Test: Minima & Maxima - 2 - Civil Engineering (CE) MCQ

Concept:

Minimum value = (4ac – b²)/4a

Calculation:

(x - 2)(x - 9)

⇒ x² - 11x + 18

We know ax² + bx + c = 0

Comparing this with above equation

⇒ a = 1, b = -11 and c = 18 

Minimum value = (4ac – b²)/4a

⇒ (4 × 1 × 18 - (-11)²)/(4 × 1)

⇒ (72 - 121)/4 

⇒ -49/4

∴ Minimum value of (x - 2)(x - 9) = -49/4

Given:

Expression: 1/(2x² + 5x + 11)

Concept:

If we get minimum value of expression (2x² + 5x + 11) then we can find maximum value of the expression 1/(2x² + 5x + 11)

Formula Used:

Minimum value of expression (ax² + bx + c) is = [4ac – b²]/4a

Calculation:

Minimum value of the expression 2x² + 5x + 11

⇒ [4 × 2 × 11 – 5²]/[4 × 2]

⇒ [88 – 25]/8

⇒ 63/8

Hence, Maximum value of the expression 1/(2x² + 5x + 11) = 1/(63/8) = 8/63

Given: 

Equation (x + 8) (7 - x)

Concept:  

When, a > 0 gives minimum value, and a < 0 gives maximum value 

For both value be using, 4ac−b² / 4a

Calculation: 

⇒ (x + 8) (7 - x) = - x² - x + 56

If we compare it with the equation ax² + bx + c

then, a = -1, b = -1, c = 56

⇒ 

Hence, the required value is 225/4.

Given:

x² + y² = 1.

x² + 4xy - y²

Concept Used:

The maximum value of a sin θ + b cos θ is .

Calculation:

Since, x² + y² = 1, we can suppose x = cos θ and y = sin θ.

Now, x² + 4xy - y²

= cos2 θ + 4 sin θ cos θ - sin2 θ

= (cos2 θ - sin2 θ) + 4 sin θ cos θ

= cos 2θ + 2 sin 2θ

Its maximum value is 

= √5.

Given:

x is the real number.

5x² - 2x + 7

Concept used:

In Quadratic Polynomial ax² + bx + c

When, a > 0

Minimum value = (4ac - b²)/4a      

Calculation:

5x² - 2x + 7

Compare the above equation with quadratic polynomial ax² + bx + c.

a = 5, b = -2 and c = 7

Now,

Minimum value = (4ac - b²)/4a 

⇒ Mininmum value = (4 × 7 × 5 - 4)/20

⇒ Minimum value = 136/20

⇒ Mininmum value = 34/5

∴ 34/5 is the minimum value of 5x2 - 2x + 7. 

Given:

We have the given expression 4x² - 5x + 1 = 0

Concept used:

The minimum value of any quadratic expression ax² + bx + c = 0, (when a > 0) is -D/4a

Where, D = b² - 4ac

Calculation:

Here, a, b and c is 4, -5 and 1 respectively

So, D = (-5)² - (4 × 4 × 1)

⇒ 25 - 16

⇒ 9

So, minimum value = (-9)/(4 × 4)

⇒ (-9)/16

∴ The minimum value of given expression is -9/16.

Given:

The Polynomial = 1/(3x² + 4x + 8)

Concept used:

If we get minimum value of expression (3x² + 4x + 8), then we can find maximum value of the expression 1/(3x² + 4x + 8)

Formula used:

If polynomial ax² + bx + c, When, a > 0

then Minimum value of polynomial = (4ac – b²)/4a

Calculation:

If we compared (3x² + 4x + 8) with ax² + bx + c, then

a = 3, b = 4 and c = 8

Minimum value of the expression 3x² + 4x + 8

⇒ {(4 × 3 × 8) - (4)²}/(4 × 3)

⇒ (96 - 16)/12

⇒ 80/12

⇒ 20/3

Maximum value of the expression 1/(2x2 + 5x + 11) = 1/(20/3) = 3/20

∴ The maximum value of the expression 1/(3x² + 4x + 8) is 3/20.

Given:

(x + y + z) = 22

Concept:

Arithmetic Mean > Geometric Mean

Calculation:

(x + y + z) = 22

⇒ (x – 5) + (y – 2) + (z – 6) = 22 – 5 – 2 – 6

⇒ (x – 5) + (y – 2) + (z – 6) = 9

Since, AM > GM

[(x – 5) + (y – 2) + (z – 6)]/3 > ³√[(x – 5)(y – 2)(z – 6)]

⇒ ³√[(x – 5)(y – 2)(z – 6)] < [(x – 5) + (y – 2) + (z – 6)]/3

⇒ ³√[(x – 5)(y – 2)(z – 6)] < 9/3

⇒ ³√[(x – 5)(y – 2)(z – 6)] < 3

⇒ [(x – 5)(y – 2)(z – 6)] < 3³

⇒ [(x – 5)(y – 2)(z – 6)] < 27

∴ The maximum value of (x – 5)(y – 2)(z – 6) will be 27.

Given:

(- x² + 3x + 7)

Concept used:

dy/dx = 0, the value of x gives the minimum value when d2y/dx2 is greater than 0, and gives the maximum value when d2y/dx2 is less than 0

Calculation:

(- x2 + 3x + 7)      ----(i)

Differentiating (i),

dy/dx = 0

⇒ - 2x + 3 = 0      ----(ii)

⇒ x = 3/2

By double differentiating equation (ii),

d²y/dx² = - 2 < 0

That means the maximum value of equation (i) is at x = 3/2

Maximum value of equation (i)

⇒ - 9/4 + 9/2 + 7

⇒ (- 9 + 18 + 28)/4

⇒ 37/4

∴ The maximum value is 37/4.

Given:

x is real.

-3x² + 5x + 10

Concept used:

For quadratic polynomial ax² + bx + c, When a < 0

Maximum value = (4ac - b²)/4a

Calculation:

-3x² + 5x + 10

Compare the above equation with quadratic polynomial ax² + bx + c.

a = -3, b = 5, c = 10

Here, a < 0 so

Maximum value = (4ac - b²)/4a

⇒ Maximum value = {4 × (-3) × 10 - 25)}/(-12)

⇒ Maximum value = {-145}/(-12)

⇒ Maximum value = 145/12

∴ 145/12 is the maximum value of -3x² + 5x + 10.