Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Minima/ maxima (statement questions) - Geometry questions
Ex 6.3, 27 (MCQ)
Example 23 Important
Example 29 Important
Example 24
Misc 9 Important
Example 25 Important
Misc 6 Important
Ex 6.3,21
Ex 6.3, 20 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3,22 Important
Misc 7
Misc 8 Important
Ex 6.3,17 You are here
Ex 6.3,18 Important
Example 36 Important
Ex 6.3,19 Important
Misc 5 Important
Ex 6.3,23 Important
Misc 12 Important
Misc 14 Important
Example 26 Important
Misc 15 Important
Minima/ maxima (statement questions) - Geometry questions
Last updated at June 12, 2023 by Teachoo
Ex 6.3, 17 A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.Let 𝑥 cm be the length of a side of the removed square Hence, Length after removing = 18 – 𝑥 – 𝑥 = 18 – 2𝑥 Breadth after removing = 18 – 𝑥 – 𝑥 = 18 – 2𝑥 Height of the box = 𝑥 We need to maximize volume of box Let V be the volume of a box Volume of a cuboid = l × b × h V(𝑥)=(18−2𝑥)(18−2𝑥)(𝑥) V(𝑥)=(18−2𝑥)^2 𝑥 Finding v’(𝑥) V’ (𝑥) = 𝑑((18−2𝑥)^2 𝑥)/𝑑𝑥 V’ (𝑥) = 𝑑((18−2𝑥)^2 )/𝑑𝑥 . 𝑥 + 𝑑(𝑥)/𝑑𝑥 . (18−2𝑥)^2 = (2(18−2𝑥).𝑑(18−2𝑥)/𝑑𝑥)𝑥+(1.(18−2𝑥)^2 ) = (2(18−2𝑥).(0−2))𝑥+ (18−2𝑥)^2 = –4(18−2𝑥)𝑥+(18−2𝑥)^2 = (18−2𝑥) [−4𝑥+(18−2𝑥)] = (18−2𝑥)(18−6𝑥) Putting V’(𝑥) = 0 So, 𝑥 = 3, 9 If 𝑥 = 9 Breadth of a box = 18 – 2𝑥 = 18 – 2(9) = 18 – 18 = 0 Since, breadth cannot be zero, ⇒ x = 9 is not possible Hence 𝑥 = 3 only Now finding V’’ (𝑥) V’(𝑥)=(18−2𝑥)(18−6𝑥) V’ (𝑥)=𝑑(18−2𝑥)/𝑑𝑥.(18−6𝑥)+𝑑(18−6𝑥)/𝑑𝑥.(18−2𝑥) = (0 −2)(18−6𝑥)+(−6)(18−2𝑥) = –2(18−6𝑥)−6(18−2𝑥) = –36 + 12x – 108 + 12x = 24x – 144 Putting 𝑥 = 3 V’’(3) = 24 × 3 – 144 = – 72 < 0 Since V’’(𝑥)<0 for 𝑥=3 ∴ 𝑥 = 3 is point of Maxima V(x) "is Maximum " at 𝑥=3 Hence, 3 cm side of the square to be cut off so that the volume of the box is maximum