CAT Question Bank  Maxima & Minima

@SoumyaRanjanPatra Answer is root 2. Explanations below
Method 1 :
Let x^2+y^2 = k = k * 1
=> x^2 + y^2 = k(x^2 + 2xy  y^2)/2
Divide throughout by xy
=> 2 (x/y) + 2 (y/x) = k (x/y) + 2k  k (y/x)
Let x/y = t
=> (2k)t^2 + 2kt +(2+k)=0
since x/y is real, so D ≥ 0
=> 4k^2  4 (2k)(2+k) ≥ 0
4k^2  4(4k^2) ≥ 0
8k^2 16 ≥ 0
k^2 ≥ 2 and k should be +ve only
so k ≥ root2
so min value= sqrt (2)
[Credits : Bruce Wayne]Method 2:
x^2 + 2xy  y^2 = 2
x = asinQ
y = acosQ
so x^2 + y^2 = a^2
a^2sin^Q + 2a^2sinQ * cosQ  a^2cos^2Q = 2
a^2((sin^2Q  cos^2Q + 2sinQ * coSQ)=2
a^2((sin2Q  cos2Q) = 2
a^2 = 2/(sin2Qcos2Q)
now max value of sin2Q  cos2Q=root2
so a^2 = 2/root2
so a^2 = root2
so min value will be root2
[Credits : @hemant_malhotra ]

2^(1/(x^2  x + 1)) will be maximum when x^2  x + 1 is minimum.
Differentiate and equate to 0 => 2x  1 = 0
x = 1/2
when x = 1/2, x^2  x + 1 = 1/4  1/2 + 1 = 3/4
maximum value is 2^(1/(3/4)) = 2^(4/3)

Method 1 :
AM ≥ GM
(p + q + r)/3 ≥ (pqr)^1/3  (1)
(1/p + 1/q + 1/r)/3 ≥ (1/pqr)^1/3  (2)
(1) * (2) > 1/9 (p + q + r)(1/p + 1/q + 1/r) ≥ 1
(p + q + r)(1/p + 1/q + 1/r) ≥ 9
So minimum value = 9Method 2 :
(p + q + r)(1/p + 1/q + 1/r)
= p/p + p/q + p/r + q/p + q/q + q/r + r/p + r/q + r/r
= 1 + 1 + 1 + (p/q + q/p) + (p/r + r/p) + (q/r + r/q)
= 3 + 2 + 2 + 2 = 9 [ @ChhabbisGyarah ]Method 3 :
(p + q + r)(1/p + 1/q + 1/r) = (p + q + r)(pq + qr + pr)/pqr is minimum when pqr is maximum.
pqr is maximum when p = q = r
(p + q + r)(1/p + 1/q + 1/r) = 3p * 3/p = 9

Maximum occurs when a^2 = b^2 = c^2 = d^2 = 25
=> a = b = c = d = +/ 5
Maximum value of a + b + c + d = 5 + 5 + 5 + 5 = 20

@rowdyrathore
min = 20
max = 20

@rowdyrathore min = 7
max = 73
cauchy method

Graph is as below

@rowdyrathore can u plz solve this?

pls post tye solutions or atleast the correct answers.

@RowdyRathore (26/3)^(1/8)