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Test: Minima & Maxima - 1 - Civil Engineering (CE) MCQ


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15 Questions MCQ Test Engineering Mathematics - Test: Minima & Maxima - 1

Test: Minima & Maxima - 1 for Civil Engineering (CE) 2024 is part of Engineering Mathematics preparation. The Test: Minima & Maxima - 1 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Minima & Maxima - 1 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Minima & Maxima - 1 below.
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Test: Minima & Maxima - 1 - Question 1

If x ∈ R, then minimum and maximum limit of  are

Detailed Solution for Test: Minima & Maxima - 1 - Question 1

Concept : 

Maximum and Minimum Values of the Quadratic Expression :

Let's suppose y = ax2 + bx + c ,

when determining the maximum and minimum values of ax2 + bx + c.

suppose x is real ,then the descriminant of equation 

ax2 + bx + c - y = 0

⇒ b2 - 4a(c - y) = 0

Calculation :

Let,

yx2 + 2xy + 9y = x2 - 2x + 9

x2 ( y - 1 ) + x ( 2y + 2 ) + 9 ( y - 1 ) = 0

Now,

D = 0

b2 - 4ac = 0

⇒ ( 2y + 2 ) - 4( y - 1 ) ×9 ( y - 1 ) = 0

4y2 + 4 + 8y - 36y2 - 36 + 72y = 0

32y2 - 80y + 32 = 0

2y2 - 5y + 2 = 0

2y ( y - 2 ) - 1 ( y - 2 ) = 0

( y - 2 ) ( 2y - 1 ) = 0

⇒ y = 2 and , y = 1/2

Thus,

maximum value of the function = 2

minimum value of the function = 12

Hence, The correct option is ( 3 ).

Test: Minima & Maxima - 1 - Question 2

If 3 ≤ x ≤ 10 and 5 ≤ y ≤ 15, then maximum value of (x / y) is-

Detailed Solution for Test: Minima & Maxima - 1 - Question 2

Given:

x = [3,10] & y = [5, 15]

Calculation:

For maximum value of x/y; we will take x maximum value and y minimum value.

∴ Max. value of (x/y) = 10/5 = 2

Test: Minima & Maxima - 1 - Question 3

A manufacturing company has its variable cost given by C(x) = x(80 - x2) where 'x' is the number of quantities produced. The price per unit is p(x) = 5x where 'x' is the number of units in demand. The ratio of marginal cost and marginal revenue when 5 units were both produced and in demand - 

Detailed Solution for Test: Minima & Maxima - 1 - Question 3

Concept:

Marginal cost is the instantaneous rate of change of cost function with respect to the production output x.

Marginal cost =

Marginal revenue is the instantaneous rate of change of revenue function with respect to the production output x.

Marginal revenue =

Revenue function R(x) = xp(x)

Calculation:

Given:

Variable cost function C(x) = x(80 - x2)

Price per unit p(x) = 5x 

Marginal cost = 

⇒ C'(x) = 80 - 3x2

⇒ Marginal cost at (x = 5) = 80 - 3(5)2 = 5

Revenue function R(x) = x p(x)

⇒ R(x) = x(5x)

⇒ R(x) = 5x2

Marginal revenue,

⇒ R'(x) = dR(x)dx 

⇒ R'(x)= d(5x2)dx=10x

⇒ Marginal revenue at (x = 5) = 10 × 5 = 50

Therefore ratio of marginal cost and marginal revenue = 5 : 50 = 1 : 10

Test: Minima & Maxima - 1 - Question 4

If f(x) = |x + 1| + |x + 10|, then find the minimum value of f(x).

Detailed Solution for Test: Minima & Maxima - 1 - Question 4

Calculation:

f(x) = |x + 1| + |x + 10|

|x + 1| = 0 then x = -1

|x + 10| = 0 then x = -10

f(x) can be zero when x = -1 or x = -10

f(-1) = |-1 + 1| + |-1 + 10| = 9

f(-10) = |-10 + 1| + |-10 + 10| = 9

∴ Minimum value of f(x) = 9 at x = -1 or x = -10

Test: Minima & Maxima - 1 - Question 5

If the minimum value of expression x2 + 5x + p = 0 is (-1/4), then find the value of p.

Detailed Solution for Test: Minima & Maxima - 1 - Question 5

Given:

We have the given expression x2 + 5x + p = 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 1, 5 and p respectively

So, D = (5)2 - (4 × 1 × p)

⇒ 25 - 4p

According to the question, we have

minimum value = -(25 - 4p)/(4 × 1)

⇒ -(25 - 4p)/4 = (-1/4)

⇒ 25 - 4p = 1

⇒ 4p = 25 - 1

⇒ p = 24/4

⇒ p = 6

∴ The required value of p is 6.

