maximize ( b = 5xy^{2} ), where ( x ) and ( y ) are positive numbers such that ( x + y^{2}=12 ).\nthe…

maximize ( b = 5xy^{2} ), where ( x ) and ( y ) are positive numbers such that ( x + y^{2}=12 ).\nthe maximum value of ( b ) is\n(simplify your answer. type an exact answer, using radicals as needed)

maximize ( b = 5xy^{2} ), where ( x ) and ( y ) are positive numbers such that ( x + y^{2}=12 ).\nthe maximum value of ( b ) is\n(simplify your answer. type an exact answer, using radicals as needed)

Answer

Explanation:

Step1: Express (x) in terms of (y)

From (x + y^{2}=12), we get (x = 12 - y^{2}).

Step2: Substitute (x) into (B)

Substitute (x = 12 - y^{2}) into (B = 5xy^{2}), then (B(y)=5(12 - y^{2})y^{2}=60y^{2}-5y^{4}).

Step3: Find the derivative of (B(y))

Using the power rule ((x^{n})^\prime=nx^{n - 1}), (B^\prime(y)=(60y^{2}-5y^{4})^\prime=120y-20y^{3}=20y(6 - y^{2})).

Step4: Find the critical points

Set (B^\prime(y)=0). Since (y>0), (20y(6 - y^{2}) = 0) gives (y^{2}=6) (because (y>0), (y=\sqrt{6})).

Step5: Find the second - derivative of (B(y))

(B^{\prime\prime}(y)=(120y - 20y^{3})^\prime=120-60y^{2}). When (y^{2}=6), (B^{\prime\prime}(\sqrt{6})=120-60\times6=120 - 360=- 240<0). So (B(y)) has a maximum at (y^{2}=6).

Step6: Find the value of (x)

When (y^{2}=6), from (x = 12 - y^{2}), we have (x=12 - 6 = 6).

Step7: Calculate the maximum value of (B)

Substitute (x = 6) and (y^{2}=6) into (B = 5xy^{2}), (B=5\times6\times6 = 180).

Answer:

(180)