- Maximize $U(x,y)=x^{\frac{1}{2}}y^{\frac{1}{2}}$
- Subject to$8x+y=16$
- Where $x,y\geq 0$
Antwoord 1 correct
Correct
Antwoord 2 optie
$U(\frac{2}{3},10\frac{2}{3})=2\frac{2}{3}$
Antwoord 2 correct
Fout
Antwoord 3 optie
$U(1\frac{1}{3},5\frac{1}{3})=2\frac{2}{3}$
Antwoord 3 correct
Fout
Antwoord 4 optie
$U(4,4)=4$
Antwoord 4 correct
Fout
Antwoord 1 optie
$U(1,8)=\sqrt{8}$
Antwoord 1 feedback
Correct: $\dfrac{U'_x(x,y)}{U'_y(x,y)}=\dfrac{\frac{1}{2}x^{-\frac{1}{2}}y^{\frac{1}{2}}}{\frac{1}{2}x^{\frac{1}{2}}y^{-\frac{1}{2}}}=\dfrac{8}{1}$ gives $8x=y$. We plug this into the restriction: $8x+8x=16$. Hence, $x=1$ and that gives $y=8$. $U(1,8)=\sqrt{8}$. We determine via the boundary points whether this is a maximum. $U(2,0)=0$ and $U(0,16)=0$ and hence $U(1,8)=\sqrt{8}$ is the maximum.
Go on.
Go on.
Antwoord 2 feedback
Wrong: $\dfrac{U'_x(x,y)}{U'_y(x,y)}=\dfrac{\frac{1}{2}x^{-\frac{1}{2}}y^{\frac{1}{2}}}{\frac{1}{2}x^{\frac{1}{2}}y^{-\frac{1}{2}}}$.
Try again.
Try again.
Antwoord 3 feedback
Wrong: $\dfrac{U'_x(x,y)}{U'_y(x,y)}=\dfrac{\frac{1}{2}x^{-\frac{1}{2}}y^{\frac{1}{2}}}{\frac{1}{2}x^{\frac{1}{2}}y^{-\frac{1}{2}}}$.
Try again.
Try again.
Antwoord 4 feedback
Wrong: $(x,y)=(4,4)$ is not allowed, because $8\cdot 4+4\neq 16$.
Try again.
Try again.