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Algebra

Inequalities

ALL MODULES

CAT 2025 Lesson : Inequalities - Max & Min for Range & Substitution

6.3 Maximum and Minimum given the range of variables

In these questions we need to take the minimum and maximum of each variable, as the case may be, to ascertain the minimum or maximum value of the expression provided.

Example 18

Given that

-1 \le v \le 1

-2 \le u \le -0.5

and

-2 \le z \le -0.5

and

w = vz/u

, then which of the following is necessarily true?
[CAT 2003L]

(1)

-0.5 \le w \le 2

(2)

-4 \le w \le 4

(3)

-4 \le w \le 2

(4)

-2 \le w \le 0.5

Solution

w = \dfrac{vz}{u}

z

and

u

are always negative,

\dfrac{z}{u}

will always be positive.

v

can take minimum and positive values.

To maximise and minimise

\dfrac{vz}{u}

, we find the largest

\dfrac{z}{u}

and multiply by the largest and smallest values of

v

respectively.

Largest value of

\dfrac{z}{u} = \dfrac{-2}{-0.5} = 4

Maximum value of

\dfrac{vz}{u} = 1 \times 4 = 4

Minimum value of

\dfrac{vz}{u} = -1 \times 4 = -4

\therefore

The range of

w

-4 \le w \le 4

Answer: (2)

-4 \le w \le 4

Example 19

Where

a, b

and

c

are real numbers such that

1 < a < 4

2 < b < 10

and

4 < c < 8

, then what is the

(I) minimum possible integral value of

\dfrac{bc}{a}

?

(II) maximum possible integral value of

\dfrac{abc - c}{a}

Solution

It is to be noted that

a, b

and

c

are real numbers and the ranges provided between integers, which are not included. Therefore, we can find the minimum or maximum possible integral values for the expressions by using these integral end points and finally adding 1 or subtracting 1 from them respectively.

Case I: To minimise

\dfrac{bc}{a}

, the numerator

bc

should be minimised while denominator

a

should be maximised.

If end points are included,

Minimum value of

bc = 2 \times 4 =8

Maximum value of

a = 4

Minimum value

\dfrac{bc}{a} = \dfrac{8}{4} = 2

When end points are not included, minimum integral value of

\dfrac{bc}{a}

=

2 + 1

=

3

Case II:

\dfrac{abc - c}{a} = bc - \dfrac{c}{a} = c \left( b - \dfrac{1}{a} \right)

To maximise this, we need to take the maximum values for

c

and

b

and take the minimum value for

\dfrac{1}{a}

(as it is being subtracted).

If end points are included,
Maximum value of

c \left( b - \dfrac{1}{a} \right) = 8 \left( 10 - \dfrac{1}{4} \right) = 8 \times \dfrac{39}{4} = 78

When end points are not included, maximum integral value of

\dfrac{abc - c}{a} = 78 - 1 = 77

Answer: (I)

3

; (II)

77

6.4 Substituting Values

Quite a few questions from this lesson will have variables in the questions and answer options. In such questions, substituting values will help save time.

Example 20

Which of the following is the largest if

p > q > r > s > x

, where

x

is a positive real number.

(1)

rs(p - x)(q - x)

(2)

sp(q - x)(r - x)

(3)

qr(s - x)(p - x)

(4)

pq(r - x)(s - x)

Solution

Let

p, q, r, s

and

x

5, 4, 3, 2

and

1

respectively.

Option (1):

rs(p - x)(q - x) = 3 \times 2 \times 4 \times 3 = 72

Option (2):

sp(q - x)(r - x) = 2 \times 5\times 3 \times 2 = 60

Option (3):

qr(s - x)(p - x) = 4 \times 3 \times 1 \times 4 = 48

Option (4):

pq(r - x)(s - x) = 5 \times 4 \times 2 \times 1 = 40

\therefore

rs(p - x)(q - x)

is the largest.

Alternatively

When a constant is subtracted from a number, then larger the number, lower the percentage reduction. Therefore,

x

should be subtracted from the largest two terms –

p

and

q

. Therefore, option (1) is the correct choice.

Answer: (1)

rs(p - x)(q - x)

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