Divide 30 into two parts such that product of one and cube of the other is maximum mcq

CBSE 12-science - Maths

pls provide solution

Asked by saisidh05 | 04 Jul, 2022, 03:04: PM

CBSE 12-science - Maths

Asked by givduf | 11 Jul, 2020, 09:00: AM

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CBSE 12-science - Maths

please answer.

Asked by Kchawla94 | 02 Apr, 2019, 06:45: PM

ANSWERED BY EXPERT

Divide 24 into 3 parts such that the product of the first, square of second and cube of the third is maximum.

Divide the number 30 into two parts such that their product is maximum.

Let the first part of 30 – x.

∴ Their product

= x(30 – x) 

= 30x –  x2

= f'(x)         ......(Say)

∴ f'(x) = `"d"/("d"x)(30 - 2x)`

= 0 – 2 × 1

= – 2

The root of the equation f'(x) = 0

i.e. 30 – 2x = 0 is x = 15

and f'(15) = – 2 < 0

∴ By the second derivative test, f is maximum at x = 15.

Hence, the required parts of 30 are 15 and 15.

Concept: Maxima and Minima

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Page 2

Let the first part of 20 be x.Then the second part is 20 – x.∴ sum of their squares 

= x2 + (20 – x)2

∴ f'(x)  `d/dx[x^2 + (20 - x)^2]`

= `2x + 2(20 - x).d/dx(20 - x)`

= 2x + 2(20 – x) x (0 – 1)= 2x – 40 + 2x= 4x – 40and

f"(x) = `d/dx(4x - 40)`

= 4 x 1 – 0= 4The root of the equation f'(x) = 0,i.e. 4x – 40 = 0 is x = 10 and f"(10) = 4 > 0∴ by the second derivative test, f is minimum at x = 10.

Hence, the required parts of 20 are 10 and 10.