Test: Minima & Maxima - 1 - Question 6

Find the minimum value of the expression 3x2 + 6x + 6.

Detailed Solution for Test: Minima & Maxima - 1 - Question 6

Concept used:

Min value of a quadratic equation ax2 + bx + c is c – (b2/4a) or –D/4a, where D = b2 – 4ac

Calculation:

Method 1

Min value = c – b2/4a

⇒ Min value = 6 – [62/(4 × 3)]

∴ Min value = 3

Method 2

3x2 + 6x + 6 = 3 × (x2 + 2x + 2)

⇒ 3 × (x2 + 2x + 1 + 1)

⇒ 3 × ((x + 1)2 + 1)

Minimum value of the expression is at x = −1

⇒ Min value = 3 × (0 + 1)

∴ Min value is 3

Test: Minima & Maxima - 1 - Question 7

A rectangle is given with a perimeter of 48 cm. If the rectangle encloses maximum area possible, then the area of the rectangle will be

Detailed Solution for Test: Minima & Maxima - 1 - Question 7

Concept:

For maximum value of A, dA/dx = 0;

Calculation:

Given rectangle perimeter = 48 cm;

Let the sides of rectangle be x, y;

⇒ 2x + 2y = 48 cm ⇒ x + y = 24 cm;

Area of the rectangle will be A = xy;

For area to be maximum, dA/dx = 0

We have y = A/x ⇒ x + A/x = 24;

⇒ x2 – 24x = A

dA/dx = 0 ⇒ 2x – 24 = 0

⇒ x = 12 cm ⇒ y = 12 cm;

Now the maximum area will be 12 × 12 = 144 cm2;

Test: Minima & Maxima - 1 - Question 8

Find the difference between the maximum value of -3x2 + 2x + 13 = 0 and the minimum value of x2 - 2x + 3 = 0.

Detailed Solution for Test: Minima & Maxima - 1 - Question 8

Given:

-3x2 + 2x + 13 = 0

x2 - 2x + 3 = 0

Concept used:

For quadratic polynomial ax2 + bx + c

1) When a > 0  

Minimum value = (4ac - b2)/4a  

2) When a < 0

Maximum value = (4ac - b2)/4a  

Calculation:

For equation -3x2 + 2x + 13 = 0

Compare the above equation with quadratic polynomial ax2 + bx + c = 0

a = -3, b = 2, c = 13

Here, a < 0

Maximum value = (4ac - b2)/4a  

⇒ Maximum value = {(4) × (-3) × 13 - 4}/(-12)

⇒ Maximum value = { -39 - 1}/(-3)

⇒ Maximum value = 40/3      ----(1)

Now for equation x2 - 2x + 3 = 0

Compare the above equation with quadratic polynomial ax2 + bx + c = 0

a = 1, b = -2 and c = 3

Here, a > 0

Minimum value = (4ac - b2)/4a  

⇒ Minimum value = (4 × 1 × 3 - 4)/4

⇒ Minimum value = 2      ----(2)

Difference between maximum and minimum value 

⇒ (40/3) - 2

⇒ 34/3

∴ The difference between the maximum and the minimum value is 34/3.

Test: Minima & Maxima - 1 - Question 9

If 2x−1/3 + 2x1/3 = 5, then find the value of x.

Detailed Solution for Test: Minima & Maxima - 1 - Question 9

Given, 2x−1/3 + 2x1/3 = 5

Npw, put x1/3 = y

∴ 2y2 – 5y + 2 = 0

⇒ (y - 2) (2y - 1)

⇒ y = 2 and y = ½

Case I

⇒ y = 2 and y = x1/3

⇒ x1/3 = 2

⇒ x = 8

Case II

⇒ y = 1/2 and y = x1/3

⇒ x1/3 = 1/2

⇒ x = 1/8

∴ The values of x are 8 and 1/8

Test: Minima & Maxima - 1 - Question 10

If x is real, then find the minimum value of (3x2 - 2x + 8).

Detailed Solution for Test: Minima & Maxima - 1 - Question 10

 

Given:

Quadratic function 3x2 - 2x + 8

where, x is the real number

Concept used:

In Quadratic Polynomial ax2 + bx + c

If, a > 0, then

Minimum value = (4ac - b2)/4a      

Calculation:

3x2 - 2x + 8

If we compared (3x2 - 2x + 8) with quadratic polynomial ax2 + bx + c.

