# If y varies jointly as the cube of x, w, and the square of z, what is the effect on w when x, y, and z are doubled?

Jun 1, 2017

$w$ decreases by a factor of 16.

That is, ${w}_{\text{new}} = \frac{1}{16} w$

#### Explanation:

Based on this description, we can create the following equation:

$y = k {x}^{3} w {z}^{2}$

Where $k$ is a constant which doesn't really matter for this particular problem.

The problem asks what happens to $w$ when the other variables are doubled, so let's solve for $w$ to see what it is originally.

$y = k {x}^{3} w {z}^{2}$

$\frac{y}{k {x}^{3} {z}^{2}} = w$

Now, the problem asks us to double everything except $w$ and see what happens.

$\frac{\left(2 y\right)}{k {\left(2 x\right)}^{3} {\left(2 z\right)}^{2}} = {w}_{\text{new}}$

$\frac{2 y}{k \left(8 {x}^{3}\right) \left(4 {z}^{2}\right)} = {w}_{\text{new}}$

$\frac{2}{8 \cdot 4} \left(\frac{y}{k {x}^{3} {z}^{2}}\right) = {w}_{\text{new}}$

$\frac{1}{16} \left(\frac{y}{k {x}^{3} {z}^{2}}\right) = {w}_{\text{new}}$

$\frac{1}{16} w = {w}_{\text{new}}$

So $w$ decreases by a factor of 16 when the other 3 variables are doubled.