Answer
width=$3.44$, height=$2.61$ $A\approx8.98unit^2$
Work Step by Step
Step 1. Assume the upper right corner of the rectangle has coordinates $(x,y)$ as shown in the figure.
Step 2. As the point is on the curve, we have $y=4cos(0.5x), 0\lt x\lt \pi$
Step 3. The area of the rectangle is given by $A=2xy=8x\ cos(0.5x)$
Step 4. To find the maximum of the area, take the derivative of the function to get $A'=8cos(0.5x)-4x\ sin(0.5x)$ Step 5. Let $A'=0$, we have $x=2cot(0.5x)$ which gives $x\approx1.721$ in the domain (use a graphing device).
Step 6. The area at this point is $A=8(1.721)\ cos(0.5(1.721))\approx 8.98unit^2$ with dimensions width=$2x=3.442unit$ and height=$y=4cos(0.5(1.307))\approx2.61unit$
Step 7. Check $A''=(-4sin(0.5x)-4sin(0.5x)-4x\ cos(0.5x))|_{x=1.721}=-10.6\lt0$, thus the function is concave down indicating a maximum of the area at this point.
