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QUESTION 3

The loading distribution, [tex]\( w \)[/tex], on a beam is given by [tex]\( w = -36x^2 + 50x \)[/tex], where [tex]\( x \)[/tex] is the distance in meters from one end of the beam. Find the value of [tex]\( x \)[/tex] at which maximum loading occurs using the second derivative test.


Sagot :

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.
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