## Algebra and Trigonometry 10th Edition

A = $2x(\sqrt {36-x^2}), 0\leq x \leq 6$
Area $A = 2xy = 2x(\sqrt {36-x^2})$ (Use the diagram and the given function to find the area of the rectangle.) $0\leq x \leq 6$ for the expression under the square root to be non-negative.