ML Aggarwal Class 12 Maths Solutions Case Study Based Questions Section C

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ML Aggarwal Class 12 Maths Solutions Case Study Based Questions Section C

Chapter 1 Application of Calculus in Commerce and Economics
Case-Study Based Questions (Solved)

Case – Study 1 :

A firm has the cost function C(x) = \(\frac{x^3}{3}\) – 7x² + 111x + 50 and demand function x = 100 – p.
.
Based on the above Information, answer the following questions:

(i) The total revenue function is
(a) R(x) = x² – 100x
(b) R(x) = 100x – x²
(c) R(x) = 100 – x
(d) none of these
Solution:
(b) R(x) = 100x – x²

Given C(x) = \(\frac{x^3}{3}\) – 7x² + 111x + 50
and demand function x = 100 – p
⇒ p = 100 – x
∴ Revenue function = R (x)
= px
= (100 – x) x

(ii) The total profit function is
(a) – \(\frac{x^3}{3}\) + 6x² – 11x – 50
(b) \(\frac{x^3}{3}\) – 6x² – 11x + 50
(c) \(\frac{x^3}{3}\) + 6x² – 11x + 50
(d) – \(\frac{x^3}{3}\) – 6x² + 11x + 50
Solution:
(a) – \(\frac{x^3}{3}\) + 6x² – 11x – 50

∴ Total profit function = P (x)
= R (x) – C (x)
= (100 x – x²) – {\(\frac{x^3}{3}\) – 7x² + 111x + 50}
= – \(\frac{x^3}{3}\) + 6x² – 11x – 50

(iii) The value of x for which profit is maximum is
(a) 8
(b) 9
(c) 10
(d) 11
Solution:
(d) 11

\(\frac{d}{d x}\) P(x) = – x² + 12x – 11
and \(\frac{d^2}{d x^2}\) P(x) = – 2x + 12
For maxima / minima,
\(\frac{d}{d x}\) P(x) = 0
⇒ – x² + 12x – 11 = 0
⇒ x² – 12x + 11 = 0
⇒ x = 1, 11
at x = 11 ;
\(\frac{d^2}{d x^2}\) P(x) = – 22 + 12 = – 10 < 0
Thus P(x) is maximum when x = 11.

(iv) The maximum profit is
(a) ₹ 131.11
(b) ₹ 113.31
(c) ₹ 111.33
(d) ₹ 133.11
Solution:
(c) ₹ 111.33

∴ Maximum profit = P (11)
= \(-\frac{11^3}{3}\)
= \(-\frac{1331}{3}\)
= \(\frac{-1331+2178-363-150}{3}\)
= \(\frac{334}{3}\)
= ₹ 111.33

Case – Study 2 :

The average cost function associated with producing and marketing x units of an item is given by
AC = 2x – 11 + \(\frac{50}{x}\).

Based on the above information, answer the following questions:

(i) The total cost function is
(a) C(x) = 2x² – 11x + 50
(b) C(x) = 2 – \(\frac{50}{x^2}\)
(c) C(x) = 2x – \(\frac{50}{x}\)
(d) C(x) = 2x² + 11x – 50
Solution:
(a) C(x) = 2x² – 11x + 50

Given Average cost function = AC
= 2x – 11 + \(\frac{50}{x}\)
∴ total cost function = x AC (x)
= x (2x – 11 + \(\frac{50}{x}\))
= 2x² – 11x + 50

(ii) The marginal cost function is
(a) MC = 4x + 11
(b) MC = 4x – 11
(c) MC = 2 + \(\frac{50}{x^2}\)
(d) MC = \(\frac{100}{x^3}\)
Solution:
(b) MC = 4x – 11

Marginal cost function MC = \(\frac{d}{d x}\) C(x)
= \(\frac{d}{d x}\) [2x² – 11x + 50]
= 4x – 11

(iii) The marginal cost when x = 4 units is
(a) 5
(b) 27
(c) 18
(d) 13
Solution:
(a) 5

When x = 4 ;
MC = 4 × 4 – 11 = ₹ 5

(iv) The range of values of x for which AC ¡s increasing is
(a) x < 5 (b) x > – 5
(c) x < – 5 (d) x > 5
Solution:
(d) x > 5

Now \(\frac{d}{d x}\) AC = 2 – \(\frac{50}{x^2}\)
Now \(\frac{d}{d x}\) AC > 2
⇒ 2 – \(\frac{50}{x^2}\) > 0
⇒ \(\frac{50}{x^2}\) < 2
⇒ x² > 25
⇒ |x| > 5
⇒ x > 5 or x < – 5 (but x > 0)
⇒ x > 5

Case – Study 3:

The marginal cost (in ₹) of a product ¡s given by MC = \(\frac{300}{\sqrt{3 x+25}}\) and the fixed cost is ₹ 5000.

