Exercise 3 | Wiskunde op Tilburg University

Antwoord 1 correct

Correct

Antwoord 2 optie

$z(1\frac{1}{3},6\frac{2}{3})=6\frac{14}{27}$

Antwoord 2 correct

Fout

Antwoord 3 optie

$z(0,64)=0$

Antwoord 3 correct

Fout

Antwoord 4 optie

$z(16,64)=-3072$

Antwoord 4 correct

Fout

Antwoord 1 optie

$z(2,36)=64$

Antwoord 1 feedback

Correct: We rewrite

$\sqrt{y}+x=8$ to

$y=(8-x)^2$ . We plug this into the objective function:

$Z(x)=x(8-x)^2-x^3=-16x^2+64x$ .

$Z'(x)=-32x+64$ . Putting equal to zero results in

$x=2$ and

$y=36$ .

$z(2,36)=64$ . We check the boundaries:

$z(8,0)=-512$ and

$z(0,64)=0$ . Hence,

$z(2,36)=64$ is a maximum.

Go on.

Antwoord 2 feedback

Wrong:

$\sqrt{y}+x=8$ cannot be rewritten as

$y=8-x$ .

Try again.

Antwoord 3 feedback

Wrong: This is not a maximum, but a minimum.

Try again.

Antwoord 4 feedback

Wrong: To find a stationary point of the function you need to put the derivative (and not the function itself) equal to zero.

See Stationary point.