The deflection \( D \) of a beam of length \( L \) is \( D=5 x^{4}-9 L x^{3}+4 L^{2} x^{2} \), where \( x \) ls the distance from one end of the beam, Find the value of \( x \) that yields the maximum deflection. Need Help? Readit SUBMT ANSWER VIEW PREVIOUS QUESTION QUESTION 18 of 20 VIEW NEXT QUESTION

Let the deflection be \[ D = 5x^4 - 9Lx^3 + 4L^2x^2, \] where \( x \) is the distance from one end of the beam. 1. **Differentiate \( D \) with respect to \( x \):** \[ \frac{dD}{dx} = \frac{d}{dx}\Bigl(5x^4 - 9Lx^3 + 4L^2x^2\Bigr) = 20x^3 - 27Lx^2 + 8L^2x. \] 2. **Factor the derivative:** Factor an \( x \) out from each term: \[ \frac{dD}{dx} = x\left(20x^2 - 27Lx + 8L^2\right). \] 3. **Set the derivative equal to zero:** \[ x\left(20x^2 - 27Lx + 8L^2\right) = 0. \] This gives the solutions: - \( x = 0 \) - and the quadratic equation \[ 20x^2 - 27Lx + 8L^2 = 0. \] 4. **Solve the quadratic equation using the quadratic formula:** For the quadratic \( ax^2 + bx + c = 0 \) with \[ a = 20, \quad b = -27L, \quad c = 8L^2, \] the solutions are \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{27L \pm \sqrt{(27L)^2 - 4(20)(8L^2)}}{40}. \] 5. **Compute the discriminant:** \[ (27L)^2 - 4(20)(8L^2) = 729L^2 - 640L^2 = 89L^2. \] Therefore, the solutions are \[ x = \frac{27L \pm L\sqrt{89}}{40}. \] 6. **Determine the maximum:** The beam's endpoints, \( x = 0 \) and \( x = L \), yield \( D = 0 \) (since substituting these values into the deflection function gives zero). Hence, the maximum deflection occurs at one of the interior critical points. To decide between \[ x_1 = \frac{27L - L\sqrt{89}}{40} = \frac{(27 - \sqrt{89})L}{40} \quad \text{and} \quad x_2 = \frac{27L + L\sqrt{89}}{40} = \frac{(27 + \sqrt{89})L}{40}, \] we can consider the second derivative or evaluate the function. A calculation of the second derivative shows that \( x_1 \) corresponds to a maximum. Thus, the value of \( x \) that yields the maximum deflection is \[ \boxed{\frac{(27-\sqrt{89})L}{40}}. \]