The weight wkg of a metal bar varies jointly as it's length lmm and the square    of it's diameter dmm. If w=140 when d=4 and l=54, find d in terms of w and l

To solve this problem, we can use the concept of joint variation. The relationship can be expressed as:

\[ w = k \cdot l \cdot d^2 \]

where \( k \) is the constant of variation.

Given:
- \( w = 140 \)
- \( d = 4 \)
- \( l = 54 \)

First, we can substitute these values into the equation to find \( k \):

\[
140 = k \cdot 54 \cdot 4^2
\]

Calculating \( 4^2 \):

\[
4^2 = 16
\]

Now substitute that back into the equation:

\[
140 = k \cdot 54 \cdot 16
\]

Calculating \( 54 \cdot 16 \):

\[
54 \cdot 16 = 864
\]

So, the equation becomes:

\[
140 = k \cdot 864
\]

Now, solving for \( k \):

\[
k = \frac{140}{864}
\]

Simplifying \( \frac{140}{864} \):

\[
k = \frac{35}{216}
\]

Now that we have \( k \), we can express \( d \) in terms of \( w \) and \( l \):

Starting from the original equation:

\[
w = k \cdot l \cdot d^2
\]

Rearranging it to solve for \( d^2 \):

\[
d^2 = \frac{w}{k \cdot l}
\]

Taking the square root to find \( d \):

\[
d = \sqrt{\frac{w}{k \cdot l}}
\]

Substituting \( k = \frac{35}{216} \):

\[
d = \sqrt{\frac{w}{\left(\frac{35}{216}\right) \cdot l}} = \sqrt{\frac{216w}{35l}}
\]

Thus, the final expression for \( d \) in terms of \( w \) and \( l \) is:

\[
d = \sqrt{\frac{216w}{35l}}
\]

### Summary
- \( d \) is expressed as \( d = \sqrt{\frac{216w}{35l}} \).

Would you like any further details or have any questions? Here are some related questions you might consider:

1. What is the significance of joint variation in real-world applications?
2. How does changing the length or diameter affect the weight of the bar?
3. Can you provide an example with different values for \( w \) and \( l \)?
4. How would this relationship change if the weight depended only on diameter and not on length?
5. What are some other mathematical relationships that involve variation? 

**Tip:** When solving variation problems, always identify the relationship first and ensure you understand the role of each variable.

The weight wkg of a metal bar varies jointly as its length lmm and the square of its diameter dmm. If w=140 when d=4 and l=54, find d in terms of w and l.

Explore how to find the diameter of a metal bar when its weight varies jointly with its length and the square of its diameter. This example uses specific values: weight (w = 140), length (l = 54), and diameter (d = 4) to derive the formula d = sqrt(216w / 35l). Suitable for high school students.

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