Slide36.JPG

Slide37.JPG
Slide38.JPG
Slide39.JPG
Slide40.JPG Slide41.JPG

Go Ad-free

Transcript

Example 36 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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo