ARITHMETICMATHSSIMPLIFICATION

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

1506846_281807341985154_437711047_n1

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

Previous post

QUESTIONS ON SIMPLIFICATION (PART-IX)

Next post

QUESTIONS ON TIME AND WORK (PART-I)

Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

1. Maha English Grammar (for Competitive Exams)
2. Maha English Practice Sets (for Competitive Exams)