- Maximize $z(x,y)=4x^{\frac{2}{3}}y^{\frac{1}{3}}$
- Subject to $\frac{1}{2}x^2+y^2=27$
- Where $x,y\geq 0$
Antwoord 1 correct
Correct
Antwoord 2 optie
$z(2\cdot \sqrt{13\frac{1}{2}}, \sqrt{13\frac{1}{2}})=23\frac{33}{100}$
Antwoord 2 correct
Fout
Antwoord 3 optie
$z(4\cdot \sqrt{3}, \sqrt{3})=17\frac{47}{103}$
Antwoord 3 correct
Fout
Antwoord 4 optie
$z(2,5)=10\frac{6}{7}$
Antwoord 4 correct
Fout
Antwoord 1 optie
$z(6,3)=4\cdot \sqrt[3]{108}$
Antwoord 1 feedback
Correct: $\dfrac{\frac{8}{3} x^{-\frac{1}{3}}y^{\frac{1}{3}}}{\frac{4}{3} x^{\frac{2}{3}}y^{-\frac{2}{3}}}=\dfrac{2y}{x}=\dfrac{x}{2y}$ and hence, $4y^2=x^2$ and therefore, $2y=x$. We plug this into the restriction: $\frac{1}{2}(2y)^2+y^2=27$, which gives $y=3$. As a consequence, $x=2\cdot 3=6$. $z(6,3)=4\cdot \sqrt[3]{108}$. We check the boundaries: $z(\sqrt{54},0)=0$ and $z(0,\sqrt{27})=0$. Hence, $z(6,3)=4\cdot \sqrt[3]{108}$ is the maximum.
Go on.
Go on.
Antwoord 2 feedback
Wrong: $(2y)^2\neq 2y^2$.
Try again.
Try again.
Antwoord 3 feedback
Wrong: $4y^2=x^2$ does not imply $4y=x$.
Try again.
Try again.
Antwoord 4 feedback
Wrong: Do no just guess.
Try (again).
Try (again).