Convert the word problem into equations. First, create an equation for the value to be maximized. Then, create an equation relating the constraints. For example, if the problem is "What is the maximum fence area that can be created from 40 meters (m) of fencing?" you would first create an equation for the area inside the fence. This yields "A = x * y," where "x" and "y" are sides of the fence. Then, create an equation using the constraints. The perimeter is 40 meters, so this yields "2x + 2y = 40."
Format the constraint equation to solve for one of the variables. In this example, the process yields "y = 20 - x."
Plug the equation from Step 2 into the maximization equation. In this example, the process yields "A = x(20-x)," or "A = 20x - x^2."
Take the derivative of the equation from Step 3. In this example, the process yields "dA/dx = 20 - 2x."
Set the derivative from Step 4 equal to zero. In this example, the process yields "0 = 20 - 2x."
Solve the equation from Step 5 to get one of the variable values. In this example, the process yields x = 10.
Plug the known variable from Step 6 into the original constraint equation to get the value of the other variable. In this example, 2(10) + 2y = 40, or y = 10. The maximum area possible is 10m * 10 m, or 100 m^2.