Why Compare These Two Rectangle Problems?
A rectangle has two dimensions, but many practical questions impose one constraint. If the perimeter is fixed, which rectangle encloses the greatest area? If the area is fixed, which rectangle uses the least perimeter? In both cases, calculus leads to the same proportion: the two side lengths must be equal.
The conclusion may seem intuitive: a square is balanced. The calculus explains precisely what “balanced” means and establishes that the square is not merely a plausible choice but the unique optimal rectangle. The two questions are related, but they use different constraints and different objective functions, so each needs its own model.
Maximum Area with a Fixed Perimeter
Suppose the perimeter is a fixed positive number \(P\). Since \(2x+2y=P\), the dimensions must satisfy \(x+y=P/2\). Solving for one side gives \(y=P/2-x\), so the area becomes a function of one variable:
The interval is open because both side lengths must be positive. The area function is a downward-opening quadratic. Differentiating gives \(A'(x)=P/2-2x\), so its only critical point is \(x=P/4\). At that point, \(y=P/2-x=P/4\): the dimensions are equal.
To see why this is the greatest area, not just a local candidate, note that \(A''(x)=-2<0\) throughout the feasible interval. Also, \(A'(x)>0\) when \(x<P/4\), and \(A'(x)<0\) when \(x>P/4\). Thus area increases up to \(P/4\) and decreases after it. At either excluded endpoint, the limiting area is zero, whereas the area at the critical point is positive.
Derivation: The Fixed-Perimeter Result
The calculation above can be written as a complete optimization argument. Begin with the constraint, express the objective using one variable, and verify that the candidate beats every other feasible choice:
From \(2x+2y=P\), obtain \(y=P/2-x\), with \(0<x<P/2\).
\(A(x)=Px/2-x^2\), so \(A'(x)=P/2-2x\). Setting \(A'(x)=0\) gives \(x=P/4\), and then \(y=P/4\).
Since \(A'(x)>0\) to the left of \(P/4\) and \(A'(x)<0\) to its right, the function rises and then falls across the entire feasible interval. The candidate is therefore the unique absolute maximum.
At \(x=y=P/4\), \(A=xy=(P/4)^2=P^2/16\).
There is also a short algebraic way to see the same balance. If \(x+y=P/2\), then \((x-y)^2\geq 0\). Expanding gives \(x^2-2xy+y^2\geq 0\), which is equivalent to \((x+y)^2\geq 4xy\). Therefore \(xy\leq (P/2)^2/4=P^2/16\). Equality holds exactly when \(x-y=0\), or \(x=y\). The derivative argument identifies the optimum by calculus; this inequality makes the equality condition especially visible.
Worked Example: Greatest Area with a Perimeter of 40 Meters
A rectangle has perimeter \(40\) meters. Find the dimensions that give the greatest area and verify the result.
Step 1: Write the constraint. From \(2x+2y=40\), divide by \(2\) to get \(x+y=20\), so \(y=20-x\). Positive dimensions require \(0<x<20\).
Step 2: Form the area function. \(A(x)=x(20-x)=20x-x^2\), measured in square meters.
Step 3: Find and test the critical point. \(A'(x)=20-2x\). Solving \(20-2x=0\) gives \(2x=20\), so \(x=10\). Then \(y=20-10=10\). Since \(A''(x)=-2<0\), and the derivative is positive for \(x<10\) and negative for \(x>10\), this point is the unique maximum.
Step 4: Calculate and check. The maximum area is \(A=10\cdot10=100\) square meters. The perimeter check is \(2(10)+2(10)=40\) meters, as required. The area function also gives \(A(10)=20(10)-10^2=200-100=100\) square meters.
Minimum Perimeter with a Fixed Area
Now suppose the area is a fixed positive number \(K\). The constraint is \(xy=K\), so \(y=K/x\), where \(x>0\). The perimeter to minimize is consequently
Differentiate: \(L'(x)=2-2K/x^2\). Setting this equal to zero gives \(x^2=K\). Since \(x>0\), the feasible solution is \(x=\sqrt K\), and the constraint then gives \(y=K/\sqrt K=\sqrt K\). The two side lengths are equal.
This critical point is the absolute minimum. For \(0<x<\sqrt K\), \(x^2<K\), so \(L'(x)<0\). For \(x>\sqrt K\), \(x^2>K\), so \(L'(x)>0\). The perimeter decreases before the critical point and increases afterward. In addition, \(L(x)\) grows without bound as \(x\) approaches zero from above or grows without bound as \(x\) increases without limit. There is no boundary value that could be smaller.
Derivation: The Fixed-Area Result
For completeness, the fixed-area optimization follows directly from the constraint and a global sign check. No other shape or constraint is being introduced: \(x\) and \(y\) remain the positive side lengths of a rectangle.
Since \(xy=K\) and \(x>0\), write \(y=K/x\).
\(L(x)=2(x+K/x)\), for all \(x>0\).
\(L'(x)=2-2K/x^2=0\) implies \(x^2=K\), hence \(x=\sqrt K\) because \(x\) is positive. Then \(y=K/x=\sqrt K\).
