Question

The variable $x$ varies directly as the square of $y$, and $y$ varies directly as the cube of $z$. If $x$ equals $-16$ when $z$ equals 2, what is the value of $x$ when $z$ equals $\frac{1}{2}$?

Answer 1

If $x$ varies directly as $y^2$, then there is some constant $a$ for which

$x=ay^2$

Similarly, there is some constant $b$ such that

$y=bz^3$

Given that $x=-16$ when $z=2$, we have

$\begin{cases}-16=ay^2\\y=8b\end{cases}\implies-16=a(8b)^2\implies ab^2=-\dfrac14$

Now when $z=\frac12$, we get

$\begin{cases}x=ay^2\\y=\frac b8\end{cases}\implies x=a\left(\dfrac b8\right)^2=\dfrac{ab^2}{64}=\dfrac{-\frac14}{64}=\boxed{-\dfrac1{256}}$