Minima and Maxima

Today, I take up another one of the sought after topics of the MBA aspirants which has been appearing in examinations regularly if not frequently. Minima - Maxima concept is primarily based upon the concept of AM >= GM and can be seen as one of the applications of AM-GM inequality.So before moving along, I will throw some light on this particular inequality but would keep it brief to suit the context of our discussion today.

AM : Arithmetic Mean 
GM : Geometric Mean
AM >= GM basically states that AM of a list of non-negative real numbers is greater than or equal to the GM of the same list while AM = GM holding true only when all numbers in the list are equal.

Now with that in mind, we can begin with the concept of maxima and minima. I will take up examples and discuss approaches to it which in turn will clear the concept as a whole.

If  sum of positive numbers is constant then product is maximum when all numbers are equal

Q1 If a + b + c = 7 where a,b,c are positive real numbers. Find the maximum value of a^2.b^3.c^4

Method 1 : AM >= GM

[(a/2 + a/2) + (b/3 + b/3 + b/3) + (c/4 + c/4 + c/4 + c/4)]/9 >= [ (a^2.b^3.c^4)/(2^2.3^3.4^4) ]^(1/9)
a^2.b^3.c^4 <= 2^2.3^3.4^4.7^9/9^9 a^2.b^3.c^4 <= 2^10.7^9/3^15


Method 2 : Making all terms equal

a/2 = a/2 = b/3 = b/3 = b/3 = c/4 = c/4 = c/4 = c/4 = 7/9
a^2.b^3.c^4 <= 2^2.3^3.4^4.(7/9)^9


Method 3 : Ratio distribution

Powers of a,b,c are in ratio 2 : 3 : 4
Divide 7 in ratio 2 : 3 : 4
a = (2/9)*7 ; b = (3/9)*7 ; c = (4/9)*7
a^2.b^3.c^4 <= 2^2.3^3.4^4.(7/9)^9

Method 4 : Recommended Method

a/(power of a) = b/(power of b) = c/(power of c) = (a + b + c)/(Sum of their powers) = 7/9
a = 2*7/9 ; b = 3*7/9 ; c = 4*7/9
Substitute in product to get :
a^2.b^3.c^4 <= 2^2.3^3.4^4.(7/9)^9


Q2 Find the maximum value of (a + x)^3 * (a - x)^4 for x < a
Let p = a + x , q = a - x
p + q = 2a. Need to find maximum of p^3 * q^4
p/3 = q/4 = 2a/7
p = 6a/7 , q = 8a/7
p^3 * q^4 has a maximum value of : (6a/7)^3 * (8a/7)^4 by Method 4.

Q3 If  2a + 5b = 7 where a and b are positive real numbers. Find the maximum value of
(a + 1)^2 * (b + 2)^3
2(a + 1) + 5(b + 2) = 7 + 2 + 10 = 19
Let p = (a + 1) , q = (b + 2)
2p/2 = 5q/3 = 19/5
p = 19/5 , q = 57/25
p^2 * q^3 has a maximum value of : (19/5)^2 * (57/25)^3


If product of positive numbers is constant then their sum is minimum when the numbers are equal

Q1 If a, b, c are positive and a × b × c = 17, find minimum value of a + b + c.

All above four methods can be applied assuming the required minimum sum as "k" and solving as above.

Method 1 :
k/3 >= abc^(1/3)
k >= 3*17^(1/3)

Method 4 :
a/1 = b/1 = c/1 = k/3
a = b = c = k/3
abc <= k^3/27
k >= 3*17^(1/3)

and similarly with other two methods.


Q2 Find the least value of 3x + 4y + 5z for positive values of x and y, subject to the condition xyz = 6.

Method 1 : (3x + 4y + 5z)/3 >= (60xyz)^(1/3)
k >= 6*45^(1/3)

Method 4 :
3x/1 = 4y/1 = 5z = k/3
x = k/9 , y = k/12 , z = k/15
k/3 >= (60xyz)^(1/3)
k >= 6*45^(1/3)

Q3 Find the least value of 4x +2 y + 3z for positive values of x , y &z subject to the condition
x^2*y^3*z^4 = 6.

Method 4 :
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
k = 6× (6 × 18^2 × 6^3 × (27/4)^4 }^1/9


Extension of the concept to integers

While considering integers, if making all numbers equal is not a possibility, we take them as close as possible to near the equality in AM >= GM.

Q1 If abc = 100 , where a,b,c are positive integers , find the maximum & minimum value of a + b + c.

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

Q2 If a + b + c =100, where a,b,c are positive integers then find maximum & minimum value of abc.

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


That will be all for this article piece on minima and maxima. Minima and Maxima also has extensions in quadratic equations and other composite forms of quadratic which we would look into some other day. Clarity on this article will also help you in other topics and hence was taken up before moving forward to other concepts.


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Cheers!
AS