- Minimize $z(x,y)=3x+2y+4$
- Subject to $x^2y=48$
- Where $x,y> 0$
Antwoord 1 correct
Correct
Antwoord 2 optie
$z(\frac{4}{3}\sqrt[3]{36},\sqrt[3]{36})=6\sqrt[3]{36}+4$
Antwoord 2 correct
Fout
Antwoord 3 optie
$z(\frac{3}{4}\sqrt[3]{85\frac{1}{3}},\sqrt[3]{85\frac{1}{3}})=4\frac{1}{4}\sqrt[3]{85\frac{1}{3}}+4$
Antwoord 3 correct
Fout
Antwoord 4 optie
None of the other answers is correct.
Antwoord 4 correct
Fout
Antwoord 1 optie
$z(4,3)=22$
Antwoord 1 feedback
Correct: The First-order condition gives $\dfrac{3}{2}=\dfrac{2xy}{x^2}=\dfrac{2y}{x}$ what results in $x=\frac{4}{3}y$. Plugging this into the restriction gives $(\frac{4}{3}y)^2y=48$, which results in $y=3$ with $x=4$. $z(4,3)=22$. There are no boundaries. Therefore, we take (for instance) $z(2,12)=34$ and $z(6,1\frac{1}{3})=24\frac{2}{3}$. From this it follows that $z(4,3)=22$ is the minimum.
Go on.
Go on.
Antwoord 2 feedback
Wrong: $(\frac{4}{3}y)^2\neq \frac{4}{3}y^2$.
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Try again.
Antwoord 3 feedback
Wrong: $3x=4y$ does not result in $x=\frac{3}{4}y$.
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Try again.
Antwoord 4 feedback
Wrong: The correct answer is among them.
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Try again.