Question

Divide the number 20 in two positive integers such that the product of square of one and cube of other is maximum.​

Answer 1

Correct option is

A

8,12

Let the ture positive numbers be x,(x,y>0)

x+y=20 - Given

We need to maximise x

2

y

3

x=20−y

f(y)=y

3

(20−y)

2

for minimum

f

′

(y)=0

3y

2

(20−y)

2

+y

3

2(20−y)(−1)=0

3y

2

(20−y)

2

−2y

3

(20−y)=0

y

2

(20−y)(3(20−y)−2y)=0

60−3y−2y=0 y



=0&y



=20

y=12 if y=20 then x=0 which does not

x=8 make the x^2y^3

maximum

here is your answer

have a nice day ahead

thanks