Maxima & Minima - Sibanand Pattnaik


  • QA/DILR Mentor | Be Legend


    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 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 = 102

    If 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


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