Determining a Maximum Area

Rectangles Enclosed by a Fixed Length of Fencing

An optimization problem asks for the best value that is still allowed by a constraint. For example, a fixed length of fencing can enclose many rectangles, but those rectangles do not all have the same area. A quadratic function models the area of every permitted rectangle in terms of one side length.

Express one dimension in terms of the other, substitute that relationship into the area formula, and then interpret the vertex of the resulting parabola. This procedure works for any rectangular fencing problem with the same type of constraint.

Why a downward opening quadratic has a maximum

A quadratic function has the form

When $a<0$, its graph opens downward. The vertex is therefore the highest point of the graph. If the vertex lies inside the feasible domain of the problem, its function value is the maximum.

Locating the vertex

For $f(x)=ax^2+bx+c$, the vertex coordinates are found from

The equivalent discriminant form is

The algebra identifies the vertex, but the context decides whether that vertex is usable. Lengths cannot be negative, a production count may need to be an integer, and time cannot fall outside the interval described by the model.

Maximum Area of a Rectangular Chicken Coop

A farmer has $20\text{ m}$ of fencing and wants to enclose a rectangular chicken coop. What dimensions produce the greatest possible area?

1. Name the dimensions. Let the length be $p\text{ m}$ and the width be $l\text{ m}$.

2. Translate the perimeter constraint.

   $$2p+2l=20$$

   $$p+l=10$$

   $$p=10-l$$

3. Build an area function in one variable.

   $$A=p\cdot l$$

   $$A(l)=(10-l)l$$

   $$A(l)=10l-l^2$$

   $$A(l)=-l^2+10l$$

4. State the feasible domain. Both side lengths must be positive, so $0<l<10$. The coefficient $a=-1$ is negative, which means the area graph opens downward.

5. Locate the vertex.

   $$l_v=-\frac{b}{2a}=-\frac{10}{2(-1)}=\frac{-10}{-2}=5$$

6. Recover the other dimension and evaluate the area.

   $$p=10-l=10-5=5$$

   $$A(5)=5\cdot5=25$$

7. Confirm that the vertex gives the maximum area. The vertex lies inside $0<l<10$. At the boundary values, the rectangle collapses and the area approaches $0$, so the interior vertex gives the global maximum.

The greatest area is therefore $25\text{ m}^2$, obtained from a $5\text{ m}\times5\text{ m}$ square.

Constraint Model Domain and Vertex

The table collects the four pieces that turn the fence problem into one variable. Each row states a part of the problem beside its mathematical form.

Applying the Same Method to Profit

A profit function also asks for the highest point of the parabola, only the coordinates now mean quantity and profit.

The same method works whenever a quantity follows a downward parabola. A profit function is one such case, and its vertex gives the quantity with the largest profit.

Production profit

Suppose the profit from selling $x$ toys, measured in thousands of rupiah, is

Because $a=-2<0$, the model has a maximum:

The model predicts a maximum profit of $\text{IDR }1\,250\,000$ when $25$ toys are sold. Here the vertex is already an integer, so no rounding comparison is needed.

Projectile height

Suppose a toy rocket's height after $t\text{ s}$ is

The physically meaningful interval follows from the times at which the height is zero:

The model therefore starts at ground level when $t=0$ and returns to ground level when $t=8$, so its physical time domain is $0\le t\le8$. Within that interval, the maximum occurs at

The rocket reaches a maximum height of $80\text{ m}$ after $4\text{ s}$.

Practice

The problem fixes the perimeter and asks for the largest rectangle, so the constraint has to replace the second variable. Write the area as a function of one variable first.

A rectangle has perimeter $60\text{ cm}$. Determine the length and width that maximize its area, and find the maximum area.

Worked solution

1. Let the length be $p\text{ cm}$ and the width be $l\text{ cm}$.

2. Convert the perimeter constraint into an expression for the length.

   $$2p+2l=60$$

   $$p+l=30$$

   $$p=30-l$$

3. Substitute into the area formula.

   $$A(l)=p\cdot l$$

   $$A(l)=(30-l)l$$

   $$A(l)=30l-l^2$$

   $$A(l)=-l^2+30l$$

4. The feasible domain is $0<l<30$, and $a=-1<0$, so the vertex gives the maximum.

5. Find the width at the vertex.

   $$l_v=-\frac{30}{2(-1)}=\frac{-30}{-2}=15\text{ cm}$$

6. Find the corresponding length and area.

   $$p=30-l=30-15=15\text{ cm}$$

   $$A(15)=15\cdot15=225\text{ cm}^2$$

The maximum area is $225\text{ cm}^2$, produced by a $15\text{ cm}\times15\text{ cm}$ square.