Exercise 4 | Wiskunde op Tilburg University

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.

Antwoord 2 feedback

Wrong:

$\textrm{slope}\neq - \dfrac{Q'_K(L,K)}{Q'_L(L,K)}$ .

See Tangent line to level curve.

Antwoord 3 feedback

Wrong:

$(a^3)^{-3}\neq a^1$ .

See Properties power functions.

Antwoord 4 feedback

Wrong:

$\dfrac{Q'_L(L,K)}{Q'_K(L,K)}\neq -\dfrac{L}{K}$ .

See Tangent line level curve.