Based on the above information, answer the following questions :

(i) The cost function is
(a) C(x) = 600 \(\sqrt{3 x+25}\) – 100
(b) C(x) = 200 \(\sqrt{3 x+25}\) + 5000
(c) C(x) = 200 \(\sqrt{3 x+25}\) + 4000
(d) C(x) = 200 \(\sqrt{3 x+25}\) – 4000
Solution:

Given MC = \(\frac{300}{\sqrt{3 x+25}}\)
∴ C(x) = ∫ MC(x) dx
= ∫ \(\frac{300}{\sqrt{3 x+25}}\) dx
= \(\frac{300(3 x+25)^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right) 3}\) + k
C(x) = 200 \(\sqrt{3 x+25}\) + k ………………(1)
When x = 0,
fixed cost i.e. C (0) = 5000
∴ from(1) ; we have
5000 = 200 × 5 + k
⇒ k = 4000
∴ from (1) ;
C(x) = 200 \(\sqrt{3 x+25}\) + 4000

(ii) The cost of producing 25 units of the product is
(a) ₹ 7000
(b) ₹ 6000
(c) ₹ 8000
(d) ₹ 6500
Solution:
(b) ₹ 6000

∴ required cost of producing 25 units = C(25)
= ₹ 200 \(\sqrt{3 \times 25+25}\) + 4000
= ₹ [200 × 10 + 4000]
= ₹ 6000

(iii) The average cost function is
(a) AC = \(\frac{200 \sqrt{3 x+25}}{x}+\frac{4000}{x}\)
(b) AC = \(\frac{200 \sqrt{3 x+25}}{x}+\frac{5000}{x}\)
(c) AC = \(\frac{100 \sqrt{3 x+25}}{x}+\frac{4000}{x}\)
(d) AC = \(\frac{200 \sqrt{3 x+25}}{x}-\frac{4000}{x}\)
Solution:
(a) AC = \(\frac{200 \sqrt{3 x+25}}{x}+\frac{4000}{x}\)

∴ Average cost function = \(\frac{C(x)}{x}\)
= \(\frac{200 \sqrt{3 x+25}+4000}{x}\)

(iv) The average cost of producing 200 units is
(a) ₹ 5
(b) ₹ 25
(c) ₹ 45
(d) ₹ 50
Solution:
(c) ₹ 45

∴ Required average cost of producing 200 units = AC (200)
= \(\frac{200 \sqrt{3 \times 200+25}+4000}{200}\)
= \(\frac{200 \times 25+4000}{200}\)
= \(\frac{9000}{200}\) = ₹ 45

Chapter 2 Linear Regression
Case – Study Based Questions (Solved)

Case – Study 1 :

For five observations of pairs (x, y) of correlated variables, the following results were obtained :
Σx = 15,
Σy = 18,
Σx² = 55,
Σy² = 74
and Σxy = 58.

Based on the above Information, answer the following questions:

(i) The regression coefficient of x on y i.e. b_xy is
(a) \(\frac{2}{5}\)
(b) \(\frac{10}{23}\)
(c) \(-\frac{2}{5}\)
(d) \(-\frac{10}{23}\)
Solution:
(b) \(\frac{10}{23}\)

(ii) The equation of the line of regression of y on x is
(a) 2x – 5y + 12 = 0
(b) 23x – 10y – 13 = 0
(c) 2x + 23y – 113 = 0
(d) 2x – 5y – 12 = 0
Solution:
(a) 2x – 5y + 12 = 0

\(\bar{x}=\frac{\Sigma x}{n}\)
= \(\frac{15}{5}\) = 3 ;
\(\bar{y}=\frac{\Sigma y}{n}\)
= \(\frac{18}{5}\) = 3.6
b_yx = \(\frac{n \Sigma x y-\Sigma x \Sigma y}{n \Sigma x^2-(\Sigma x)^2}\)
= \(\frac{5 \times 58-15 \times 18}{5 \times 55-15^2}\)
= \(\frac{20}{50}=\frac{2}{5}\)
Thus repression line ofy on x be given by
⇒ y – \(\vec{y}\) = b (x – \(\vec{x}\))
⇒ \(\left(y-\frac{18}{5}\right)=\frac{2}{5}(x-3)\)
⇒ 5y – 18 = 2x – 6
⇒ 2x – 5y + 12 = 0

(iii) The most likely value of y when x = 8 is
(a) 6.4
(b) 6.5
(c) 5.6
(d) 5.8
Solution:
(c) 5.6

from (1) ;
5y = 2x + 12
⇒ y = \(\frac{1}{5}\) (2x + 12)
∴ y(8) = \(\frac{1}{5}\) (2 × 8 + 12)
= \(\frac{28}{5}\) = 5.6

(iv) The coefficient of correlation between x and y is
(a) 3.41
(b) 0.174
(c) – 0.417
(d) 0.417
Solution:
(d) 0.417

Since b_xy = \(\frac{10}{3}\) > 0
and b_yx = \(\frac{2}{5}\) > 0
and r = + \(\sqrt{b_{x y} \cdot b_{y x}}\)
= \(\sqrt{\frac{10}{23} \times \frac{2}{5}}\)
= \(\frac{2}{\sqrt{23}}\)
= 0.4170

Case – Study 2 :

For a given bivariate data, the two regression lines x + y – 8 = 0 and 4x + 9y – 57 = 0, and the standard deviation of y is 2.

