Answer
a. $y=1-x$
b. $A=2x-2x^2$, $0\leq x\leq 1$
c. $A=\frac{1}{2}$, dimensions $1\times\frac{1}{2}$
Work Step by Step
a. Using the figure given in the Exercise, the coordinates of A and B are $A(1,0),B(0,1)$. Thus the line equation for AB is $\frac{y-0}{1-0}=\frac{x-1}{0-1}$ or $y=-x+1$. For point $P(x,y)$, we have $y=1-x$ and $P(x,1-x)$ where $0\leq x\leq 1$.
b. The area of the rectangle is given by $A=2xy=2x(1-x)=2x-2x^2$
c. Take the derivative of $A$ to get $A'=2-4x$ and the maximum of the area happens when $A'=0$ or $x=1/2$, which gives $A=2\times\frac{1}{2}-2(\frac{1}{2})^2=\frac{1}{2}$ with dimensions of $1\times\frac{1}{2}$
