       1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Miscellaneous

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 𝑥 be the length & 𝑦 be the breadth of the tank We know that Volume of tank = l × b × h 8 = 2 × 𝑥 × 𝑦 4 = 𝑥𝑦 𝑥𝑦 = 4 𝑦 = 4/𝑥 Given Building tank cost Rs. 70 per sq. meter for base Area of base = 𝑙 × b Area of base = 𝑥𝑦 Cost of base = 70(𝑥𝑦) Also given Cost is 45 per square meter for sides Area of sides = 2(ℎ𝑙+ℎ𝑏) …(1) = 2(2𝑥+2𝑦) = 2 × 2(𝑥+𝑦) = 4(𝑥+𝑦) Cost of making sides = 45[4(𝑥+𝑦)]= 180 (𝑥+𝑦) Let C be the total cost of tank C (𝑥) = Cost of Base + Cost of Sides C (𝑥) = 70(𝑥𝑦) + 180 (𝑥+𝑦) C (𝑥) = 70(4) + 180 (𝑥+4/𝑥) C (𝑥) = 280 + 180 (𝑥+4/𝑥) C (𝑥) = 280 + 180 (𝑥+4𝑥^(−1) ) (From (1): y = 4/𝑥) We need to minimize cost of tank Finding C(x)’ C(𝑥) = 280 + 180 (𝑥+4𝑥^(−1) ) Differentiating w.r.t x C’(𝑥) = 𝑑(280 + 180(𝑥 + 4𝑥^(−1) ))/𝑑𝑥 C’(𝑥)=0+180(1+(−1)4𝑥^(−1−1) ) C’(𝑥)=180(1−4𝑥^(−2) ) C’(𝑥)=180(1−4/𝑥^2 ) Putting C’(𝒙)=𝟎 180 (1−4/𝑥^2 )=0 (1−4/𝑥^2 )=0 ((𝑥^2 − 4))/𝑥^2 =0 𝑥2 – 4 = 0 (𝑥−2)(𝑥+2)=0 So, 𝑥 = 2 or 𝑥 = –2 So, 𝑥 = 2 only Finding C’’(𝒙) C’(𝑥)=180(1−4𝑥^(−2) ) Differentiating w.r.t 𝑥 C’’(𝑥)=𝑑(180(1 − 4𝑥^(−2) ))/𝑑𝑥 (Since length of side cannot be negative) = 180 𝑑(1−4𝑥^(−2) )/𝑑𝑥 = 180(0−4(−2) 𝑥^(−2−1) ) = 180 (8𝑥^(−3) ) = 1440 𝑥-3 = 1440/𝑥^3 Putting 𝒙 = 2 C’’(2)=1440/(2)^3 > 0 ⇒ 𝑥 = 2 is point of minima ⇒ C’(𝑥) is least at 𝑥 = 2 Thus, Least cost of construction C (𝑥) = 280 + 180 (2+4/2) = 280 + 180 (2+2) = 280 + 180 (4) = 280 + 720 = 1000 Hence, least cost of construction is Rs.1000

Miscellaneous 