Find the area of the largest rectangle that can be inscribed in a right triangle with legs of length 3 cm and 4 cm if two sides of the rectangle lie along the legs. A. 6 cm2 C. 4 cm2 B. 5 cm2 D. 3 cm2 Solution Click here to… Jhun Vert Thu, 06/20/2024 - 13:22 Solution Click here to expand or collapse this section Area of largest rectangle: $A_{max} = \frac{1}{2}A_{tri}$ $A_{max} = \frac{1}{2} \times \frac{1}{2}(4)(3)$ $A_{max} = 3 ~ \text{cm}^2$ Detailed Solution Click here to expand or collapse this section Area of the rectangle : $A = xy$ From the figure above: $\dfrac{y}{4 - x} = \dfrac{3}{4}$ $y = \frac{3}{4}(4 - x)$ Hernce, $A = x \cdot \frac{3}{4}(4 - x)$ $A = \frac{3}{4}(4x - x^2)$ For largest rectangle: $\dfrac{dA}{dx} = \frac{3}{4}(4 - 2x) = 0$ $x = 2$ Largest rectangle: $A_{max} = \frac{3}{4}\left[ 4(2) - 2^2 \right]$ $A_{max} = 3 ~ \text{cm}^2$ Log in or register to post comments Log in or register to post comments
Solution Click here to… Jhun Vert Thu, 06/20/2024 - 13:22 Solution Click here to expand or collapse this section Area of largest rectangle: $A_{max} = \frac{1}{2}A_{tri}$ $A_{max} = \frac{1}{2} \times \frac{1}{2}(4)(3)$ $A_{max} = 3 ~ \text{cm}^2$ Detailed Solution Click here to expand or collapse this section Area of the rectangle : $A = xy$ From the figure above: $\dfrac{y}{4 - x} = \dfrac{3}{4}$ $y = \frac{3}{4}(4 - x)$ Hernce, $A = x \cdot \frac{3}{4}(4 - x)$ $A = \frac{3}{4}(4x - x^2)$ For largest rectangle: $\dfrac{dA}{dx} = \frac{3}{4}(4 - 2x) = 0$ $x = 2$ Largest rectangle: $A_{max} = \frac{3}{4}\left[ 4(2) - 2^2 \right]$ $A_{max} = 3 ~ \text{cm}^2$ Log in or register to post comments
Solution Click here to…