Example 4 from Algebra and Trigonometry, 5.8 Modeling Using Variation

Solving Problems Involving Joint Variation

A quantity $x$ varies directly with the square of $y$ and inversely with the cube root of $z .$ If $x = 6$ when $y = 2$ and $z = 8,$ find $x$ when $y = 1$ and $z = 27.$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Begin by writing an equation to show the relationship between the variables.

$x = \frac{k y^{2}}{\sqrt[3]{z}}$

Substitute $x = 6,$ $y = 2,$ and $z = 8$ to find the value of the constant $k .$

$\begin{matrix} 6 & = & \frac{k 2^{2}}{\sqrt[3]{8}} \\ 6 & = & \frac{4 k}{2} \\ 3 & = & k \end{matrix}$

Now we can substitute the value of the constant into the equation for the relationship.

$x = \frac{3 y^{2}}{\sqrt[3]{z}}$

To find $x$ when $y = 1$ and $z = 27,$ we will substitute values for $y$ and $z$ into our equation.

$\begin{matrix} x & = & \frac{3 {(1)}^{2}}{\sqrt[3]{27}} \\ = & 1 \end{matrix}$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

A quantity $x$ varies directly with the square of $y$ and inversely with $z .$ If $x = 40$ when $y = 4$ and $z = 2,$ find $x$ when $y = 10$ and $z = 25.$

How did it go?

Solution

Your turn

Solution

Your turn