Maxima & Minima  Sibanand Pattnaik

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 : 64a + b + c =19, find maximum value of a.b.c
Ans 19^3/27a + 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)^9Method 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)^9Method 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)^9Method 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)^9If 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 = 125If 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 * 7/3Find 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 /7If 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 * 3^3/5^8Concept: 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 : 15If a, b, c are positive and a × b × c = 17, find minimum value of a + b + c.
Ans: 3*(17)^1/3Find 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)^1/3Find 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)}^1/9Method 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)}^1/9Method 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)}^1/9Find 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 = 6If abc = 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 = 102If a + b + c =100, where a,b,c are positive integers then find maximum & minimum value of abc
We know that “if sum is constant ,then product is maxiimum 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