Barn-Side Garden: Quadratic Area Maximization
A landscape club is building a rectangular vegetable garden against the straight side of a barn. The barn forms one side of the garden and needs no fencing. The club has 40 feet of fencing to enclose the other three sides. Let x be the length of each of the two sides perpendicular to the barn, and let y be the length of the side parallel to the barn. The club wants to choose x and y to maximize the garden area. The club also must keep y at least 12 feet to fit a wheelchair-accessible path along the barn side. The club cannot use more than 40 feet of fencing, but it may use less. The garden must be rectangular, and the barn side is longer than 20 feet.
1. Write the garden area A as a function of x. Show how the fencing constraint produces the equation.
Read the worked answer
The two sides perpendicular to the barn each have length x, and the side parallel to the barn has length y. The fencing covers three sides, so 2x + y = 40 if all fencing is used. Solving for y gives y = 40 - 2x. Area is length times width: A(x) = x(40 - 2x) = 40x - 2x^2. This is a quadratic function opening downward, so it has a maximum.
2. Find the value of x that maximizes the area, and find the corresponding y and area.
Read the worked answer
For A(x) = -2x^2 + 40x, the vertex occurs at x = -b/(2a) = -40/(2(-2)) = 10. Then y = 40 - 2(10) = 20. The maximum area is A(10) = 10(20) = 200 square feet. The dimensions are 10 feet by 20 feet, with the 20-foot side along the barn.
3. Does the requirement that y be at least 12 feet change the optimal dimensions? Explain.
Read the worked answer
No. At the unconstrained optimum, y = 20 feet, which is greater than 12 feet. The constraint is satisfied. To see the boundary, if y = 12, then 2x + 12 = 40, so x = 14 and the area is 14(12) = 168 square feet. That is less than 200, so the maximum remains at x = 10 and y = 20.