Determining a Minimum Area

Finding the Smallest Value within the Allowed Domain

Some optimization problems ask for the smallest possible value. A manufacturer may want the lowest cost, or a design may need the least total area under a fixed geometric condition. The objective changes, but the modeling process is the same: translate the constraint, build a function in one variable, and interpret its vertex.

Finding a minimum does not mean choosing zero automatically. The original problem determines which values are permitted, and the minimum must lie within that feasible domain.

Why an upward opening quadratic has a minimum

A quadratic function has the form

When $a>0$, its graph opens upward. The vertex is therefore the lowest point of the graph. If the vertex lies inside the feasible domain, its function value is the minimum.

Locating the vertex

For $f(x)=ax^2+bx+c$, use

The equivalent discriminant form is

Always return to the context after calculating the vertex. Check the domain, units, and whether the variable must be an integer.

Minimum Area from One Wire and Two Squares

A $40\text{ cm}$ wire is cut into two pieces. Each piece is bent into a square. Where should the wire be cut so that the sum of the two square areas is as small as possible?

1. Name the side lengths. Let the first square have side $x\text{ cm}$ and the second have side $y\text{ cm}$.

2. Translate the wire-length constraint.

   $$4x+4y=40$$

   $$x+y=10$$

   $$y=10-x$$

3. Build the total-area function.

   $$A=x^2+y^2$$

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

   $$A(x)=x^2+(100-20x+x^2)$$

   $$A(x)=2x^2-20x+100$$

4. State the feasible domain. Both squares must have positive side lengths, so $0<x<10$. Because $a=2>0$, the graph opens upward.

5. Locate the vertex.

   $$x_v=-\frac{-20}{2(2)}=\frac{20}{4}=5$$

6. Recover the second side length and evaluate the total area.

   $$y=10-x=10-5=5$$

   $$A(5)=5^2+5^2=50$$

7. Interpret the cut. Each square needs perimeter $4(5)=20\text{ cm}$, so the wire is cut into two $20\text{ cm}$ pieces. The full $40\text{ cm}$ wire is still used. Only the sum of the enclosed areas is minimized.

The minimum total area is $50\text{ cm}^2$, obtained when both squares have side length $5\text{ cm}$.

Constraint Model Domain and Vertex

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

Another Application with Production Cost

Suppose the cost of producing $x$ items, measured in thousands of rupiah, is

Because $a=3>0$, the model has a minimum:

The model predicts a minimum cost of $\text{IDR }200\,000$ at $10$ items. A real production decision would also check whether the model's stated domain and operational constraints allow that quantity.

Practice

The task asks for two numbers with a fixed sum and the largest possible product, so the sum replaces one variable. Write the product as a function of a single number first.

The sum of two positive numbers is $16$. Determine the two numbers that minimize the sum of their squares, and find that minimum sum.

Worked solution

1. Let the first number be $x$ and the second be $y$.

2. Translate the sum constraint.

   $$x+y=16$$

   $$y=16-x$$

3. Build the objective function.

   $$S=x^2+y^2$$

   $$S(x)=x^2+(16-x)^2$$

   $$S(x)=x^2+(256-32x+x^2)$$

   $$S(x)=2x^2-32x+256$$

4. The feasible domain is $0<x<16$, and $a=2>0$, so the vertex gives the minimum.

5. Find the first number at the vertex.

   $$x_v=-\frac{-32}{2(2)}=\frac{32}{4}=8$$

6. Find the second number and evaluate the objective.

   $$y=16-x=16-8=8$$

   $$S(8)=8^2+8^2=128$$

The two numbers are $8$ and $8$, and the minimum sum of their squares is $128$.