- Maximalize $z(x,y)=x^2+2xy+y^2+16y$
- Subject to $x+3y=6$
- Where $x,y\geq 0$
Antwoord 1 correct
Correct
Antwoord 2 optie
$z(3,1)=32$
Antwoord 2 correct
Fout
Antwoord 3 optie
$z(\frac{3}{4},1\frac{3}{4})=34\frac{1}{4}$
Antwoord 3 correct
Fout
Antwoord 4 optie
$z(-6,4)=68$
Antwoord 4 correct
Fout
Antwoord 1 optie
None of the other answers is correct.
Antwoord 1 feedback
Correct: We rewrite $x+3y=6$ as $x=6-3y$ and plug this into the objective function: $Z(y)=(6-3y)^2+2(6-3y)y+y^2+16y=4y^2-8y+36$. $Z'(y)=8y-8$. Hence, the stationary point is $y=1$, with corresponding $x=3$. $z(3,1)=32$. We check the boundaries: $z(0,2)=16$, but $z(6,0)=36$. Hence, $z(6,0)=36$ is the maximum.
Go on.
Go on.
Antwoord 2 feedback
Wrong: Do not forget the boundaries.
Try again.
Try again.
Antwoord 3 feedback
Wrong: $(6-3y)^2 \neq 6-3y^2$.
Try again.
Try again.
Antwoord 4 feedback
Wrong: $x=-6$ is outside the domain of the function.
See Constrained optimization functions of two variables.
See Constrained optimization functions of two variables.