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Exercises Polynomials and factoring

Example 3 from Algebra and Trigonometry, 5.8 Modeling Using Variation

Solving an Inverse Variation Problem

A quantity $y$ varies inversely with the cube of $x .$ If $y = 25$ when $x = 2,$ find $y$ when $x$ is 6.

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
The general formula for inverse variation with a cube is $y = \frac{k}{x^{3}} .$ The constant can be found by multiplying $y$ by the cube of $x .$

$\begin{matrix} k & = & x^{3} y \\ = & 2^{3} \cdot 25 \\ = & 200 \end{matrix}$
Now we use the constant to write an equation that represents this relationship.

$\begin{matrix} y & = & \frac{k}{x^{3}}, k = 200 \\ y & = & \frac{200}{x^{3}} \end{matrix}$

Substitute $x = 6$ and solve for $y .$

$\begin{matrix} y & = & \frac{200}{6^{3}} \\ = & \frac{25}{27} \end{matrix}$

Your turn

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

A quantity $y$ varies inversely with the square of $x .$ If $y = 8$ when $x = 3,$ find $y$ when $x$ is 4.

How did it go?

PreviousWriting a Formula for an Inversely Proportional Relationship Next Solving Problems Involving Joint Variation

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 5.8, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.