HomeRectangle Optimization: Maximize Area and Minimize Perimeter Free original lessons
Rectangle Optimization: Maximize Area and Minimize Perimeter
Learn to solve rectangle optimization problems by defining variables, writing constraints and objectives, then finding the optimum with clear reasoning.
Free lesson
Four exercises
Worked answers
Algebra and Function Practice
Rectangle optimization problems ask you to choose side lengths that best meet a goal, such as enclosing the largest area with a fixed fence or using the least fence for a fixed area. The key is to translate the word problem into two equations: a constraint that limits your choices and an objective that you want to maximize or minimize. Then express the objective in terms of one variable, find the vertex of the resulting quadratic, and interpret the answer in context. This lesson builds a repeatable reasoning template and gives you practice with self-checking methods.
Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.
Define Variables and Write the Constraint
Start by naming what you control. For a rectangle, let the length be L and the width be W, both in the same unit. Write the constraint as an equation. If a fence of fixed length P surrounds the rectangle, then 2L + 2W = P. If the area is fixed at A, then L times W = A. Keep units consistent and state them explicitly.
A common mistake is mixing units or forgetting that the constraint uses perimeter while the objective uses area. Write the constraint first, then the objective, so you do not confuse them.
- Choose variable names for length and width.
- Write the constraint equation from the problem statement.
- State the units for each quantity.
- Solve the constraint for one variable if possible.
Write the Objective and Substitute
The objective is the quantity to maximize or minimize. For area, the objective is A = L times W. For perimeter, the objective is P = 2L + 2W. Use the constraint to eliminate one variable. For example, if 2L + 2W = 20, then W = 10 - L. Substitute into the objective to get a function of one variable.
After substitution, the objective becomes a quadratic. For area with fixed perimeter, A = L(10 - L) = 10L - L squared. This opens downward, so the maximum occurs at the vertex.
- Write the objective formula.
- Substitute the expression from the constraint.
- Simplify to a single-variable quadratic.
- Identify whether it opens upward or downward.
Find the Optimum and Interpret
For a quadratic in the form y = a x squared + b x + c, the vertex x-coordinate is -b divided by 2a. Compute it carefully, then find the other side length from the constraint. Check that both sides are positive and that the answer makes sense in context.
A useful check: for a fixed perimeter, the maximum area rectangle is a square. For a fixed area, the minimum perimeter rectangle is also a square. Use this as a quick sanity check, but still show the algebra.
- Compute the vertex value using -b divided by 2a.
- Find the paired side length from the constraint.
- Verify positivity and units.
- State the maximum or minimum value clearly.
Check with a Table or Symmetry
A table of nearby values can confirm the vertex. For A = 10L - L squared, try L = 4, 5, 6. The areas are 24, 25, 24, so the maximum is at L = 5. Symmetry about the vertex also helps: values equidistant from the vertex give equal results.
Always answer the question asked. If the problem asks for dimensions, give both length and width. If it asks for the maximum area, give the area with units.
- Test one value below and one above the vertex.
- Compare results to confirm the optimum.
- Write the final answer with units and context.
Worked example
A farmer has 24 meters of fencing to enclose a rectangular pen. What dimensions give the maximum area, and what is that area?
Worked answer
The maximum area is 36 square meters, with length 6 meters and width 6 meters.
Four practice exercises
Exercise 1
A rectangle has a fixed perimeter of 20 meters. Find the dimensions that maximize area and state the maximum area.
Show answer and check
Length 5 meters, width 5 meters, maximum area 25 square meters.
Why: Constraint: 2L + 2W = 20, so W = 10 - L. Objective: A = L(10 - L) = 10L - L squared. This quadratic opens downward. Vertex at L = -10 divided by (2 times -1) = 5. Then W = 5. Area = 25.
Check: Test L = 4 and L = 6: areas are 24 and 24, both less than 25, confirming the maximum at L = 5.
Exercise 2
A rectangle must have an area of 36 square centimeters. Find the dimensions that minimize perimeter and state the minimum perimeter.
Show answer and check
Length 6 centimeters, width 6 centimeters, minimum perimeter 24 centimeters.
Why: Constraint: L times W = 36, so W = 36 divided by L. Objective: P = 2L + 2W = 2L + 72 divided by L. By symmetry or by testing, the minimum occurs at L = 6, giving W = 6 and P = 24. Alternatively, for fixed area the minimum perimeter rectangle is a square.
Check: Test L = 4 gives W = 9 and P = 26; L = 9 gives W = 4 and P = 26; both exceed 24.
Exercise 3
A rectangular garden uses a wall as one side and 30 meters of fencing for the other three sides. Find the dimensions that maximize area and state the maximum area.
Show answer and check
Length 15 meters (parallel to the wall), width 7.5 meters, maximum area 112.5 square meters.
Why: Constraint: L + 2W = 30, so L = 30 - 2W. Objective: A = L times W = (30 - 2W)W = 30W - 2W squared. This opens downward. Vertex at W = -30 divided by (2 times -2) = 7.5. Then L = 15. Area = 112.5.
Check: Test W = 7 gives L = 16 and area 112; W = 8 gives L = 14 and area 112; both less than 112.5.
Exercise 4
A rectangle has a perimeter of 16 meters. Find the dimensions that maximize area and state the maximum area.
Show answer and check
Length 4 meters, width 4 meters, maximum area 16 square meters.
Why: Constraint: 2L + 2W = 16, so W = 8 - L. Objective: A = L(8 - L) = 8L - L squared. Vertex at L = -8 divided by (2 times -1) = 4. Then W = 4. Area = 16.
Check: Test L = 3 gives area 15; L = 5 gives area 15; both less than 16.
Common mistakes
- Using the constraint equation as the objective. Label the constraint and the objective separately before substituting.
- Forgetting to substitute and trying to optimize with two variables. Solve the constraint for one variable and replace it in the objective.
- Mixing units such as meters and centimeters. Convert all lengths to one unit before writing equations.
- Reporting only one side length when the problem asks for dimensions. Use the constraint to find the paired side and state both.
Self-check rubric
- Defines variables with units.
- Writes the constraint equation correctly.
- Writes the objective formula correctly.
- Substitutes to get a single-variable quadratic.
- Computes the vertex accurately.
- Finds both side lengths and checks positivity.
- States the final answer with units and context.
- Uses a check such as a table or symmetry.
Questions and answers
Why is the best rectangle often a square?
For a fixed perimeter, area is maximized when length equals width. For a fixed area, perimeter is minimized when length equals width. The algebra shows the vertex occurs at equal sides.
Can I use calculus instead of the vertex formula?
Yes. The derivative of the quadratic gives the same result. The vertex formula is faster for quadratics and is sufficient here.
What if the constraint is not a full perimeter?
Write the constraint exactly as stated. For example, if one side uses a wall, the fence length may be L + 2W = P. Substitute accordingly.
How do I know my answer is a maximum and not a minimum?
Check the sign of the quadratic coefficient. If it is negative, the parabola opens downward and the vertex is a maximum. If positive, it is a minimum.