{"id":170407,"date":"2024-05-21T16:05:36","date_gmt":"2024-05-21T10:35:36","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170407"},"modified":"2024-05-21T16:09:32","modified_gmt":"2024-05-21T10:39:32","slug":"ml-aggarwal-class-12-maths-solutions-case-study-based-questions-section-c","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-12-maths-solutions-case-study-based-questions-section-c\/","title":{"rendered":"ML Aggarwal Class 12 Maths Solutions Case Study Based Questions Section C"},"content":{"rendered":"

The availability of step-by-step ML Aggarwal Class 12 ISC Solutions<\/a> Case Study Based Questions Section C can make challenging problems more manageable.<\/p>\n

ML Aggarwal Class 12 Maths Solutions Case Study Based Questions Section C<\/h2>\n

Chapter 1 Application of Calculus in Commerce and Economics<\/span>
\nCase-Study Based Questions (Solved)<\/span><\/p>\n

Case – Study 1 :<\/p>\n

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

(i) The total revenue function is
\n(a) R(x) = x2<\/sup> – 100x
\n(b) R(x) = 100x – x2<\/sup>
\n(c) R(x) = 100 – x
\n(d) none of these
\nSolution:
\n(b) R(x) = 100x – x2<\/sup><\/p>\n

Given C(x) = \\(\\frac{x^3}{3}\\) – 7x2<\/sup> + 111x + 50
\nand demand function x = 100 – p
\n\u21d2 p = 100 – x
\n\u2234 Revenue function = R (x)
\n= px
\n= (100 – x) x<\/p>\n

(ii) The total profit function is
\n(a) – \\(\\frac{x^3}{3}\\) + 6x2<\/sup> – 11x – 50
\n(b) \\(\\frac{x^3}{3}\\) – 6x2<\/sup> – 11x + 50
\n(c) \\(\\frac{x^3}{3}\\) + 6x2<\/sup> – 11x + 50
\n(d) – \\(\\frac{x^3}{3}\\) – 6x2<\/sup> + 11x + 50
\nSolution:
\n(a) – \\(\\frac{x^3}{3}\\) + 6x2<\/sup> – 11x – 50<\/p>\n

\u2234 Total profit function = P (x)
\n= R (x) – C (x)
\n= (100 x – x2<\/sup>) – {\\(\\frac{x^3}{3}\\) – 7x2<\/sup> + 111x + 50}
\n= – \\(\\frac{x^3}{3}\\) + 6x2<\/sup> – 11x – 50<\/p>\n

(iii) The value of x for which profit is maximum is
\n(a) 8
\n(b) 9
\n(c) 10
\n(d) 11
\nSolution:
\n(d) 11<\/p>\n

\\(\\frac{d}{d x}\\) P(x) = – x2<\/sup> + 12x – 11
\nand \\(\\frac{d^2}{d x^2}\\) P(x) = – 2x + 12
\nFor maxima \/ minima,
\n\\(\\frac{d}{d x}\\) P(x) = 0
\n\u21d2 – x2<\/sup> + 12x – 11 = 0
\n\u21d2 x2<\/sup> – 12x + 11 = 0
\n\u21d2 x = 1, 11
\nat x = 11 ;
\n\\(\\frac{d^2}{d x^2}\\) P(x) = – 22 + 12 = – 10 < 0
\nThus P(x) is maximum when x = 11.<\/p>\n

(iv) The maximum profit is
\n(a) \u20b9 131.11
\n(b) \u20b9 113.31
\n(c) \u20b9 111.33
\n(d) \u20b9 133.11
\nSolution:
\n(c) \u20b9 111.33<\/p>\n

\u2234 Maximum profit = P (11)
\n= \\(-\\frac{11^3}{3}\\)
\n= \\(-\\frac{1331}{3}\\)
\n= \\(\\frac{-1331+2178-363-150}{3}\\)
\n= \\(\\frac{334}{3}\\)
\n= \u20b9 111.33<\/p>\n

\"ML<\/p>\n

Case – Study 2 :<\/p>\n

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

Based on the above information, answer the following questions:<\/p>\n

(i) The total cost function is
\n(a) C(x) = 2x2<\/sup> – 11x + 50
\n(b) C(x) = 2 – \\(\\frac{50}{x^2}\\)
\n(c) C(x) = 2x – \\(\\frac{50}{x}\\)
\n(d) C(x) = 2x2<\/sup> + 11x – 50
\nSolution:
\n(a) C(x) = 2x2<\/sup> – 11x + 50<\/p>\n

Given Average cost function = AC
\n= 2x – 11 + \\(\\frac{50}{x}\\)
\n\u2234 total cost function = x AC (x)
\n= x (2x – 11 + \\(\\frac{50}{x}\\))
\n= 2x2<\/sup> – 11x + 50<\/p>\n

