Question

divide 30 into two parts(not necessarily integers) such that the product of one part with the square of the other shall be a maximum

Answer 1

Answer:

10 and 20

Step-by-step explanation:

To find-----> Divide 30 in to two parts such that the product of one part with square of other shall be maximum.

Solution---> Let two parts of 30 be x and ( 30 - x )

Now , Let product of square of one part and other part be p .

p = x² ( 30 - x )

= 30 x² - x³

Now , differentiating with respect to x .

=> dp / dx = d / dx ( 30 x² - x³ )

= 30 ( 2x ) - 3 x²

=> dp / dx = 60x - 3x²

Differentiating with respect to x again , we get,

=> d²p / dx² = d / dx ( 60x ) - d / d ( 3x² )

= 60 ( 1 ) - 3 ( 2x )

=> d²p / dx² = 60 - 6x

Now , for maximum p

dp / dx = 0

=> 60 x - 3 x² = 0

=> 3 x ( 20 - x ) = 0

x ≠ 0 So ,

( 20 - x ) = 0

=> x = 20

Now , ( d²p / dx² ) at x = 20

= 60 - 6 ( 20 )

= 60 - 120

= - 60 ( negative )

So , p is maximum at x = 20 .

So one part is 20 and other part is

( 30 - x ) = 30 - 20 = 10 .

So parts of 30 are 10 and 20 .