KBAT Problems
KBAT Problem: Best Use of a Fixed Length
A KBAT problem about fencing a maximum area with a fixed length, optimisation that leads back to a quadratic.
Express area with one variable
A fixed length of fencing encloses a rectangle against a wall. Write the area as a function of one side using the fixed length, and it becomes a quadratic whose maximum is the turning point.
Read the maximum
Complete the square to find the turning point, which gives the side length for the largest area and the area itself. Check the answer fits the fixed length, a sanity check that catches arithmetic slips.
Understand the problem
An original optimisation. A farmer has 60 m of fencing to enclose a rectangular pen, using a straight wall as one long side so no fence is needed there.
The task is the largest possible area. Fixed length with maximum area points to writing area as a quadratic in one variable, then finding its highest point.
Plan and solve
- Let the two sides perpendicular to the wall each be x m; the side parallel to the wall is (60 − 2x) m.
- Area A = x(60 − 2x) = 60x − 2x².
- This quadratic opens downward, so its maximum is at x = −b ÷ 2a = −60 ÷ (2 × −2) = 15.
- Then the parallel side = 60 − 2(15) = 30 m, and A = 15 × 30 = 450 m².
Check and a variant
Check the fencing used: 15 + 15 + 30 = 60 m, exactly what the farmer has, and 450 m² is the area of a 15 by 30 rectangle. A twist: remove the wall so all four sides need fencing.
Then 2(x + y) = 60, x + y = 30, and the biggest area is a 15 by 15 square = 225 m². Leaning on the wall doubles the usable area, a neat result worth stating.
Frequently asked questions
Why does the area become a quadratic?
Because area is length times width, and the fixed fencing lets you write one dimension in terms of the other. Substituting turns 'length × width' into an expression with a squared term, like 60x − 2x².
Quadratics have a single highest or lowest point, which is exactly the maximum area the question is chasing.
How do I find the maximum without calculus?
Use the vertex of the quadratic. For A = ax² + bx + c the turning point is at x = −b ÷ 2a, and because the x² coefficient here is negative the curve peaks there.
Substitute that x back to get the maximum area. Completing the square gives the same turning point if you prefer that method.
Does using the wall really change the answer?
Yes, a lot. With one side free you spend all the fencing on three sides, so the enclosed area is larger, 450 m² here versus 225 m² for a four-sided pen with the same 60 m.
Recognising that the wall removes one side from the perimeter equation is the key modelling step.