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

Marginal cost function MC = \\(\\frac{d}{d x}\\) C(x)
\n= \\(\\frac{d}{d x}\\) [2x2<\/sup> – 11x + 50]
\n= 4x – 11<\/p>\n

(iii) The marginal cost when x = 4 units is
\n(a) 5
\n(b) 27
\n(c) 18
\n(d) 13
\nSolution:
\n(a) 5<\/p>\n

When x = 4 ;
\nMC = 4 \u00d7 4 – 11 = \u20b9 5<\/p>\n

(iv) The range of values of x for which AC \u00a1s increasing is
\n(a) x < 5 (b) x > – 5
\n(c) x < – 5 (d) x > 5
\nSolution:
\n(d) x > 5<\/p>\n

Now \\(\\frac{d}{d x}\\) AC = 2 – \\(\\frac{50}{x^2}\\)
\nNow \\(\\frac{d}{d x}\\) AC > 2
\n\u21d2 2 – \\(\\frac{50}{x^2}\\) > 0
\n\u21d2 \\(\\frac{50}{x^2}\\) < 2
\n\u21d2 x2<\/sup> > 25
\n\u21d2 |x| > 5
\n\u21d2 x > 5 or x < – 5 (but x > 0)
\n\u21d2 x > 5<\/p>\n

\"ML<\/p>\n

Case – Study 3:<\/p>\n

The marginal cost (in \u20b9) of a product \u00a1s given by MC = \\(\\frac{300}{\\sqrt{3 x+25}}\\) and the fixed cost is \u20b9 5000.<\/p>\n

Based on the above information, answer the following questions :<\/p>\n

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

Given MC = \\(\\frac{300}{\\sqrt{3 x+25}}\\)
\n\u2234 C(x) = \u222b MC(x) dx
\n= \u222b \\(\\frac{300}{\\sqrt{3 x+25}}\\) dx
\n= \\(\\frac{300(3 x+25)^{-\\frac{1}{2}+1}}{\\left(-\\frac{1}{2}+1\\right) 3}\\) + k
\nC(x) = 200 \\(\\sqrt{3 x+25}\\) + k ………………(1)
\nWhen x = 0,
\nfixed cost i.e. C (0) = 5000
\n\u2234 from(1) ; we have
\n5000 = 200 \u00d7 5 + k
\n\u21d2 k = 4000
\n\u2234 from (1) ;
\nC(x) = 200 \\(\\sqrt{3 x+25}\\) + 4000<\/p>\n

(ii) The cost of producing 25 units of the product is
\n(a) \u20b9 7000
\n(b) \u20b9 6000
\n(c) \u20b9 8000
\n(d) \u20b9 6500
\nSolution:
\n(b) \u20b9 6000<\/p>\n

\u2234 required cost of producing 25 units = C(25)
\n= \u20b9 200 \\(\\sqrt{3 \\times 25+25}\\) + 4000
\n= \u20b9 [200 \u00d7 10 + 4000]
\n= \u20b9 6000<\/p>\n

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

\u2234 Average cost function = \\(\\frac{C(x)}{x}\\)
\n= \\(\\frac{200 \\sqrt{3 x+25}+4000}{x}\\)<\/p>\n

(iv) The average cost of producing 200 units is
\n(a) \u20b9 5
\n(b) \u20b9 25
\n(c) \u20b9 45
\n(d) \u20b9 50
\nSolution:
\n(c) \u20b9 45<\/p>\n

\u2234 Required average cost of producing 200 units = AC (200)
\n= \\(\\frac{200 \\sqrt{3 \\times 200+25}+4000}{200}\\)
\n= \\(\\frac{200 \\times 25+4000}{200}\\)
\n= \\(\\frac{9000}{200}\\) = \u20b9 45<\/p>\n

\"ML<\/p>\n

Chapter 2 Linear Regression<\/span>
\nCase – Study Based Questions (Solved)<\/span><\/p>\n

Case – Study 1 :<\/p>\n

For five observations of pairs (x, y) of correlated variables, the following results were obtained :
\n\u03a3x = 15,
\n\u03a3y = 18,
\n\u03a3x2<\/sup> = 55,
\n\u03a3y2<\/sup> = 74
\nand \u03a3xy = 58.<\/p>\n

Based on the above Information, answer the following questions:<\/p>\n

(i) The regression coefficient of x on y i.e. bxy<\/sub> is
\n(a) \\(\\frac{2}{5}\\)
\n(b) \\(\\frac{10}{23}\\)
\n(c) \\(-\\frac{2}{5}\\)
\n(d) \\(-\\frac{10}{23}\\)
\nSolution:
\n(b) \\(\\frac{10}{23}\\)<\/p>\n

