Exercise 4 | Wiskunde op Tilburg University

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.

Antwoord 2 feedback

Wrong: Do not forget the boundaries.

Try again.

Antwoord 3 feedback

Wrong:

$(6-3y)^2 \neq 6-3y^2$ .

Try again.

Antwoord 4 feedback

Wrong:

$x=-6$ is outside the domain of the function.

See Constrained optimization functions of two variables.