Ex 6.3,6 - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6 You are here
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12 Important
Ex 6.3,13
Ex 6.3,14 Important
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19 Important
Ex 6.3, 20 Important
Ex 6.3,21
Ex 6.3,22 Important
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Last updated at April 16, 2024 by Teachoo
Ex 6.3, 6 Find the maximum profit that a company can make, if the profit function is given by š(š„) = 41 ā 72š„ ā 18š„2The profit function is given by p(x) = 41 ā 72x ā 18x2 pā(x) = ā72 ā 36 x Putting pā (x) = 0 ā72 ā 36x = 0 ā36x = 72 š„ =(ā72)/36= ā2 Now, Pā(x) = ā36 Since Pā (x) < 0 š„=ā2 is the maxima ā“ Maximum profit = p(ā2) p(x) = 41 ā 72x ā 18x2 š(ā2)=41ā72 (ā2)ā18 (ā2)^2 =41+144ā18 (4) =41+144ā72 =113 Hence, the maximum profit is 113 unit.