# 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*=

*x*

^{2}– 5

*x*+ 6.

*Graphical method*

As we have seen in the concepts of quadratic equation, graph of

So, the minimum value of

*y*=*ax*^{2}+*bx*+*c*= 0 will have its minimum value at*x*= and*y*_{min}= .So, the minimum value of

*y*=*x*^{2}– 5*x*+ 6 will be*y*_{min}= =*Quadratic equation method*

Suppose we have find the minimum value of
We can find the minimum value by breaking this equation into the sum of a whole square and whatever is left i.e., P

*y*=*x*^{2 }– 5*x*+6^{2}+ Q.*x*^{2}– 5*x*+ 6 = [*x*^{2}– 2.*x*.5/2 + (5/2)^{2}] – ¼=

We know that the minimum value ofis equal to zero, so the minimum value of

*y*=Example-1

Find the minimum value of

*y*=(*x*−*a*) (*x*−*b*).- ab
- 0

Solution

*x*

^{2}–

*x*(

*a*+

*b*) +

*ab*

=

=

Now we know that the minimum value of any perfect square is 0.

So, the minimum value of

*y*=

# 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
Maximum value Obviously, the maximum value can be extended upto +∝.
Minimum value In this case, all the three parts of
To obtain critical points, put all the three components |

*y*=*f*(*x*) = |*x*+ 3| + |*x*− 5| + |*x*− 7|*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.*x*+ 3|, |*x*− 5| and*x*− 7| one by one equal to zero.|

*x*+ 3| = 0, so,*x*= −3|

*x*− 5| = 0, so,*x*= 5and |

Now putting these values in

*x*− 7| = 0, so,*x*= 7Now putting these values in

*f*(*x*) gives us the following result:At

*x*= −3,*f*(*x*) = 18At

*x*= 5,*f*(*x*) = 10At

*x*= 7,*f*(*x*) = 12So, the minimum value of

*f*(

*x*) = 10

Alternatively, drawing the graph gives us the clear picture regarding the movement of the graph which gives us the minimum value of
The dotted lines in the above graph show the movement of the graph. Obviously, at
The inspection of the graph gives us the maximum value of

*f*(*x*).*x*= 5, graph is at its lowest point which is the minimum value of the graph.*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- 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.- 8
- 16
- 20
- 24

Solution

The maximum value of

*xy*will occur when*x*=*y*= 4So, the maximum value of

*xy*= 4 × 4 = 16To 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*= −∝- 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*(*x*) =*x*> 0?Solution

Since

Now, the product of all the terms

Now, the product of all the terms

*x*^{2}, 1, 1/*x*and 1/*x*is 1, the sum of these terms will have the minimum value for*x*^{2}= 1 = 1/*x*i.e., for*x*= 1. Hence, the minimum value of*f*(*x*) is 4.Example-4

If

*ab*^{2}*c*^{3}= 27 × 2^{8}, then find the minimum value of*a*+*b*+*c*(given that*a, b, c*> 0).Solution

To find the minimum value of
Assuming
Thus,
Now,
Corollary For any positive value of

*a*+*b*+*c*, we need to know the product of*a, b*and*c*.*a*=*x*,*b*= 2*y*and*c*= 3*z*, we have*x*× 4*y*^{2}× 27*z*^{3}= 27 × 2^{8}. So now we have to calculate the minimum value of*x*+ 2*y*+ 3*z*.*ab*^{2}*c*^{3}= 27 × 2^{8}⇒*x*×*y*^{2}×*z*^{3}= 2^{6}.*x*+ 2*y*+ 3*z*=*x*+*y*+*y*+*z*+*z*+*z*and we know that*x*×*y*×*y*×*z*×*z*×*z*= 2^{6}. So, the least value will occur when all are equal to 2. Thus, the least sum will be 12.*x*, the minimum value of the sum of*x*and its reciprocal will be 2.Example-5

If

*a*> 1,*b*> 1 then the minimum value of log_{b}*a*+ log_{a}*b*is- 0
- 1
- 2
- None of these

Solution

Assume log
So, the minimum value of (log

_{a}*b*=*N*, so log_{b}*a*= 1/*N*_{b}*a*+ log_{a}*b*) = 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- AM ≥ GM ≥ HM
- (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*)?- 0 ≤ N ≤ 1
- 1 ≤ N ≤ 2
- 2 ≤
*N*≤ 3 - 0 ≤
*N*≤ 1

Solution

Using AM ≥ GM

[(

*a + b*) + (*c + d*)] ≥ [(*a + b*)]^{1/2 }= (2) ≥ (*N*)^{1/2}Also, (

So, 0 ≤ *a*+*b*) (*c*+*d*) ≥ 0*N*≤ 1