# MAXIMUM & MINIMUM VALUES OF ALGEBRAIC EXPRESSIONS/EQUATIONS

## MAXIMUM & MINIMUM VALUES OF ALGEBRAIC EXPRESSIONS/EQUATIONS

#### QUERY 1

**The minimum value of (x-2)(x-9) is?**

A) -11/4

B) 49/4

C) 0

D) -49/4

**MAHA GUPTA**

(x – 2)(x – 9) = x² – 11x + 18

We see that this expression is a quadratic one.

Note that in a quadratic expression when the coefficient of x² is greater than 0 the minimum value of it is (-D/4a) and maximum is infinitive & when the coefficient of x² is smaller than 0 the minimum value of it is infinitive and maximum is ( -D/4a)

Here, as the coefficient of x² is greater than 0, the minimum value of it

= ( -D/4a)

= -(b² – 4ac)/4a

= -(121 – 4*1*18)/4*1

= -49/4 (option ‘D’)

#### QUERY 2

A) 0

B) -1

C) -3

D) -2

**MAHA GUPTA**

You can do this sum with logical thinking just; means without paper pen. If you see only and only x = 0 will give you the minimum value of the given expression. (option ‘D’)

#### QUERY 3

**If x + y + z = 7 and xy + yz + zx = 10; find the maximum value of x?**

A) 2

B) 7

C) √29

D) 5

**MAHA GUPTA**

x + y + z = 7

Squaring both sides

(x + y + z)² = (7)²

=> x² + y² + z² + 2xy + 2yz +2zx = 49

=> x² + y² + z² +2(xy + yz + zx) = 49

=> x² + y² + z² +2(10) = 49 —-Replacing given value of xy + yz + zx

=> x² + y² + z² = 29

To know the maximum value of x², we must know the minimum values of y² and z². As both of them are in squares, they must not be less than ‘0’ each. So minimum value of y² and z² is ‘0’ each.

So, for maximum value, x²= 29

=> x = √29 or x < √29

But √29 is not satisfying the given equations, therefore x must be smaller than this.

We see only 5 among the given options satisfies this, so the answer (option ‘D’)

SHORT

Try options you’ll get 5 only

#### QUERY 4

**If a + b + c = 13. Find the maximum value of (a – 3)(b – 2)(c – 1)**

A) 27/343

B) 433/27

C) 344/27

D) 343/27

**KUMAR SAURABH**

First we convert the terms of the given equation as are in the find expression

Now, a + b + c = 13

=> (a – 3) + (b – 2) + (c- 1) = 13 – 6

=> (a – 3) + (b – 2) + (c – 1) = 7

Now terms are same, we know that the arithmetic mean of non-negative real numbers is greater than or equal to the geometric mean if terms/items of two things are same, and geometric mean of three numbers = the cubic root of their product

So by above, {(a – 3) + (b – 2) + (c – 1)}/3 > or = {(a – 3)(b – 2)(c – 1)}^{1/3}

=> 7/3 > or = {(a – 3)(b – 2)(c – 1)}^{1/3 }—putting (a – 3) + (b – 2) + (c – 1) = 7

=> {(a – 3)(b – 2)(c – 1)} < or = (7/3)³

Hence the maximum value of the find expression = (7/3)³ = 343/27 (option ‘D’)

#### QUERY 5

**x + y = 9; if x and y are real numbers find the maximum value of xy**

A) 20

B) 81/4

C) 81

D) 21

**MAHA GUPTA**

Though this question can be solved by AM, GM method also, here we do need not that. If sum of two numbers is there, the maximum value of their multiplication can only be when if both of them are equal. Here the sum is 9; so both of them should be 9/2.

Therefore maximum value of xy = (9/2)*(9/2) = 81/4 (option ‘B’)

ADDITIONAL

If the numbers are integers, they must be 4 and 5 for xy to be the maximum as the numbers needs to have as near as possible. So xy = 4*5 = 20