Find the area of the largest rectangle that can be inscribed in the ellipse x^2/25 + y^2/16 = 1. Find the area of the largest rectangle that can be inscribed in the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$. A. 35 C. 45 B. 40 D. 50 Log in or register to post comments Solution Click here to… Jhun Vert Sat, 06/29/2024 - 15:13 Solution Click here to expand or collapse this section For largest rectangle $x = \dfrac{a}{\sqrt{2}}$ and $y = \dfrac{b}{\sqrt{2}}$ $A_{max} = 4xy$ $A_{max} = 4 \left( \dfrac{5}{\sqrt{2}} \right) \left( \dfrac{4}{\sqrt{2}} \right)$ $A_{max} = 40 ~ \text{unit}^2$ Detailed Solution Click here to expand or collapse this section $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ $y^2 = 16\left( 1 - \dfrac{x^2}{25} \right)$ $y^2 = 16\left( \dfrac{25 - x^2}{25} \right)$ $y^2 = \frac{16}{25}(25 - x^2)$ $y = \frac{4}{5} \sqrt{25 - x^2}$ Area of rectangle inscribed in the ellipse $A = 4xy$ $A = 4x \left( \frac{4}{5} \sqrt{25 - x^2} \right)$ $A = \frac{16}{5}x \sqrt{25 - x^2}$ For the largest rectangle $\dfrac{dA}{dx} = \frac{16}{5} \left[ x \cdot \dfrac{-2x}{2\sqrt{25 - x^2}} + \sqrt{25 - x^2} \cdot 1 \right] = 0$ $x \cdot \dfrac{-2x}{2\sqrt{25 - x^2}} + \sqrt{25 - x^2} \cdot 1 = 0$ $\dfrac{-x^2}{\sqrt{25 - x^2}} + \sqrt{25 - x^2} = 0$ $\sqrt{25 - x^2} = \dfrac{x^2}{\sqrt{25 - x^2}}$ $25 - x^2 = x^2$ $2x^2 = 25$ $x = \frac{5}{\sqrt{2}}$ Area of the largest rectangle $A_{max} = \frac{16}{5} \cdot \frac{5}{\sqrt{2}} \cdot \sqrt{25 - \frac{25}{2}}$ $A_{max} = 40 ~ \text{unit}^2$ Log in or register to post comments

Solution Click here to… Jhun Vert Sat, 06/29/2024 - 15:13 Solution Click here to expand or collapse this section For largest rectangle $x = \dfrac{a}{\sqrt{2}}$ and $y = \dfrac{b}{\sqrt{2}}$ $A_{max} = 4xy$ $A_{max} = 4 \left( \dfrac{5}{\sqrt{2}} \right) \left( \dfrac{4}{\sqrt{2}} \right)$ $A_{max} = 40 ~ \text{unit}^2$ Detailed Solution Click here to expand or collapse this section $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ $y^2 = 16\left( 1 - \dfrac{x^2}{25} \right)$ $y^2 = 16\left( \dfrac{25 - x^2}{25} \right)$ $y^2 = \frac{16}{25}(25 - x^2)$ $y = \frac{4}{5} \sqrt{25 - x^2}$ Area of rectangle inscribed in the ellipse $A = 4xy$ $A = 4x \left( \frac{4}{5} \sqrt{25 - x^2} \right)$ $A = \frac{16}{5}x \sqrt{25 - x^2}$ For the largest rectangle $\dfrac{dA}{dx} = \frac{16}{5} \left[ x \cdot \dfrac{-2x}{2\sqrt{25 - x^2}} + \sqrt{25 - x^2} \cdot 1 \right] = 0$ $x \cdot \dfrac{-2x}{2\sqrt{25 - x^2}} + \sqrt{25 - x^2} \cdot 1 = 0$ $\dfrac{-x^2}{\sqrt{25 - x^2}} + \sqrt{25 - x^2} = 0$ $\sqrt{25 - x^2} = \dfrac{x^2}{\sqrt{25 - x^2}}$ $25 - x^2 = x^2$ $2x^2 = 25$ $x = \frac{5}{\sqrt{2}}$ Area of the largest rectangle $A_{max} = \frac{16}{5} \cdot \frac{5}{\sqrt{2}} \cdot \sqrt{25 - \frac{25}{2}}$ $A_{max} = 40 ~ \text{unit}^2$ Log in or register to post comments

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