Then, a = 3, b = -2 and c = 8

Minimum value = (4ac - b2)/4a 

⇒ Mininmum value = {4 × 3 × 8 - (-2)2}/(4 × 3)

⇒ Minimum value = 92/12

⇒ Mininmum value = 23/3

∴ The minimum value of the quadratic polynomial ax2 + bx + c is 23/3.

Test: Minima & Maxima - 1 - Question 11

Find the minimum value of (2x + 1)(x + 3).

Detailed Solution for Test: Minima & Maxima - 1 - Question 11

Given:

(2x + 1)(x + 3)

Concept used:

In Quadratic Polynomial ax2 + bx + c

When, a > 0

Minimum value = (4ac - b2)/4a    

Calculation:

(2x + 1)(x + 3)

⇒ 2x2 + 6x + x + 3

⇒ 2x2 + 7x + 3

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

a = 2, b = 7 and c = 3

Here, a > 0

Minimum value = (4ac - b2)/4a    

⇒ Minimum value = ( 4 × 2 × 3 - 49)/8

⇒ Minimum value = (24 - 49)/8

⇒ Minimum value = -25/8

∴ -25/8 is the minimum value of (2x + 1)(x + 3).

Test: Minima & Maxima - 1 - Question 12

Find the minimum value of (2x - 3)(x - 5).

Detailed Solution for Test: Minima & Maxima - 1 - Question 12

Given:

(2x - 3)(x - 5)

Concept used:

For a polynomial ax2 + bx + c

If, a > 0, then

Minimum value = (4ac - b2)/4a    

Calculation:

(2x - 3)(x - 5)

⇒ 2x2 - 10x - 3x + 15

⇒ 2x2 - 13x + 15

If we compared (2x2 - 13x + 15) with ax2 + bx + c, then

a = 2, b = -13, c = 15

According to the concept,

Minimum value = (4ac - b2)/4a    

⇒ Minimum value = {(4 × 2 × 15) - (-13)2}/8

⇒ Minimum value = (120 - 169)/8

⇒ Minimum value = -49/8

∴ The minimum value of the quadratic function (2x - 3)(x - 5) is -49/8.

Test: Minima & Maxima - 1 - Question 13

If x is real, find the minimum value of (2x2 + 5x + 7)

Detailed Solution for Test: Minima & Maxima - 1 - Question 13

Given:

(2x2 + 5x + 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:

(2x2 + 5x + 7)      ----(i)

Differentiating (i),

dy/dx = 0

⇒ 4x + 5 = 0      ----(ii)

⇒ x = - 5/4

By double differentiating equation (ii),

d2y/dx2 = 4 > 0

That means the minimum value of equation (i) is at x = - 5/4

Minimum value of equation (i)

⇒ 25/8 - 25/4 + 7

⇒ (25 - 50 + 56)/8

⇒ 31/8

∴ The minimum value is 31/8.

Test: Minima & Maxima - 1 - Question 14

If N is a four digit number formed by digits x1, x2, x3 and x4, then maximum value of  is-

Detailed Solution for Test: Minima & Maxima - 1 - Question 14

Calculation:

Here,

Number will be maximum when denominator minimum.

∴ Minimum value of denominator = 1

∴ x1 = 1, x2 = 0, x3 = 0, x4 = 0

∴ 1000/(1 + 0 + 0 + 0) = 1000

Test: Minima & Maxima - 1 - Question 15

A manufacturing firm has its variable cost given by C(x) = x(12 - x3) where 'x' is the number of quantities produced. The price per unit is p(x) = 6x - 2 where 'x' is the number of units in demand. The ratio of marginal cost and marginal revenue when 3 units were both produced and in demand - 

Detailed Solution for Test: Minima & Maxima - 1 - Question 15

Concept:

Marginal cost is the instantaneous rate of change of cost function with respect to the production output x.

Marginal cost = C'(x) = dC(x) / dx 

Marginal revenue is the instantaneous rate of change of revenue function with respect to the production output x.

Marginal revenue = R'(x) = dR(x) / dx 

Revenue function R(x) = x p(x)

Calculation:

Given:

Variable cost function, C(x) = x(12 - x3)

Price per unit p(x) = 6x - 2 

Revenue function R(x) = x p(x)

⇒ R(x) = x(6x - 2)

⇒ R(x) = 6x2 - 2x

  • Marginal revenue,

⇒ R'(x) = dR(x) / dx 

⇒ R'(x)= d(6x2 − 2x) / dx = 12x − 2

⇒ Marginal revenue at (x = 3) = 12 × 3 - 2 = 34 Rs

Therefore the marginal revenue when 3 units were both produced and in demand will be 34 Rs.

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