Example 50 - Chapter 6 Class 12 Application of Derivatives

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 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﷐4𝑥3 – 22𝑥2 + 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(3﷐𝑥﷮2﷯ − 11𝑥 + 6﷯﷮𝑑𝑥﷯ 𝑉’’﷐𝑥﷯ = 4 ﷐3×2𝑥−11﷯ 𝑉’’﷐𝑥﷯ = 4 ﷐6𝑥−11﷯ Putting x = ﷐2﷮3﷯ in 𝑉’’﷐𝑥﷯ 𝑉’’﷐﷐2﷮3﷯﷯=4﷐6﷐﷐2﷮3﷯﷯−11﷯ = 4 ﷐4−11﷯= –28 < 0 ∴ 𝑉’’﷐𝑥﷯ < 0 at 𝑥 = ﷐2﷮3﷯ ⇒ 𝑥 = ﷐2﷮3﷯ is point of maxima ⇒ 𝑉﷐𝑥﷯ is largest when 𝑥 = ﷐2﷮3﷯ Largest volume is 𝑉﷐𝑥﷯ = x﷐3−2𝑥﷯ ﷐8−2𝑥﷯ at x = ﷐2﷮3﷯ 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