# x varies directly as the square of y and x = 12 when y = 4. What is the value of x when y = 6?

## Question:

x varies directly as the square of y and x = 12 when y = 4. What is the value of x when y = 6?

## Variation:

(i) If x varies directly as y then {eq}x=my {/eq}, where 'm' is a proportionality constant.

(ii) If x varies inversely as y then {eq}x= \dfrac{n}{y} {/eq}, where 'n' is a proportionality constant.

The problem says, "{eq}x {/eq} varies directly as the square of {eq}y {/eq}.

So by the definition of direct variation,

$$x= k y^2 \,\,\,\,\,\,\rightarrow (1)$$

Here, {eq}k {/eq} is the proportionality constant.

Substitute the first set of given values {eq}x=12 {/eq} and {eq}y=4 {/eq} in (1) and solve for {eq}k {/eq}:

$$12 = k (4)^2 \\ 12= 16k \\ \text{Dividing both sides by 16}, \\ k= \dfrac{12}{16}= \dfrac{3}{4}$$

Substitute this and another given value {eq}y=6 {/eq} in (1) and solve for {eq}x {/eq}:

$$x= \dfrac{3}{4} (6)^2 = \dfrac{3}{4}(36) = \boxed{\mathbf{27}}$$