MinMax02 Enclosing a rectangular area.

A rectangular area is to be enclosed with 320 feet of fence. What dimensions of rectangle give the maximum area?

The quantity to be maximized is the area, the product of the lengths of the two sides of the rectangle. The defining equation, the A equals… equation, is A = ab. Before maximizing the area (taking the derivative of A), the product ab must be written in terms of one variable. This requires a "constraint" equation relating a to b. The constraint in the problem is that the total length of fence 2a + 2b must equal 320. With this constraint equation A can be written in terms of a or b, it makes no difference.

Solve the constraint equation for a, and substitute in the area equation.

2a + 2b = 320 so a = 160 - b and A = (160 - b)b = 160b - b²

The maximum occurs when the graph of A vs. b goes through a maximum. A maximum is defined, in calculus, as slope zero and second derivative negative.

Solving dA / db = 160 - 2b and setting this equal to zero we get 160 - 2b = 0 and b = 80.

The second derivative d²A / db² = -2 confirming b = 80 as a maximum.

Go back to the constraint equation and note that for b = 80, a = 80. The area is maximum for a square.

Often max-min problems can be done with the first and second derivative. If you fell a little insecure, sketch the graph of the function. All the information, and then some, is already available for sketching the graph.

The original equation A = 160b - b² is a parabola that opens down and goes through the points b = 0 and b = 160 with symmetry line at b = 80. If you had any trouble with that last sentence go back to the graphing of parabolas and review the procedure. The calculus tells us that the slope is zero at b = 80 and that the curve goes through a maximum at that point. This confirms what we already know from algebra analysis. The curve is sketched in Fig 5-3.