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.

Answer and Explanation:

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}:

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


Learn more about this topic:

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Solving Equations of Direct Variation

from Algebra I: High School

Chapter 17 / Lesson 9
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