      1. Chapter 6 Class 12 Application of Derivatives (Term 1)
2. Serial order wise
3. Examples

Transcript

Example 50 An open topped box is to be constructed by removing equal squares from each corner of a 3 meter by 8 meter rectangular sheet of aluminum and folding up the sides. Find the volume of the largest such box.Let 𝒙 m be the length of a side of the removed square Hence, Length after removing = 8 – 𝑥 – 𝑥 = 8 – 2𝒙 Breadth after removing = 3 – 𝑥 – 𝑥 = 3 – 2𝒙 Height of the box = 𝒙 We need to maximize volume of box Let V be the volume of a box V = Length × Breadth × Height) = (8−2𝑥)(3−2𝑥)(𝑥) = (8−2𝑥)(3𝑥−2𝑥2) = 8(3𝑥−2𝑥2) – 2x (3𝑥−2𝑥2) = 24𝑥 – 16x2 – 6𝑥2 + 4𝑥3 = 4𝒙3 – 22𝒙2 + 24𝒙 Now, 𝑉(𝑥) = 4𝑥3 – 22𝑥2 + 24𝑥 Diff w.r.t. x 𝑉′(𝑥) = 𝑑(4𝑥^3 − 22𝑥^2 + 24𝑥)/𝑑𝑥 𝑉′(𝑥) = 4 × 3x2 – 22 × 2𝑥 + 24 𝑉′(𝑥) = 12𝑥2 – 44𝑥 + 24 𝑉′(𝑥) = 4(3𝑥2−11𝑥+6) Putting 𝑽′(𝒙) = 0 4(3𝑥2−11𝑥+6) = 0 3𝑥2−11𝑥+6 = 0 3𝑥2 –9𝑥 – 2𝑥 + 6 = 0 3𝑥(𝑥−3) –2 (𝑥−3) = 0 (3𝑥−2)(𝑥−3)= 0 So, 𝒙=𝟐/𝟑 & 𝒙=𝟑 If 𝒙 = 3 Breadth of a box = 3 – 2𝑥 = 3 – 2(3) = 3 – 6 = –3 Since, breadth cannot be negative, ∴ x = 3 is not possible Hence, 𝒙 = 𝟐/𝟑 only Finding 𝑽’’(𝒙) 𝑉’(𝑥) = 4(3𝑥2−11𝑥+6) Diff w.r.t 𝑥 𝑉’’(𝑥) = 𝑑(4(3𝑥^2 − 11𝑥 + 6)/𝑑𝑥 𝑉’’(𝑥) = 4 (3×2𝑥−11) 𝑉’’(𝑥) = 4 (6𝑥−11) Putting x = 𝟐/𝟑 𝑽’’(𝟐/𝟑)=4(6(2/3)−11) = 4 (4−11)= –28 < 0 Since 𝑉’’(𝑥) < 0 at 𝑥 = 2/3 ∴ 𝑥 = 2/3 is point of maxima Hence, 𝑽(𝒙) is largest when 𝒙 = 𝟐/𝟑 Largest volume is 𝑉(𝑥) = x(3−2𝑥) (8−2𝑥) 𝑽(𝟐/𝟑) = 2/3 (3−2(2/3)) (8−2(2/3)) = 2/3 (3−4/3)(8−4/3) = 2/3 ((9 − 4)/3)((24 − 4)/3) = 2/3 (5/3)(20/3) = 200/27 Since dimension of volume is m3 Largest volume is 𝟐𝟎𝟎/𝟐𝟕 m3 