 1 − a are concave functions, so their sum is, too. Also, f ( a ) is symmetric about a = 2 1 ; these two facts mean its maximum value can only occur at a = 0 (and a = 1 ), or at a = 2 1 . It's easy to check that the largest of these is when a = 2 1 , giving c = 2 .
Alternatively, if we say a = x 2 and b = y 2 , this problem becomes "maximise x + y subject to x 2 + y 2 = 1 ", which is a slightly more familiar formulation and can be solved in a number of different ways (for instance, geometrically - plotting in the x − y plane, the constraint is a circle and the function to be maximised a straight line).
We know that s a + b ≤ c . Squaring both sides of the equation (they are both positive) we get s a + b + 2 a b ≤ c . From AM-GM (geometric mean is always less or equal the arithmetic one) we know that a b ≤ 2 a + b which means 2 a b ≤ a + b = s . Therefore s a + b + 2 a b ≤ s a + b + s = s 2 s = 2 ≤ c , so c = 2 (because c is the maximum and this expression cannot be greater than 2).
this problem is equivalent to minimizing r a + b for the eqn a 2 + b 2 = r 2 .
Geometrically, this is the equivalent of finding the line tangent to a circle with the eqn a + b = k , i.e a 4 5 ∘ line.
this is an isosceles right triangle, meaning we would have r r 2 + r 2 be the minimum, i.e 2
Hint :
Use the power mean inequality, i.e.
p n a 1 p + a 2 p + . . . + a n p ≥ q n a 1 q + a 2 q + . . . + a n q
where, a i is a positive real number and p > q
No restriction over p and q? Will the inequality hold irrespective of whether p is less or more than q?
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Ohh srry I forgot to mention that p > q
Substitute x = a and y = b to the original expression yielding x 2 + y 2 x + y .
Now by the QM-AM inequality x 2 + y 2 x + y = 2 x + y x 2 + y 2 2 2 ≤ 2 x 2 + y 2 x 2 + y 2 2 2 = 2 ,
where equality is attained when x = y .
Hence c = 2 .
a = u²
b = v²
s = a + b = u² + v² = r²
u = r cos(t)
v = r sin(t)
y = cos(t) + sin(t)
y' = -sin(t) + cos(t) = 0
t = pi/4 en y = sqrt(2)
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By Cauchy-Schwarz inequality , we have:
s a + b ≤ s ( 1 + 1 ) ( a + b ) = s 2 s = 2
Therefore, c = 2 .