When two quantities vary inversely, their products are always equal to a constant, which we can call k. If the square of x and the cube of y vary inversely, this means that the product of the square of x and the cube of y will equal k. We can represent the square of x as x² and the cube of y as y³. Now, we can write the equation for inverse variation.

x²y³ = k

We are told that when x = 8, y = 8. We can substitute these values into our equation for inverse variation and then solve for k.

8²(8³) = k

k = 8²(8³)

Because this will probably be a large number, it might help just to keep it in exponent form. Let's apply the property of exponents which says that a^ba^c = a^b+c.

k = 8²(8³) = 8²⁺³ = 8⁵.

Next, we must find the value of y when x = 1. Let's use our equation for inverse variation equation, substituting 8⁵ in for k.

x²y³ = 8⁵

(1)²y³ = 8⁵

y³ = 8⁵

In order to solve this, we will have to take a cube root. Thus, it will help to rewrite 8 as the cube of 2, or 2³.

y³ = (2³)⁵

We can now apply the property of exponents that states that (a^b)^c = a^bc.

y³ = 2^3•5 = 2¹⁵ 

In order to get y by itself, we will have the raise each side of the equation to the 1/3 power.

(y³)^(1/3) = (2¹⁵)^(1/3)

Once again, let's apply the property (a^b)^c = a^bc.

y^{(3 • 1/3)} = 2^{(15 • 1/3)}

y = 2⁵ = 32

The answer is 32. 

$\displaystyle x$ varies directly as $\displaystyle A$ and inversely as $\displaystyle y$, so for some constant of variation $\displaystyle K$,

$\displaystyle \frac{xy}{A} =K$.

We can square both sides to obtain:

$\displaystyle \left (\frac{xy}{A} \right ) ^{2}=K^{2}$

$\displaystyle \frac{x^{2}y^{2}}{A^{2}}= K^{2}$

$\displaystyle z = y ^{2}$.

$\displaystyle B = \sqrt{A}$, so $\displaystyle B^{4} =\left ( \sqrt{A} \right )^{4} = A^{2}$.

By substitution,

$\displaystyle \frac{x^{2}z}{B^{4}}= K^{2}$.

Using $\displaystyle K^{2}$ as the constant of variation, we see that $\displaystyle z$ varies inversely as the square of $\displaystyle x$ and  directly as the fourth power of $\displaystyle B$.

The volume of a cylinder can be calculated from its height and the radius of its base using the formula:

$\displaystyle \pi r^{2} h = V$

$\displaystyle t = \sqrt{r}$, so $\displaystyle t ^{2}= r$;

$\displaystyle v = \sqrt[4]{h}$, so $\displaystyle h = v^{4}$.

The volume is 1,000, and by substitution, using the other equations:

$\displaystyle \pi (t^{2})^{2} v^{4} = 1,000$

$\displaystyle \pi t^{4} v^{4} = 1,000$

$\displaystyle \frac{\pi t^{4} v^{4}}{\pi} = \frac{1,000}{\pi}$

$\displaystyle t^{4} v^{4} = \frac{1,000}{\pi}$

$\displaystyle \sqrt[4]{t^{4} v^{4} }= \sqrt[4]{\frac{1,000}{\pi}}$

$\displaystyle tv= \sqrt[4]{\frac{1,000}{\pi}}$

If we take $\displaystyle K= \sqrt[4]{\frac{1,000}{\pi}}$as the constant of variation, we get

$\displaystyle tv=K$,

meaning that $\displaystyle t$ varies inversely as $\displaystyle v$.

$\displaystyle s$ varies directly as the square of $\displaystyle y$ and the cube root of $\displaystyle t$, and inversely as the fourth root of $\displaystyle w$, so, for some constant of variation $\displaystyle K$,

$\displaystyle \frac{s \sqrt[4]{w}}{y^{2}\sqrt[3]{t}} = K$

We take the reciprocal of both sides, then extract the square root:

$\displaystyle \frac{y^{2}\sqrt[3]{t}}{s \sqrt[4]{w}} =\frac{1} { K}$

$\displaystyle \frac{y^{2}t^{\frac{1}{3}}}{s w ^{\frac{1}{4}}} =\frac{1} { K}$

$\displaystyle \left ( \frac{y^{2}t^{\frac{1}{3}}}{s w ^{\frac{1}{4}}} \right )^{\frac{1}{2}} =\left (\frac{1} { K} \right )^{\frac{1}{2}}$

$\displaystyle \frac{y t^{\frac{1}{6}}}{s ^{\frac{1}{2}} w ^{\frac{1}{8}}}=\left (\frac{1} { K} \right )^{\frac{1}{2}}$

$\displaystyle \frac{y\sqrt[6]{t}}{\sqrt{s} \cdot \sqrt[8]{w}}=\left (\frac{1} { K} \right )^{\frac{1}{2}}$

Taking $\displaystyle \left (\frac{1} { K} \right )^{\frac{1}{2}}$as the constant of variation, we see that $\displaystyle y$ varies directly as the square root of $\displaystyle s$ and the eighth root of $\displaystyle w$, and inversely as the sixth root of $\displaystyle t$.

The formula for inverse variation is as follows: $\displaystyle y=\frac{k}{x}$

Use the x and y values from the first part of the sentence to find k. 

$\displaystyle 14=\frac{k}{3}$

$\displaystyle k=42$

Then use that k value and the given x value to find y.

$\displaystyle y=\frac{42}{28}$

$\displaystyle y=1\frac{1}{2}$