Maximize $B=4 x y^{2}$, where $x$ and $y$ are positive numbers such that $x+y^{2}=6$.
The maximum value of $B$ is
(Simplify your answer. Type an exact answer, using radicals as needed.)

Step 1 ：Define the function to be maximized as \(B=4xy^2\) and the constraint as \(x+y^2=6\).

Step 2 ：Set up the Lagrangian function as \(L(x, y, λ) = 4xy^2 - λ(x+y^2-6)\).

Step 3 ：Find the partial derivatives of \(L\) with respect to \(x\), \(y\), and \(λ\), and set them equal to zero. This gives us the system of equations: \(4y^2 - λ = 0\), \(8xy - 2yλ = 0\), and \(-x - y^2 + 6 = 0\).

Step 4 ：Solve the system of equations to find the values of \(x\), \(y\), and \(λ\). The solutions are \((3, -\sqrt{3}, 12)\), \((3, \sqrt{3}, 12)\), and \((6, 0, 0)\).

Step 5 ：Since \(x\) and \(y\) are positive numbers, discard the solution \((3, -\sqrt{3}, 12)\). We are left with two solutions: \((3, \sqrt{3}, 12)\) and \((6, 0, 0)\).

Step 6 ：Substitute these solutions back into the function \(B=4xy^2\) to find which one gives the maximum value. For \((3, \sqrt{3}, 12)\), \(B=36\). For \((6, 0, 0)\), \(B=0\).

Step 7 ：\(\boxed{36}\) is the maximum value of \(B\).