Find all the possible values of `b >0,` so that the area of the bounded region enclosed between the parabolas `y=x-b x^2a n dy=(x^2)/b` is maximum.

Answer

b = 1

Explanation

Eliminating y from `y=(x^(2))/(b)` and `y=x-bx^(2)`, we get
`x^(2)=bx-b^(2)x^(2)`
` rArr x=0, (b)/(1+b^(2))`

Thus, the area enclosed between the parabolas
`A=int_(0)^(b(1+b)^(2))(x-bx^(2)-(x^(2))/(b))dx `
`=[(x^(2))/(2)-(x^(3))/(3)*(1+b^(2))/(b)]_(0)^(b(1+b)^(2))=(1)/(6)*(b^(2))/((1+b^(2))^(2))`
On differentiating w.r.t. b, we get
`(dA)/(db)=(1)/(6)*((1+b^(2))^(2)*2b-2b^(2)*(1+b^(2))*2b)/((1+b^(2))^(4))`
`=(1)/(3)*(b(1-b^(2)))/((1+b^(2))^(3))`
For maximum value of A, put `(dA)/(db)=0`
` rArr b=-1,0,1, " since "b gt 0 `
` therefore " wo oonsider only " b=1.`
Sign scheme for ` (dA)/(db) " around " b=1 ` is as shown below:

From sign scheme, it is clear that A is maximum at ` b=1.`