For \(x<\sqrt K\), \(L'(x)<0\); for \(x>\sqrt K\), \(L'(x)>0\). Thus \(L\) decreases to the critical point and increases afterward. Its values also tend to infinity at both extremes of the domain.
At \(x=y=\sqrt K\), \(L=2\sqrt K+2\sqrt K=4\sqrt K\).
An algebraic check also explains the balance. Because \(xy=K\), \((x+y)^2=(x-y)^2+4xy=(x-y)^2+4K\geq 4K\). Both side lengths are positive, so \(x+y\geq 2\sqrt K\), and therefore \(2(x+y)\geq 4\sqrt K\). Equality requires \(x-y=0\), so it occurs exactly for the square. This agrees with the derivative result and shows why no rectangle with unequal sides can use less perimeter at the same area.
Worked Example: Least Perimeter for an Area of 144 Square Meters
A rectangle must enclose \(144\) square meters. Find the dimensions that require the least perimeter, and verify the result.
Step 1: Express one side using the area constraint. Let the sides be \(x\) and \(y\). Then \(xy=144\), so \(y=144/x\), with \(x>0\).
Step 2: Form and differentiate the perimeter. \(L(x)=2x+2(144/x)=2x+288/x\). Thus \(L'(x)=2-288/x^2\).
Step 3: Find the critical point. Solve \(2-288/x^2=0\): \(2=288/x^2\), so \(2x^2=288\), \(x^2=144\), and \(x=12\), using \(x>0\). Then \(y=144/12=12\).
Step 4: Verify the minimum and calculate the perimeter. \(L'(x)<0\) for \(0<x<12\), and \(L'(x)>0\) for \(x>12\), so \(x=12\) gives the unique absolute minimum. The perimeter is \(2(12)+2(12)=48\) meters. Substituting in the area constraint gives \(12\cdot12=144\) square meters, as required.
Using the Proportion in a Comparison
Worked Example: Comparing a Two-to-One Rectangle with the Best Rectangle
A rectangle has perimeter \(36\) meters, and its longer side is twice its shorter side. Find its dimensions and compare its area with the greatest possible area for that perimeter.
Step 1: Represent the ratio. Let the shorter side be \(t\) meters and the longer side be \(2t\) meters. The perimeter is \(2(t+2t)=6t\).
Step 2: Apply the constraint. \(6t=36\), so \(t=6\). The dimensions are \(6\) meters by \(12\) meters. Their perimeter check is \(2(6)+2(12)=12+24=36\) meters.
Step 3: Calculate the area and compare. The rectangle's area is \(6\cdot12=72\) square meters. For perimeter \(36\), the greatest-area rectangle is a square with side \(36/4=9\) meters, so its area is \(9\cdot9=81\) square meters. The difference is \(81-72=9\) square meters. The two-to-one rectangle therefore has less area, consistent with the fixed-perimeter maximum.
Step 4: Check both candidates. The square's perimeter is \(2(9)+2(9)=36\) meters, and its area is \(81\) square meters. Both rectangles satisfy the same perimeter constraint, so the comparison isolates the effect of their proportions.
Common Pitfall: Losing the Constraint
A frequent error is to confuse the sum of the side lengths with the perimeter. For example, if the perimeter is \(40\) meters, writing \(x+y=40\) would lead to \(x=y=20\). But that candidate has perimeter \(2(20)+2(20)=80\) meters, so it is not feasible.
The correct equation is \(2x+2y=40\), which simplifies to \(x+y=20\). The balanced dimensions are then \(10\) meters and \(10\) meters. They satisfy \(2(10)+2(10)=40\) meters and give area \(100\) square meters.
A related mistake is to find a critical point and stop without checking whether it is an absolute optimum. For the fixed-perimeter model, the derivative changes from positive to negative, so the area increases and then decreases. For the fixed-area model, the perimeter derivative changes from negative to positive, so the perimeter decreases and then increases. These checks establish global optimality on the stated feasible domains.
Why Both Optima Are Squares
The two problems reverse which quantity is held fixed, but their underlying balance is the same. When the sum \(x+y\) is fixed, the product \(xy\) is greatest when the two numbers are equal. When the product \(xy\) is fixed, the sum \(x+y\) is least when the two numbers are equal. Since rectangle area is a product and half its perimeter is a sum, both optimization problems favor equal side lengths.
The formulas also provide useful checks. A fixed perimeter \(P\) gives maximum area \(P^2/16\), while a fixed area \(K\) gives minimum perimeter \(4\sqrt K\). Their units are consistent: \(P^2\) has square units, and \(\sqrt K\) has units of length. These formulas apply to positive, nondegenerate rectangles and are attained by a square.
Check Your Understanding
Use the constraint in each question and justify whether the candidate is an absolute optimum.
- For a rectangle with perimeter \(64\) meters, what side length gives the greatest area, and what is that area?
- For a rectangle with area \(225\) square centimeters, what dimensions give the least perimeter?
- For fixed perimeter, why is the area function increasing before its critical point and decreasing afterward?
- For fixed area \(K>0\), what is the sign of the perimeter derivative when \(0<x<\sqrt K\)? What does that sign mean?
- A proposed fixed-perimeter model uses \(x+y=P\). What is wrong with it when \(P\) denotes the rectangle's full perimeter?
