maximize q = 3xy², where x and y are positive numbers such that x + y² = 1. x = 1/2 y = (type an exact…
maximize q = 3xy², where x and y are positive numbers such that x + y² = 1. x = 1/2 y = (type an exact answer, using radicals as needed.)
Answer
Explanation:
Step1: Express $y^2$ in terms of $x$
Given $x + y^2=1$, then $y^2 = 1 - x$.
Step2: Substitute $y^2$ into $Q$
Substitute $y^2 = 1 - x$ into $Q = 3xy^2$, we get $Q(x)=3x(1 - x)=3x-3x^{2}$.
Step3: Find the derivative of $Q(x)$
$Q^\prime(x)=\frac{d}{dx}(3x - 3x^{2})=3-6x$.
Step4: Set the derivative equal to 0
Set $Q^\prime(x)=0$, so $3 - 6x=0$. Solving for $x$ gives $x=\frac{1}{2}$.
Step5: Find $y$
Since $y^2=1 - x$ and $x = \frac{1}{2}$, then $y^2=1-\frac{1}{2}=\frac{1}{2}$. As $y>0$, $y=\frac{\sqrt{2}}{2}$.
Answer:
$\frac{\sqrt{2}}{2}$