# Maxima & Minima - Sibanand Pattnaik - Part 1

• Concept : If the sum of positive numbers are constant then their product is maximum when all numbers are equal.
Suppose that x, y, z . . . w are n positive variables and that c is a constant, then:If x + y + z + ... + w = c, the value of xyz ... w is greatest when x = y = z = ... = w = c/n.

a + b + c = 12 , find maximum value of a.b.c

Ans : 64

a + b + c =19, find maximum value of a.b.c

Ans 19^3/27

a + b + c = 7 ; find Maximum value of a^2 * b^3 * c^4

Method 1:
a/2 + a/2 + b/3 + b/3 +b/3 + c/4 + c/4 + c/4 + c/4 = 7
now a/2 = a/2 = b/3 = b/3 = b/3 = c/4 = c/4 = c/4 = c/4 = 7/9
===> a/2 × a/2 × b/3× b/3 × b/3 × c/4 × c/4 × c/4 × c/4 = (7/9)^9
===> a^2. b^3. c^4 = 2^2 × 3^3 ×4^4 × (7/9)^9

Method 2 :
AM ≥ GM
a/2 + a/2 + b/3 + b/3 +b/3 + c/4 + c/4 + c/4 + c/4 = 7
{a/2 + a/2 + b/3 + b/3 +b/3 + c/4 + c/4 + c/4 + c/4 } /9 = { a/2 × a/2 × b/3× b/3 × b/3 × c/4 × c/4 × c/4 × c/4 }^1/9
Now on simplification we will get a^2 * b^3 * c^4 ≤ 2^2 × 3^3 ×4^4 × (7/9)^9

Method 3( preferable)
Here Powers of a, b,c are 2,3,4 respectively
Divide 7 in the ratio 2:3:4
===> a =2/9 × 7; b= 3/9 × 7; c = 4/9 × 7
Now on substitution you will get a^2 * b^3 * c^4 = 2^2 × 3^3 ×4^4 × (7/9)^9

Method 4:
To find the greatest value a^m * b^n * c^p . . . when a + b + c + ... is constant; m, n, p, ... being positive integers.
Hence a^m *b^n * c^p ... will be greatest when the factors a/m , b/n , c/p are all equal, that is, when
a/m = b/n = c/p = ... = (a+b+c+…….)/(m +n +p+….)
here a/2 = b/3 =c/4 = 7/9
now substitute the value & u will get
a^2 * b^3 * c^4 = 2^2 × 3^3 ×4^4 × (7/9)^9

If a, b and c are positive variables and a + b + c = 12, find maximum value of (a + 1) × (b + 2) × c.

Here we need to find maximum value of (a + 1) × (b + 2) × c ; so we must know (a + 1) + (b + 2) + c
Now (a + 1) + (b + 2) + c = 15 ; so each term (a + 1) = (b + 2) = c = 5
So (a + 1) × (b + 2) × c = 125

If a + 2b + 3c = 9 find maximum value of a × b × c. (Given a, b and c are positive).

a = 2b =3c = 9/3 =3 ; so a * 2b * 3c = 27 ==> abc = 27/6

If a + 2b + 3c = 63, find maximum value of a^3 × b^5 × c. (Given a, b and c are positive).

a/3 = 2b /5 = 3c /1 = (a+2b+3c)/(3+5+1) = 63/9 = 7
==> a = 21; b= 35/2 ; c= 7/3
So a^3 × b^5 × c = 21^3 * (35/2)^ 5 * 7/3

Find the greatest value of (a + x)^3 * (a - x)^ 4 for any real value of x numerically less than a.

