The availability of step-by-step ML Aggarwal Class 12 ISC Solutions Case Study Based Questions Section C can make challenging problems more manageable.

## 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. bxy 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
byx = $$\frac{n \Sigma x y-\Sigma x \Sigma y}{n \Sigma x^2-(\Sigma x)^2}$$
= $$\frac{5 \times 58-15 \times 18}{5 \times 55-15^2}$$
= $$\frac{20}{50}=\frac{2}{5}$$
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 bxy = $$\frac{10}{3}$$ > 0
and byx = $$\frac{2}{5}$$ > 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
∴ byx = – 1
Then 4x + 9y – 57 = 0 be a regression line of x on y.
⇒ x = $$-\frac{9}{4} y+\frac{57}{4}$$
∴ bxy = – $$\frac{9}{4}$$
Here bxy . byx = (- 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 = $$-\frac{4}{9} x+\frac{57}{9}$$
bxy = – $$\frac{4}{9}$$ < 0
and regression line of x on y be x + y – 8 = 0
⇒ x = – y + 8
∴ bxy = – 1 < 0
∴ r = – $$\sqrt{b_{x y} b_{y x}}$$
= $$-\sqrt{-\frac{4}{9} \times(-1)}$$
= $$-\frac{2}{3}$$
[∵ bxy, byx , 0
⇒ r < 0]

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

Given σy = 2
We know that
bxy = r $$\frac{\sigma_x}{\sigma_y}$$
– 1 = $$-\frac{2}{3} \times \frac{\sigma_x}{2}$$
⇒ σx = 3
⇒ Var (x) = σx2 = 9

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

Cov (x, y) = byx σx2
= – $$\frac{4}{9}$$ × 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 Zmin = – 46
and obtained at B (6, 16)
i.e. x = 6
and y = 16.