Exercise 2 | Wiskunde op Tilburg University

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.

Antwoord 2 feedback

Wrong:

$(2y)^2\neq 2y^2$ .

Try again.

Antwoord 3 feedback

Wrong:

$4y^2=x^2$ does not imply

$4y=x$ .

Try again.

Antwoord 4 feedback

Wrong: Do no just guess.

Try (again).