minimize ( q = 2x^{2}+4y^{2} ), where ( x + y = 6 )
( x=)
( y=)
(simplify your answer. type an exact answer, using radicals as needed. use integers or fractions for any numbers in the expression )

Question

Accepted Answer

# Explanation:
## Step1: Express $y$ in terms of $x$
From $x + y=6$, we get $y = 6 - x$.
## Step2: Substitute $y = 6 - x$ into $Q$
$$
\begin{align*}
Q&=2x^{2}+4(6 - x)^{2}\
&=2x^{2}+4(36-12x+x^{2})\
&=2x^{2}+144-48x + 4x^{2}\
&=6x^{2}-48x + 144
\end{align*}
$$
## Step3: Find the derivative of $Q$ with respect to $x$
$Q^\prime(x)=12x-48$
## Step4: Set the derivative equal to zero and solve for $x$
$12x-48 = 0$
$12x=48$
$x = 4$
## Step5: Find $y$
Since $y=6 - x$, when $x = 4$, $y=6-4=2$

# Answer:
$x = 4$
$y = 2$

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QUESTION IMAGE

Question

Explanation:

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

Step2: Substitute \(y = 6 - x\) into \(Q\)

Step3: Find the derivative of \(Q\) with respect to \(x\)

Step4: Set the derivative equal to zero and solve for \(x\)

Step5: Find \(y\)

Answer: