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=-\frac14`

Now when
`z=\frac12`, we get

`\begin{cases}x=ay^2\\y=\frac b8\end{cases}\implies x=a\left(\frac b8\right)^2=(ab^2)/(64)=(-\frac14)/(64)=\boxed{-\frac1{256}}`