Chapter 4: Applications of Derivatives - Section 4.5 - Applied Optimization - Exercises 4.5 - Page 222: 7

$80000m^2$, $200m\times400m$

Step 1. As shown in the figure, assume the side next to the river has a length of $x$ meters; we have the other two sides as $x$ and $800-2x$ (as the total length is 800m). Step 2. The area of the rectangle is given by $A=x(800-2x)=800x-2x^2$ Step 3. Take the derivative to get $A'=800-4x$ and an extreme will happen when $A'=0$, which gives $x=200m$ and $A=200(800-400)=80000m^2$. Step 4. Test the end points: when $x=0, A=0$ and when $x=400, A=0$, thus the extreme is a maximum. Step 5. The area will reach a maximum of $80000m^2$ when the dimensions are $200m\times400m$.