Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 4 Applications of Derivatives Ex 4.3 Questions and Answers.
Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.3
Solution & Step-by-Step Answer:
f(x) = 2x3 – 21x2 + 36x – 20 ∴ f'(x) = (2x3 – 21x2 + 36x – 20) = 2 × 3x2 – 21 × 2x + 36 × 1 – 0 = 6x2 – 42x + 36 and f”(x) = (6x2 – 42x + 36) = 6 × 2x – 42 × 1 + 0 = 12x – 42 f'(x) = 0 gives 6x2 – 42x + 36 = 0. ∴ x2 – 7x + 6 = 0 ∴ (x – 1)(x – 6) = 0 ∴ the roots of f'(x) = 0 are x1 = 1 and x2 = 6. For x = 1, f”(1) = 12(1) – 42 = -30 < 0 ∴ by the second derivative test, f has maximum at x = 1 and maximum value of f at x = 1 f(1) = 2(1)3 – 21(1)2 + 36(1) – 20 = 2 – 21 + 36 – 20 = -3 For x = 6, f”(6) = 12(6) – 42 = 30 > 0 ∴ by the second derivative test, f has minimum at x = 6 and minimum value of f at x = 6 f(6) = 2(6)3 – 21(6)2 + 36(6) – 20 = 432 – 756 + 216 – 20 = -128 Hence, the function f has maximum value -3 at x = 1 and minimum value -128 at x = 6.
(ii) f(x) = x. log x
Solution:
f(x) = x. log x
f'(x) = (x.log x)
= x.(log x) + log x.(x)
= x × + (logx) × 1
= 1 + log x
and f”(x) = (1 + logx)
= 0 +
=
Now, f'(x) = 0, if 1 + log x = 0
i.e. if log x = -1 = -log e
i.e. if log x = log(e-1) = log
i.e. if x =
When x = , f”(x) = = e > 0
∴ by the second derivative test,
f is minimum at x =
Minimum value of f at x =
= log()
= log(e-1)
= (-1) log e
= ……..[∵ log e = 1]
Hence, the function f has minimum at x = and minimum value is .
(iii) f(x) = x2+
Solution:
f'(x) = 0 gives 2x – = 0
∴ 2x3– 16 = 0
∴ x3= 8
∴ x = 2
For x = 2, f”(2) = 2 + = 6 > 0
∴ by the second derivative test, f has minimum at x = 2 and minimum value of f at x = 2
f(2) = (2)2+
= 4 + 8
= 12
Hence, the function f has a minimum at x = 2 and a minimum value is 12.
Solution & Step-by-Step Answer:
Let the first part of 20 be x. Then the second part is 20 – x. ∴ their product = x(20 – x) = 20x – x2 = f(x) …..(Say) ∴ f'(x) = (20x – x2) = 20 × 1 – 2x = 20 – 2x and f”(x) = (20 – 2x) = 0 – 2 × 1 = -2 The root of the equation f'(x) = 0 i.e. 20 – 2x = 0 is x = 10 and f”(10) = -2 < 0 ∴ by the second derivative test, f is maximum at x = 10. Hence, the required parts of 20 are 10 and 10.
Solution & Step-by-Step Answer:
Let x cm and y cm be the length and breadth of the rectangle. Then its perimeter is 2(x + y) = 36 ∴ x + y = 18 ∴ y = 18 – x Area of the rectangle = xy = x(18 – x) Let f(x) = x(18 – x) = 18x – x2 Then f'(x) = (18x – x2) = 18 × 1 – 2x = 18 – 2x and f”(x) = (18 – 2x) = 0 – 2 × 1 = -2 Now, f(x) = 0, if 18 – 2x = 0 i.e. if x = 9 and f”(9) = -2 < 0 ∴ by the second derivative test, f has maximum value at x = 9 When x = 9, y = 18 – 9 = 9 Hence, the rectangle is a square of side 9 cm.
Solution & Step-by-Step Answer:
Let the number of units sold be x. Then profit = S.P. – C.P. ∴ P(x) = (180 – x)x – (x2 + 60x + 50) ∴ P(x) = 180x – x2 – x2 – 60x – 50 ∴ P(x) = 120x – 2x2 – 50 P'(x) = (120x – 2x2 – 50) = 120 × 1 – 2 × 2x – 0 = 120 – 4x and P”(x) = (120 – 4x) = 0 – 4 × 1 = -4 P'(x) = 0 if 120 – 4x = 0 i.e. if x = 30 and P”(30) = -4 < 0 ∴ by the second derivative test, P(x) is maximum when x = 30. Hence, the number of units sold for maximum profit is 30.