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Minima/ maxima (statement questions) - Geometry questions
Ex 6.5, 27 (MCQ)
Example 35 Important
Example 41 Important
Example 36
Misc 12 Important
Example 37 Important
Misc 9 Important
Ex 6.5,21
Ex 6.5, 20 Important
Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5,26 Important
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Misc 10
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Ex 6.5,17 You are here
Ex 6.5,18 Important
Example 50 Important
Ex 6.5,19 Important
Misc 8 Important
Ex 6.5,23 Important
Misc 15 Important
Misc 17 Important
Example 38 Important
Misc 18 Important
Minima/ maxima (statement questions) - Geometry questions
Last updated at April 15, 2021 by Teachoo
Ex 6.5, 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 Using product rule . as (𝑢𝑣)^′=𝑢^′ 𝑣+𝑣^′ 𝑢 = (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 18 – 2𝑥 = 0 2𝑥 = 18 𝑥 = 18/2= 9 18 – 6𝑥 = 0 6𝑥 = 18 𝑥 = 18/6= 3 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) Using product .rule as (𝑢𝑣)^′=𝑢^′ 𝑣+𝑣^′ 𝑢 = –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 v