To determine the value of [tex]\( x \)[/tex] at which the maximum loading occurs for the given loading distribution function [tex]\( w = -36x^2 + 50x \)[/tex], we can follow these steps:
1. Find the First Derivative:
First, calculate the first derivative of [tex]\( w \)[/tex] with respect to [tex]\( x \)[/tex] to determine the critical points. The first derivative, denoted as [tex]\( w' \)[/tex], is given by:
[tex]\[
w' = \frac{d}{dx}(-36x^2 + 50x)
\][/tex]
Using the power rule of differentiation, we get:
[tex]\[
w' = -72x + 50
\][/tex]
2. Find the Critical Points:
Next, set the first derivative equal to zero to find the critical points:
[tex]\[
-72x + 50 = 0
\][/tex]
Solving for [tex]\( x \)[/tex], we get:
[tex]\[
72x = 50 \implies x = \frac{50}{72} = \frac{25}{36}
\][/tex]
So, the critical point we have is [tex]\( x = \frac{25}{36} \)[/tex].
3. Find the Second Derivative:
To use the second derivative test, we need to find the second derivative of [tex]\( w \)[/tex] with respect to [tex]\( x \)[/tex]. The second derivative, denoted as [tex]\( w'' \)[/tex], is given by:
[tex]\[
w'' = \frac{d}{dx}(-72x + 50)
\][/tex]
Since the derivative of a constant is zero, we get:
[tex]\[
w'' = -72
\][/tex]
4. Apply the Second Derivative Test:
The second derivative test states that if [tex]\( w''(x) < 0 \)[/tex] at a critical point, then [tex]\( w \)[/tex] has a local maximum at that point.
Since [tex]\( w'' = -72 \)[/tex] which is less than zero, it confirms that the function [tex]\( w \)[/tex] has a local maximum at [tex]\( x = \frac{25}{36} \)[/tex].
Therefore, the value of [tex]\( x \)[/tex] at which the maximum loading occurs is [tex]\( \frac{25}{36} \)[/tex] meters.
