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Maxima and Minima of Quadratic Equation

The standard methods of finding the maxima and the minima of all the quadratic equations, are as follows:

Let us discuss these methods with respect to a particular equation y = x2 – 5x + 6.
  1. Graphical method
As we have seen in the concepts of quadratic equation, graph of y = ax2 + bx + c = 0 will have its minimum value at x = Description: 3184.png and ymin = Description: 3193.png.
So, the minimum value of y = x2 – 5x + 6 will be ymin = Description: 3202.png=Description: 3206.png
  1. Quadratic equation method
Suppose we have find the minimum value of y = x2 – 5x +6
 
We can find the minimum value by breaking this equation into the sum of a whole square and whatever is left i.e., P2 + Q.
 
x2 – 5x + 6 = [x2 – 2.x.5/2 + (5/2)2] – ¼ 
Description: 3215.png
 
We know that the minimum value ofDescription: 3224.pngis equal to zero, so the minimum value of y = Description: 3233.png
 
Example-1
Find the minimum value of y=(xa) (xb).
  1. ab
  2. Description: 3242.png
  3. 0
  4. Description: 3251.png
Solution
x2 – x(a + b) + ab
=Description: 3260.png 
Description: 3269.png
Now we know that the minimum value of any perfect square is 0.
So, the minimum value of y = Description: 3278.png
 

Maxima and Minima of any Modulus

We have seen above how to find the maxima and the minima of any modulus. Here we will discuss some more situations involving modulus. Let us see this with the help of an example:
 
Finding the maximum and the minimum value of y = f(x) = |x + 3| + |x − 5| + |x − 7|
 
Maximum value Obviously, the maximum value can be extended upto +.
 
Minimum value In this case, all the three parts of f(x) i.e., |x + 3|, |x − 5| and |x − 7| can not be simultaneously equal to zero. So, we are required to find the critical points here at first, and then these points in f(x) to see that which one gives us the minimum value.
 
To obtain critical points, put all the three components |x + 3|, |x − 5| and x − 7| one by one equal to zero.
|x + 3| = 0, so, x = −3
|x − 5| = 0, so, x = 5
and |x − 7| = 0, so, x = 7
 

Now putting these values in
f(x) gives us the following result:
At x = −3, f(x) = 18
At x = 5, f(x) = 10
At x = 7, f(x) = 12
 

So, the minimum value of f(x) = 10
 
Description: 11-11.tif
 
Alternatively, drawing the graph gives us the clear picture regarding the movement of the graph which gives us the minimum value of f(x).
 
The dotted lines in the above graph show the movement of the graph. Obviously, at x = 5, graph is at its lowest point which is the minimum value of the graph.
 
The inspection of the graph gives us the maximum value of f(x) which is +.

Finding the Maximum or the Minimum Value of the Product

Finding the maximum or the minimum value of the product/sum of two or more than two variables when the sum/product of these variables is given
  1. If sum of two or more than two variables is given, then there product will be the maximum when the value of all the variables are equal.
Example-2
Find the maximum and the minimum value of xy subject to x + y = 8.
  1. 8
  2. 16
  3. 20
  4. 24
Solution
The maximum value of xy will occur when x = y = 4
 
So, the maximum value of xy = 4 × 4 = 16
 
To obtain the minimum value, we can take either x or y to be negative, and then the product of xy will be negative. And this process continues till −∝. So, the minimum value of xy = −∝
 
  1. If the product of two or more than two positive variables is given, then there sum will be the minimum when the value of all the variables are equal.

Example-3
What is the minimum value of f(xDescription: 3287.pngx > 0?
Solution
Since Description: 3296.png
Now, the product of all the terms x2, 1, 1/x and 1/x is 1, the sum of these terms will have the minimum value for x2 = 1 = 1/x i.e., for x = 1. Hence, the minimum value of f(x) is 4.
 
 
Example-4
If ab2c3 = 27 × 28, then find the minimum value of a + b + c (given that a, b, c > 0).
Solution
To find the minimum value of a + b + c, we need to know the product of a, b and c.
 
Assuming a = xb = 2y and c = 3z, we have x × 4y2 × 27z3 = 27 × 28. So now we have to calculate the minimum value of x + 2y + 3z.
 
Thus, ab2c3 = 27 × 28 ⇒ x × y2 × z3 = 26.
 
Now, x + 2y + 3z = x + y + y + z + z + z and we know that x × y × y × z × z × z = 26. So, the least value will occur when all are equal to 2. Thus, the least sum will be 12.
 
Corollary For any positive value of x, the minimum value of the sum of x and its reciprocal will be 2.
Description: 3305.png 
 

Example-5
If a > 1, b > 1 then the minimum value of logba + logab is
  1. 0
  2. 1
  3. 2
  4. None of these
Solution
Assume logab = N, so logba = 1/N
 
So, the minimum value of (logba + logab) = minimum value of (N + 1/N) = 2
 

Using Arithmetic Mean, Geometric Mean

Using arithmetic mean, geometric mean and harmonic mean to find the maxima and the minima
  1. AM ≥ GM ≥ HM
  2. (GM)2 = AM × HM (for two numbers only)
Example-6
If a, b, c and d are positive real numbers such that a + b + c + d = 2, then which of the following is true regarding the values of N = (a + b) (c + d)?
  1. 0 ≤ N ≤ 1
  2. 1 ≤ N ≤ 2
  3. 2 ≤ N ≤ 3
  4. 0 ≤ N ≤ 1
Solution
Using AM ≥ GM
 
Description: 3314.png[(a + b) + (c + d)] ≥ [(a + b)]1/2 Description: 3323.png(2) ≥ (N)1/2
 
Also, (a + b) (c + d) ≥ 0
 
So, 0 ≤ N ≤ 1
 





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