(ii) The equation of the line of regression of y on x is
\n(a) 2x – 5y + 12 = 0
\n(b) 23x – 10y – 13 = 0
\n(c) 2x + 23y – 113 = 0
\n(d) 2x – 5y – 12 = 0
\nSolution:
\n(a) 2x – 5y + 12 = 0<\/p>\n

\\(\\bar{x}=\\frac{\\Sigma x}{n}\\)
\n= \\(\\frac{15}{5}\\) = 3 ;
\n\\(\\bar{y}=\\frac{\\Sigma y}{n}\\)
\n= \\(\\frac{18}{5}\\) = 3.6
\nbyx<\/sub> = \\(\\frac{n \\Sigma x y-\\Sigma x \\Sigma y}{n \\Sigma x^2-(\\Sigma x)^2}\\)
\n= \\(\\frac{5 \\times 58-15 \\times 18}{5 \\times 55-15^2}\\)
\n= \\(\\frac{20}{50}=\\frac{2}{5}\\)
\nThus repression line ofy on x be given by
\n\u21d2 y – \\(\\vec{y}\\) = b (x – \\(\\vec{x}\\))
\n\u21d2 \\(\\left(y-\\frac{18}{5}\\right)=\\frac{2}{5}(x-3)\\)
\n\u21d2 5y – 18 = 2x – 6
\n\u21d2 2x – 5y + 12 = 0<\/p>\n

(iii) The most likely value of y when x = 8 is
\n(a) 6.4
\n(b) 6.5
\n(c) 5.6
\n(d) 5.8
\nSolution:
\n(c) 5.6<\/p>\n

from (1) ;
\n5y = 2x + 12
\n\u21d2 y = \\(\\frac{1}{5}\\) (2x + 12)
\n\u2234 y(8) = \\(\\frac{1}{5}\\) (2 \u00d7 8 + 12)
\n= \\(\\frac{28}{5}\\) = 5.6<\/p>\n

(iv) The coefficient of correlation between x and y is
\n(a) 3.41
\n(b) 0.174
\n(c) – 0.417
\n(d) 0.417
\nSolution:
\n(d) 0.417<\/p>\n

Since bxy<\/sub> = \\(\\frac{10}{3}\\) > 0
\nand byx<\/sub> = \\(\\frac{2}{5}\\) > 0
\nand r = + \\(\\sqrt{b_{x y} \\cdot b_{y x}}\\)
\n= \\(\\sqrt{\\frac{10}{23} \\times \\frac{2}{5}}\\)
\n= \\(\\frac{2}{\\sqrt{23}}\\)
\n= 0.4170<\/p>\n

\"ML<\/p>\n

Case – Study 2 :<\/p>\n

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.<\/p>\n

Based on the above information, answer the following questions :<\/p>\n

(i) The line of regression of y on x is
\n(a) x + y – 80
\n(b) 4x + 9y – 57 = 0
\n(c) x + y + 80
\n(d) 4x + 9y + 57 = 0
\nSolution:<\/p>\n

Let us assume x + y – 8 = 0 be the regression line of y on x.
\n\u2234 y = – x + 8
\n\u2234 byx<\/sub> = – 1
\nThen 4x + 9y – 57 = 0 be a regression line of x on y.
\n\u21d2 x = \\(-\\frac{9}{4} y+\\frac{57}{4}\\)
\n\u2234 bxy<\/sub> = – \\(\\frac{9}{4}\\)
\nHere bxy<\/sub> . byx<\/sub> = (- 1) (- \\(\\frac{9}{4}\\))
\n= \\(\\frac{9}{4}\\) > 1
\nwhich is not possible.
\nThus our assumption is wrong.
\nHence line of regression of y on x be 4x + 9y – 57 = 0
\nand line of regression of x on y be x + y – 8 = 0<\/p>\n

(ii) The coefficient of correlation between x and y is
\n(a) \\(\\frac{1}{3}\\)
\n(b) – \\(\\frac{1}{2}\\)
\n(c) \\(\\frac{2}{3}\\)
\n(d) \\(-\\frac{2}{3}\\)
\nSolution:
\n(d) \\(-\\frac{2}{3}\\)<\/p>\n

The regression line of y on x be 4x + 9y – 57 = 0
\n\u21d2 9y = – 4x + 57
\n\u21d2 y = \\(-\\frac{4}{9} x+\\frac{57}{9}\\)
\nbxy<\/sub> = – \\(\\frac{4}{9}\\) < 0
\nand regression line of x on y be x + y – 8 = 0
\n\u21d2 x = – y + 8
\n\u2234 bxy<\/sub> = – 1 < 0
\n\u2234 r = – \\(\\sqrt{b_{x y} b_{y x}}\\)
\n= \\(-\\sqrt{-\\frac{4}{9} \\times(-1)}\\)
\n= \\(-\\frac{2}{3}\\)
\n[\u2235 bxy<\/sub>, byx<\/sub> , 0
\n\u21d2 r < 0]<\/p>\n

