maximize ( b = 6xy^{2} ), where ( x ) and ( y ) are positive numbers such that ( x + y^{2}=11 ). the maximum…

Explanation:

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

From (x + y^{2}=11), we get (x = 11 - y^{2}).

Step2: Substitute (x) into (B)

Substitute (x = 11 - y^{2}) into (B = 6xy^{2}), then (B(y)=6(11 - y^{2})y^{2}=66y^{2}-6y^{4}).

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

Using the power rule ((x^{n})^\prime=nx^{n - 1}), (B^\prime(y)=(66y^{2}-6y^{4})^\prime=132y-24y^{3}=12y(11 - 2y^{2})).

Step4: Find the critical points

Set (B^\prime(y)=0). Since (y>0), (12y(11 - 2y^{2}) = 0) gives (11-2y^{2}=0). Solving for (y), we have (y^{2}=\frac{11}{2}), so (y=\sqrt{\frac{11}{2}}) (because (y>0)).

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

(B^{\prime\prime}(y)=(132y - 24y^{3})^\prime=132-72y^{2}). Substitute (y^{2}=\frac{11}{2}) into (B^{\prime\prime}(y)): (B^{\prime\prime}(\sqrt{\frac{11}{2}})=132-72\times\frac{11}{2}=132 - 396=- 264<0). So (B(y)) has a maximum at (y^{2}=\frac{11}{2}).

Step6: Find the value of (x)

When (y^{2}=\frac{11}{2}), from (x = 11 - y^{2}), we get (x=11-\frac{11}{2}=\frac{11}{2}).

Step7: Calculate the maximum value of (B)

Substitute (x=\frac{11}{2}) and (y^{2}=\frac{11}{2}) into (B = 6xy^{2}). Then (B = 6\times\frac{11}{2}\times\frac{11}{2}=\frac{726}{4}=\frac{363}{2}).

Answer:

(\frac{363}{2})