The deflection $$ D $$ of a beam of length $$ \angle $$ is $$ D=15 x^{4}\|4\| 9 x^{3}+44 L^{2} x^{2} $$, where $$ x $$ is the distance from one end of the beam. Find the value of $$ x $$ that yields the maximum defiection. $$ x=\square $$

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To find the value of $$ x $$ that yields the maximum deflection $$ D $$, we need to take the derivative of the deflection function $$ D $$ with respect to $$ x $$ and set it to zero.

The deflection function is given as:

$$
D = 15 x^{4} + 9 x^{3} + 44 L^{2} x^{2}
$$

Let's first compute the derivative $$ \frac{dD}{dx} $$ and then solve for $$ x $$ when $$ \frac{dD}{dx} = 0 $$.

The derivative is:

$$
\frac{dD}{dx} = \frac{d}{dx}(15 x^{4} + 9 x^{3} + 44 L^{2} x^{2})
$$

Now, let's calculate the derivative.
Simplify the expression by following steps:
- step0: Solution:
  $$15\times 4x^{3}+9\times 3x^{2}+44\times 2L^{2}x$$
- step1: Multiply the terms:
  $$60x^{3}+9\times 3x^{2}+44\times 2L^{2}x$$
- step2: Multiply the terms:
  $$60x^{3}+27x^{2}+44\times 2L^{2}x$$
- step3: Multiply the terms:
  $$60x^{3}+27x^{2}+88L^{2}x$$
The derivative of the deflection function $$ D $$ is:

$$
\frac{dD}{dx} = 60x^{3} + 27x^{2} + 88L^{2}x
$$

Next, we set the derivative equal to zero to find the critical points:

$$
60x^{3} + 27x^{2} + 88L^{2}x = 0
$$

We can factor out $$ x $$:

$$
x(60x^{2} + 27x + 88L^{2}) = 0
$$

This gives us one solution $$ x = 0 $$. To find the other solutions, we need to solve the quadratic equation:

$$
60x^{2} + 27x + 88L^{2} = 0
$$

Let's calculate the roots of this quadratic equation using the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
$$

where $$ a = 60 $$, $$ b = 27 $$, and $$ c = 88L^{2} $$.

Now, let's compute the discriminant and the roots.
Simplify the expression by following steps:
- step0: Solution:
  $$27^{2}-4\times 60\times 88L^{2}$$
- step1: Multiply the terms:
  $$27^{2}-21120L^{2}$$
- step2: Evaluate the power:
  $$729-21120L^{2}$$
Calculate or simplify the expression $$ -27 + \sqrt(27^2 - 4 * 60 * 88 * L^{2}) $$.
Simplify the expression by following steps:
- step0: Solution:
  $$-27+\sqrt{27^{2}-4\times 60\times 88L^{2}}$$
- step1: Multiply the terms:
  $$-27+\sqrt{27^{2}-21120L^{2}}$$
- step2: Evaluate the power:
  $$-27+\sqrt{729-21120L^{2}}$$
Calculate or simplify the expression $$ -27 - \sqrt(27^2 - 4 * 60 * 88 * L^{2}) $$.
Simplify the expression by following steps:
- step0: Solution:
  $$-27-\sqrt{27^{2}-4\times 60\times 88L^{2}}$$
- step1: Multiply the terms:
  $$-27-\sqrt{27^{2}-21120L^{2}}$$
- step2: Evaluate the power:
  $$-27-\sqrt{729-21120L^{2}}$$
The discriminant of the quadratic equation is:

$$
729 - 21120L^{2}
$$

The roots of the quadratic equation are given by:

$$
x = \frac{-27 \pm \sqrt{729 - 21120L^{2}}}{2 \cdot 60}
$$

This results in two potential solutions for $$ x $$:

1. $$ x = \frac{-27 + \sqrt{729 - 21120L^{2}}}{120} $$
2. $$ x = \frac{-27 - \sqrt{729 - 21120L^{2}}}{120} $$

Since $$ x $$ represents a distance, we only consider the positive root. Therefore, the value of $$ x $$ that yields the maximum deflection is:

$$
x = \frac{-27 + \sqrt{729 - 21120L^{2}}}{120}
$$

Thus, the final answer is:

$$
x = \frac{-27 + \sqrt{729 - 21120L^{2}}}{120}
$$
The value of $$ x $$ that yields the maximum deflection is $$ x = \frac{-27 + \sqrt{729 - 21120L^{2}}}{120} $$.

The deflection \( D \) of a beam of length \( \angle \) is \( D=15 x^{4}\|4\| 9 x^{3}+44 L^{2} x^{2} \), where \( x \) is the distance from one end of the beam. Find the value of \( x \) that yields the maximum defiection. \( x=\square \)

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