(iii) The variance of x is
\n(a) 4
\n(b) 3
\n(c) 9
\n(d) 6
\nSolution:
\n(c) 9<\/p>\n

Given \u03c3y<\/sub> = 2
\nWe know that
\nbxy<\/sub> = r \\(\\frac{\\sigma_x}{\\sigma_y}\\)
\n– 1 = \\(-\\frac{2}{3} \\times \\frac{\\sigma_x}{2}\\)
\n\u21d2 \u03c3x<\/sub> = 3
\n\u21d2 Var (x) = \u03c3x<\/sub>2<\/sup> = 9<\/p>\n

(iv) Covariance between x and y is
\n(a) – 4
\n(b) 4
\n(c) – 9
\n(d) 9
\nSolution:
\n(a) – 4<\/p>\n

Cov (x, y) = byx<\/sub> \u03c3x<\/sub>2<\/sup>
\n= – \\(\\frac{4}{9}\\) \u00d7 9 = – 4
\n= \\(\\sqrt{\\frac{10}{23} \\times \\frac{2}{5}}\\)
\n= \\(\\frac{2}{\\sqrt{23}}\\)
\n= 0.4170<\/p>\n

\"ML<\/p>\n

Chapter 3 Linear Programming<\/span>
\nCase-Study Based Questions (Solved)<\/span><\/p>\n

Case – Study 1 :<\/p>\n

The feasible region for an L.P.P. is shown in the adjoining figure. The line CB is parallel to OA.<\/p>\n

\"ML<\/p>\n

Based on the above information, answer the following questions :<\/p>\n

(i) The equation of the line OA is
\n(a) x – 2y = 0
\n(b) y – 2x = 0
\n(c) x + 2y = 0
\n(d) 2x + y = 0
\nSolution:
\n(b) y – 2x = 0<\/p>\n

The eqn. of line OA be given by
\ny – 0 = \\(\\frac{12-0}{6-0}\\) (x – 0)
\n\u21d2 y = 2x<\/p>\n

(ii) The equation of the line BC is
\n(a) y – 2x = 4
\n(b) x – 2y = 4
\n(c) y + 2x = 4
\n(d) x + 2y = 4
\nSolution:
\n(a) y – 2x = 4<\/p>\n

The eqn. of line CB be given by
\ny – 4 = \\(\\frac{16-4}{6-0}\\) (x – 0)
\n\u21d2 y – 4 = 2x
\n\u21d2 y = 2x + 4<\/p>\n

(iii) The constraints for the L.P.P. re
\n(a) y \u2265 2x, y – 2x \u2264 4, x \u2264 6, x \u2265 0, y \u2265 0
\n(b) y \u2264 2x, y – 2x \u2264 4, x \u2264 6, x \u2265 0, y \u2265 0
\n(c) y \u2265 2x, y – 2x \u2265 4, x \u2264 6, x \u2265 0, y \u2265 0
\n(d) y \u2264 2x, y – 2x \u2265 4, x \u2264 6, x \u2265 0, y \u2265 0
\nSolution:
\n(a) y \u2265 2x, y – 2x \u2264 4, x \u2264 6, x \u2265 0, y \u2265 0<\/p>\n

Clearly (0, 0) satisfies y – 2x \u2264 4 and (0, 0) satisfies y \u2265 2x.
\nAlso (0, 0) satisfies x \u2264 6.
\nThus the associated constraints for given L.P.P is as under:
\ny – 2x \u2264 4 ;
\ny \u2265 2x ;
\nx \u2264 6 ;
\nx \u2265 0, y \u2265 0
\nPoint B be the point of intersection of lines x = 6 and y – 2x = 4
\n\u2234 Coordinates of B are (6, 16).<\/p>\n

(iv) The minimum value of the objective function Z = 3x – 4y is
\n(a) 0
\n(b) – 16
\n(c) – 30
\n(d) – 46
\nSolution:
\n(d) – 46<\/p>\n

The bounded shaded region OCBAO represents the feasible region with corner points
\nO (0, 0) ; C (0, 4) ; B (6, 16) and A (6, 12).<\/p>\n

\"ML<\/p>\n

Clearly Zmin<\/sub> = – 46
\nand obtained at B (6, 16)
\ni.e. x = 6
\nand y = 16.<\/p>\n","protected":false},"excerpt":{"rendered":"

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 …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170407"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=170407"}],"version-history":[{"count":1,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170407\/revisions"}],"predecessor-version":[{"id":170410,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170407\/revisions\/170410"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170407"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170407"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170407"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}