Question

maximize ( b = 6xy^{2} ), where ( x ) and ( y ) are positive numbers such that ( x + y^{2}=11 ). the maximum value of ( b ) is (simplify your answer. type an exact answer, using radicals as needed)

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:

\(\frac{363}{2}\)