Here a + x = p; a- x =q ;
==> p + q = 2a & we r supposed to find p^3 * q^4
So p/3 = q/4 = 2a/7
==> p = 6a/7 & q = 8a /7

If 2a + 5b = 7; find the maximum value of (a+1)^2 * (b+2)^3

2a + 5b = 2(a+1) + 5 (b+2) = 2a+5b+12 = 19
2x + 5y = 19 & we need to find x^2 * y ^3
2x/2 = 5y/3 = 19/5 --
x = 19/5 & y = 57 /25
x^2 * y ^3 = (19/5)^2 * (57/25)^3
= 19^5 * 3^3/5^8

Concept: If the product of positive numbers are constant then sum is minimum when all numbers are equal.
If xyz...w = c, the value of x + y + ... w is least when x = y =... = w , so that the least value of x + y + ... + w is nc^1/n.

If a, b, c are positive and a × b × c = 125, find minimum value of a + b + c ?

Ans : 15

If a, b, c are positive and a × b × c = 17, find minimum value of a + b + c.

Ans: 3*(17)^1/3

Find the least value of 3x + 4y + 5z for positive values of x and y, subject to the condition xyz = 6.

To find the minimum value of 3x + 4y + 5z ; we must know 3x × 4y × 5z
So 3x × 4y × 5z = 3 * 4 * 5 * xyz = 360
So 3x + 4y + 5z = 3 × (360)^1/3 = 6×(45)^1/3

Find the least value of x + y + z for positive values of x, y & z subject to the condition x^2 * y^3 * z^4 = 6

Method 1:
x + y + z = x/2 + x/2 + y/3 +y/3 +y/3+ z/4+z/4+z/4+z/4 = ?
so we must know the product of
x/2 × x/2 × y/3 × y/3 ×y/3× z/4 × z/4 × z/4 × z/4 = (x^2×y^3×z^4)/ (2^2×3^3×4^4) = 6/(2^2×3^3×4^4)
so x +y + z = 9 ×{ 6/(2^2×3^3×4^4)}^1/9

Method 2:
AM ≥ GM
(x/2 + x/2 + y/3 +y/3 +y/3+ z/4+z/4+z/4+z/4)/9 ≥ (x/2 × x/2 × y/3 × y/3 ×y/3× z/4 × z/4 × z/4 × z/4)^1/9
So , x + y + z ≥ 9 ×{ 6/(2^2×3^3×4^4)}^1/9

Method 3
If x +y +z =k & x^2 * y^3 * z^4 = 6
So x/2 =y/3=z/4 = k/9
So x = 2k/9; y =3k/9; z = 4k/9
Now substitute the value of x,y & z in x^2 * y^3 * z^4 = 6
K = 9 ×{ 6/(2^2×3^3×4^4)}^1/9

Find the least value of 4x + 2y + 3z for positive values of x , y & z subject to the condition x^2 * y^3 * z^4 = 6.

If 4x +2y +3z =k & x^2 * y^3 * z^4 = 6
So 4x/2 =2y/3=3z/4 = k/9
So x = k/18; y =k/6; z = 4k/27
Now substitute the value of x,y & z in x^2 * y^3 * z^4 = 6

If a * b * c = 100 , where a,b,c are positive integers , find the maximum & minimum value of a + b + c.

We know that “if product is constant ,then sum is minimum when all no. are equal”
But here if u take all equal then it won’t be an integer.
So for the question based on integer --- if they cannot be equal then they must be very very close to each other , so 100 = 4 * 5 * 5
So minimum value of x + y + z = 4 + 5 + 5 = 14;
Now for maximum they must be far away from each other; so 100 = 1 * 1 * 100
So max of x + y + z = 1 + 1 + 100 = 102

If a + b + c = 100, where a, b and c are positive integers then find maximum & minimum value of a * b * c

We know that “if sum is constant ,then product is maximum when all no. are equal”
But here if u take all equal then it won’t be an integer.
So for the question based on integer --- if they cannot be equal then they must be very very close to each other , so 100 = 34 + 33 + 33
So maximum value of xyz = 34 * 33 * 33;
Now for minimum they must be far away from each other; so 100 = 1+1+98
So min of xyz = 1 * 1 * 98 =98

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