# Misc 9 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 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 is h = 2m & volume of tank is 8m3 Let ๐ฅ be the length & ๐ฆ be the breadth of the tank We know that Volume of tank = l ร b ร h 8 = 2 ร ๐ฅ ร ๐ฆ 4 = ๐ฅ๐ฆ ๐ฅ๐ฆ = 4 ๐ฆ = 4/๐ฅ Given Building tank cost Rs. 70 per sq. meter for base Area of base = ๐ ร b Area of base = ๐ฅ๐ฆ Cost of base = 70(๐ฅ๐ฆ) Also given Cost is 45 per square meter for sides Area of sides = 2(โ๐+โ๐) โฆ(1) = 2(2๐ฅ+2๐ฆ) = 2 ร 2(๐ฅ+๐ฆ) = 4(๐ฅ+๐ฆ) Cost of making sides = 45[4(๐ฅ+๐ฆ)]= 180 (๐ฅ+๐ฆ) Let C be the total cost of tank C (๐ฅ) = Cost of Base + Cost of Sides C (๐ฅ) = 70(๐ฅ๐ฆ) + 180 (๐ฅ+๐ฆ) C (๐ฅ) = 70(4) + 180 (๐ฅ+4/๐ฅ) C (๐ฅ) = 280 + 180 (๐ฅ+4/๐ฅ) C (๐ฅ) = 280 + 180 (๐ฅ+4๐ฅ^(โ1) ) (From (1): y = 4/๐ฅ) 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(๐ฅ)=180(1โ4/๐ฅ^2 ) 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 So, ๐ฅ = 2 only Finding Cโโ(๐) Cโ(๐ฅ)=180(1โ4๐ฅ^(โ2) ) Differentiating w.r.t ๐ฅ Cโโ(๐ฅ)=๐(180(1 โ 4๐ฅ^(โ2) ))/๐๐ฅ (Since length of side cannot be negative) = 180 ๐(1โ4๐ฅ^(โ2) )/๐๐ฅ = 180(0โ4(โ2) ๐ฅ^(โ2โ1) ) = 180 (8๐ฅ^(โ3) ) = 1440 ๐ฅ-3 = 1440/๐ฅ^3 Putting ๐ = 2 Cโโ(2)=1440/(2)^3 > 0 โ ๐ฅ = 2 is point of minima โ Cโ(๐ฅ) is least at ๐ฅ = 2 Thus, Least cost of construction C (๐ฅ) = 280 + 180 (2+4/2) = 280 + 180 (2+2) = 280 + 180 (4) = 280 + 720 = 1000 Hence, least cost of construction is Rs.1000

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Misc 4

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Misc 7

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Misc 9 Important You are here

Misc 10

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Misc 12 Important

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.