# CAT Question Bank - Maxima & Minima

• @Soumya-Ranjan-Patra 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
=> (2-k)t^2 + -2kt +(2+k)=0
since x/y is real, so D ≥ 0
=> 4k^2 - 4 (2-k)(2+k) ≥ 0
4k^2 - 4(4-k^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/(sin2Q-cos2Q)
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 = 9

Method 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 [ @Chhabbis-Gyarah ]

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

• @rowdy-rathore
min = -20
max = 20

• @rowdy-rathore min = -7
max = 73
cauchy method

• Graph is as below

• @rowdy-rathore can u plz solve this?

• pls post tye solutions or atleast the correct answers.

• @Rowdy-Rathore (26/3)^(1/8)

45

47

1

58

107

63

102