Exercise 1

Correct: $y=\dfrac{10}{x+1}$.

$Z(x)=\dfrac{30x^2}{x+1}$.

$Z'(x)=\dfrac{30x(x+2)}{(x+1)^2}$. Hence, $Z'(x)=0$ for $x=0$ (outside domain) or $x=-2$. Via a sign chart you can determine that $x=-2$ is a maximum location.

$y(-2)=\dfrac{10}{-2+1}=-10$.

Hence, $z(-2,-10)=-120$ is the solution.

Go on.

Wrong: $(x,y)=(0,10)$ is not part of the domain.

Try again.

Wrong: You cannot find the zeros of $\dfrac{30x^2+60x}{(x+1)^2}$ by putting the nominator and denominator equal to each other. Moreover, you cannot round intermediate solutions ($\sqrt{3252}$).

Try again.

Wrong: $z(-5,-2\frac{1}{2})\neq187\frac{1}{2}$, because $(-5)^2\neq -25$.

Try again.