The function $Q(L,K)$ is given by $Q(L,K)=(2L+3KL)^3$, $(L,K\geq 0)$. Determine the point $(L,K)$ where the line with slope $-\frac{8}{3}$ is tangent to the level curve of the function $Q(L,K)$ with $Q$-value 512.
Antwoord 1 correct
Correct
Antwoord 2 optie
$(L,K)=(3,\frac{11}{24})$
Antwoord 2 correct
Fout
Antwoord 3 optie
$(L,K)=(\sqrt{\dfrac{512^{-3}}{8}},\dfrac{24\sqrt{\dfrac{512^{-3}}{8}}-6}{9})$
Antwoord 3 correct
Fout
Antwoord 4 optie
$(L,K)=(8,3)$
Antwoord 4 correct
Fout
Antwoord 1 optie
$(L,K)=(1,2)$
Antwoord 1 feedback
Correct: $$\begin{align*}
\dfrac{Q'_L(L,K)}{Q'_K(L,K)}&= \dfrac{3(2L+3KL)^2\cdot(2+3K)}{3(2L+3KL)^2\cdot3L}\\
& = \dfrac{2+3K}{3L}
\end{align*}$$
$$\begin{align*}
-\frac{8}{3}&=- \dfrac{2+3K}{3L},
\end{align*}$$
and hence, $K=\dfrac{24L-6}{9}$.
Plugging in $Q(L,K)$:
$$\begin{align*}
(2L+3\dfrac{24L-6}{9}L)^3&=512\\
2L+3\dfrac{24L-6}{9}L&=8\\
2L+\dfrac{24L^2-6L}{3}&=8\\
2L+8L^2-2L&=8\\
8L^2&=8\\
L&=1
\end{align*}$$
Hence, $K=\dfrac{24\cdot 1-6}{9}=2$.
Go on.
\dfrac{Q'_L(L,K)}{Q'_K(L,K)}&= \dfrac{3(2L+3KL)^2\cdot(2+3K)}{3(2L+3KL)^2\cdot3L}\\
& = \dfrac{2+3K}{3L}
\end{align*}$$
$$\begin{align*}
-\frac{8}{3}&=- \dfrac{2+3K}{3L},
\end{align*}$$
and hence, $K=\dfrac{24L-6}{9}$.
Plugging in $Q(L,K)$:
$$\begin{align*}
(2L+3\dfrac{24L-6}{9}L)^3&=512\\
2L+3\dfrac{24L-6}{9}L&=8\\
2L+\dfrac{24L^2-6L}{3}&=8\\
2L+8L^2-2L&=8\\
8L^2&=8\\
L&=1
\end{align*}$$
Hence, $K=\dfrac{24\cdot 1-6}{9}=2$.
Go on.
Antwoord 2 feedback
Antwoord 3 feedback
Antwoord 4 feedback