Last updated at April 19, 2021 by Teachoo

Transcript

Misc 9 A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs Rs 70 per sq. meters for the base and Rs 45 per square meter for sides. What is the cost of least expensive tank?Given Depth of tank = h = 2 m & Volume of tank = V = 8 m3 Let Length of Tank = ๐ m Breadth of Tank = ๐ m We know that Volume of tank = Length ร Breadth ร Height 8 = 2 ร ๐ฅ ร ๐ฆ 4 = ๐ฅ๐ฆ ๐ = ๐/๐ Given Building tank costs Rs. 70 per sq. meter for base Area of base = Length ร Breadth Area of base = ๐ฅ๐ฆ โด Cost of base = 70๐ฅ๐ฆ Also given Cost is 45 per square meter for sides Area of sides = 2(โ๐+โ๐) = 2(2๐ฅ+2๐ฆ) = 2 ร 2(๐ฅ+๐ฆ) = 4(๐+๐) Therefore, Cost of making sides = 45[4(๐ฅ+๐ฆ)] = 180 (๐+๐) Let C be the total cost of tank C (๐) = Cost of Base + Cost of Sides C (๐ฅ) = 70๐ฅ๐ฆ + 180 (๐ฅ+๐ฆ) C (๐ฅ) = 70 ร ๐ฅ ร4/๐ฅ + 180 (๐ฅ+4/๐ฅ) C (๐ฅ) = 280 + 180 (๐ฅ+4/๐ฅ) C (๐ฅ) = 280 + 180 (๐+๐๐^(โ๐) ) We need to minimize cost of tank Finding C(x)โ C(๐ฅ) = 280 + 180 (๐ฅ+4๐ฅ^(โ1) ) Differentiating w.r.t x Cโ(๐ฅ) = ๐(280 + 180(๐ฅ + 4๐ฅ^(โ1) ))/๐๐ฅ Cโ(๐ฅ)=0+180(1+(โ1)4๐ฅ^(โ1โ1) ) Cโ(๐ฅ)=180(1โ4๐ฅ^(โ2) ) Cโ(๐)=๐๐๐(๐โ๐/๐^๐ ) Putting Cโ(๐)=๐ 180 (1โ4/๐ฅ^2 )=0 (1โ4/๐ฅ^2 )=0 ((๐ฅ^2 โ 4))/๐ฅ^2 =0 ๐ฅ2 โ 4 = 0 (๐ฅโ2)(๐ฅ+2)=0 So, ๐ฅ = 2 or ๐ฅ = โ2 Since length of side cannot be negative โด ๐ = 2 only Finding Cโโ(๐) Cโ(๐ฅ)=180(1โ4๐ฅ^(โ2) ) Differentiating w.r.t ๐ฅ Cโโ(๐)=๐(180(1 โ 4๐ฅ^(โ2) ))/๐๐ฅ = 180(0โ4(โ2) ๐ฅ^(โ2โ1) ) = 180 (8๐ฅ^(โ3) ) = 1440 ๐ฅ-3 = ๐๐๐๐/๐^๐ Putting ๐ = 2 Cโโ (2) =1440/(2)^3 > 0 Since ๐^โฒโฒ > 0 for x = 2 Thus, C is minimum at x = 2 Thus, Least cost of construction = C(2) = 280 + 180 (๐ฅ+4๐ฅ^(โ1) ) = 280 + 180 (2+4/2) = 280 + 180 (2+2) = 280 + 180 (4) = 280 + 720 = 1000 Hence, least cost of construction is Rs 1,000

Miscellaneous

Misc 1 (a)
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Misc 1 (b) Important Deleted for CBSE Board 2022 Exams

Misc 2 Important

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Misc 4

Misc 5 Important

Misc 6 Important

Misc 7

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Misc 10

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Misc 16

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Misc 20 (MCQ) Important

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Misc 24 (MCQ) Important

Chapter 6 Class 12 Application of Derivatives (Term 1)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.