Based on the above information, answer the following questions :

(i) The line of regression of y on x is
(a) x + y – 80
(b) 4x + 9y – 57 = 0
(c) x + y + 80
(d) 4x + 9y + 57 = 0
Solution:

Let us assume x + y – 8 = 0 be the regression line of y on x.
∴ y = – x + 8
∴ b_yx = – 1
Then 4x + 9y – 57 = 0 be a regression line of x on y.
⇒ x = \(-\frac{9}{4} y+\frac{57}{4}\)
∴ b_xy = – \(\frac{9}{4}\)
Here b_xy . b_yx = (- 1) (- \(\frac{9}{4}\))
= \(\frac{9}{4}\) > 1
which is not possible.
Thus our assumption is wrong.
Hence line of regression of y on x be 4x + 9y – 57 = 0
and line of regression of x on y be x + y – 8 = 0

(ii) The coefficient of correlation between x and y is
(a) \(\frac{1}{3}\)
(b) – \(\frac{1}{2}\)
(c) \(\frac{2}{3}\)
(d) \(-\frac{2}{3}\)
Solution:
(d) \(-\frac{2}{3}\)

The regression line of y on x be 4x + 9y – 57 = 0
⇒ 9y = – 4x + 57
⇒ y = \(-\frac{4}{9} x+\frac{57}{9}\)
b_xy = – \(\frac{4}{9}\) < 0
and regression line of x on y be x + y – 8 = 0
⇒ x = – y + 8
∴ b_xy = – 1 < 0
∴ r = – \(\sqrt{b_{x y} b_{y x}}\)
= \(-\sqrt{-\frac{4}{9} \times(-1)}\)
= \(-\frac{2}{3}\)
[∵ b_xy, b_yx , 0
⇒ r < 0]

(iii) The variance of x is
(a) 4
(b) 3
(c) 9
(d) 6
Solution:
(c) 9

Given σ_y = 2
We know that
b_xy = r \(\frac{\sigma_x}{\sigma_y}\)
– 1 = \(-\frac{2}{3} \times \frac{\sigma_x}{2}\)
⇒ σ_x = 3
⇒ Var (x) = σ_x² = 9

(iv) Covariance between x and y is
(a) – 4
(b) 4
(c) – 9
(d) 9
Solution:
(a) – 4

Cov (x, y) = b_yx σ_x²
= – \(\frac{4}{9}\) × 9 = – 4
= \(\sqrt{\frac{10}{23} \times \frac{2}{5}}\)
= \(\frac{2}{\sqrt{23}}\)
= 0.4170

Chapter 3 Linear Programming
Case-Study Based Questions (Solved)

Case – Study 1 :

The feasible region for an L.P.P. is shown in the adjoining figure. The line CB is parallel to OA.

Based on the above information, answer the following questions :

(i) The equation of the line OA is
(a) x – 2y = 0
(b) y – 2x = 0
(c) x + 2y = 0
(d) 2x + y = 0
Solution:
(b) y – 2x = 0

The eqn. of line OA be given by
y – 0 = \(\frac{12-0}{6-0}\) (x – 0)
⇒ y = 2x

(ii) The equation of the line BC is
(a) y – 2x = 4
(b) x – 2y = 4
(c) y + 2x = 4
(d) x + 2y = 4
Solution:
(a) y – 2x = 4

The eqn. of line CB be given by
y – 4 = \(\frac{16-4}{6-0}\) (x – 0)
⇒ y – 4 = 2x
⇒ y = 2x + 4

(iii) The constraints for the L.P.P. re
(a) y ≥ 2x, y – 2x ≤ 4, x ≤ 6, x ≥ 0, y ≥ 0
(b) y ≤ 2x, y – 2x ≤ 4, x ≤ 6, x ≥ 0, y ≥ 0
(c) y ≥ 2x, y – 2x ≥ 4, x ≤ 6, x ≥ 0, y ≥ 0
(d) y ≤ 2x, y – 2x ≥ 4, x ≤ 6, x ≥ 0, y ≥ 0
Solution:
(a) y ≥ 2x, y – 2x ≤ 4, x ≤ 6, x ≥ 0, y ≥ 0

Clearly (0, 0) satisfies y – 2x ≤ 4 and (0, 0) satisfies y ≥ 2x.
Also (0, 0) satisfies x ≤ 6.
Thus the associated constraints for given L.P.P is as under:
y – 2x ≤ 4 ;
y ≥ 2x ;
x ≤ 6 ;
x ≥ 0, y ≥ 0
Point B be the point of intersection of lines x = 6 and y – 2x = 4
∴ Coordinates of B are (6, 16).

(iv) The minimum value of the objective function Z = 3x – 4y is
(a) 0
(b) – 16
(c) – 30
(d) – 46
Solution:
(d) – 46

The bounded shaded region OCBAO represents the feasible region with corner points
O (0, 0) ; C (0, 4) ; B (6, 16) and A (6, 12).

Clearly Z_min = – 46
and obtained at B (6, 16)
i.e. x = 6
and y = 16.