# Nakafa Framework: LLM

URL: /en/subject/high-school/10/mathematics/quadratic-function/quadratic-function-minimum-area
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---

export const metadata = {
    title: "Determining Minimum Area",
    description: "Minimize areas and costs using quadratic functions with practical examples, efficient solutions, and business applications. Master cost optimization techniques.",
    authors: [{ name: "Nabil Akbarazzima Fatih" }],
    date: "04/19/2025",
    subject: "Quadratic Functions",
};

## What is Minimum Area?

Besides finding the largest value, quadratic functions can also be used to find the _smallest_ value (minimum). When might we need to find the minimum area? For example, when we have limited materials but need to make something with a certain area, and we want to use as little material as possible.

## Getting to Know Quadratic Functions Again

Remember, quadratic functions can look like a smile (<InlineMath math="ax^2 + bx + c" /> when <InlineMath math="a > 0" />) or a frown (<InlineMath math="ax^2 + bx + c" /> when <InlineMath math="a < 0" />).

Now, if we want to find the _smallest_ value (minimum), we use the smile-shaped function, so the value of <InlineMath math="a" /> is positive (<InlineMath math="a > 0" />).

The general form remains the same:

<BlockMath math="f(x) = ax^2 + bx + c" />

(<InlineMath math="a" />, <InlineMath math="b" />, and <InlineMath math="c" /> are numbers, <InlineMath math="a \neq 0" />).

## How to Find the Lowest Point (Valley)

The smallest value is at the lowest point of the smiling graph. This point is also called the vertex (but it's at the bottom).

The formula to find it is exactly the same as finding the highest point!

To find the _position_ of the valley (the <InlineMath math="x" /> value):

<BlockMath math="x_p = -\frac{b}{2a}" />

To find the _smallest value_ (the <InlineMath math="y" /> or <InlineMath math="f(x)" /> value):

<BlockMath math="y_p = f(x_p) = a(x_p)^2 + b(x_p) + c" />

Or use the discriminant shortcut formula:

<BlockMath math="y_p = -\frac{D}{4a}" />

where <InlineMath math="D = b^2 - 4ac" />.

## Example to Understand

Let's say we have a 40 cm long wire. We want to cut this wire into two parts. The first part will be formed into a square, and the second part will also be formed into a square. What should the cuts be so that the _total area_ of both squares is as small as possible (minimum)?

1.  **Name Things:** Let's say the side of the first square is <InlineMath math="x" /> cm, and the side of the second square is <InlineMath math="y" /> cm.
2.  **Wire Length Relationship:**

    - Wire for the first square: Perimeter is <InlineMath math="4x" /> cm.
    - Wire for the second square: Perimeter is <InlineMath math="4y" /> cm.
    - Total wire length:

      <BlockMath math="4x + 4y = 40" />

      Simplify (divide by 4):

      <BlockMath math="x + y = 10" />

      This means:

      <BlockMath math="y = 10 - x" />

3.  **Total Area Formula:** The total area of both squares is <InlineMath math="A = \text{Area}_1 + \text{Area}_2" />.

    <BlockMath math="A(x) = x^2 + y^2" />

    Replace <InlineMath math="y" /> with <InlineMath math="10 - x" />:

    <div className="flex flex-col gap-4">
      <BlockMath math="A(x) = x^2 + (10 - x)^2" />
      <BlockMath math="A(x) = x^2 + (100 - 20x + x^2)" />
      <BlockMath math="A(x) = 2x^2 - 20x + 100" />
    </div>

4.  **Quadratic Function Form:** We now have <InlineMath math="A(x) = 2x^2 - 20x + 100" />.

    This is a quadratic function with <InlineMath math="a = 2" />, <InlineMath math="b = -20" />, <InlineMath math="c = 100" />. Since <InlineMath math="a" /> is positive, the graph is a smile, so there is a minimum value.

5.  **Find Side x for Minimum Area:** Use the formula <InlineMath math="x_p = -b / (2a)" />:

    <BlockMath math="x_p = -\frac{-20}{2 \times 2} = \frac{20}{4} = 5" />

    So, the side of the first square must be 5 cm for the sum of areas to be minimal.

6.  **Find Side y and Minimum Area:**

    - Side of the second square: <InlineMath math="y = 10 - x = 10 - 5 = 5" /> cm.
    - Minimum Total Area: Plug <InlineMath math="x = 5" /> into <InlineMath math="A(x)" />:

      <BlockMath math="A(5) = 2(5)^2 - 20(5) + 100 = 2(25) - 100 + 100 = 50" />

      The minimum total area is <InlineMath math="50 \text{ cm}^2" />.

7.  **Conclusion:** For the total area of both squares to be minimal (<InlineMath math="50 \text{ cm}^2" />), the wire must be cut so that both squares have the same side length, which is <InlineMath math="5 \text{ cm}" />. (This means the wire is cut into two equal parts, <InlineMath math="20 \text{ cm}" /> and <InlineMath math="20 \text{ cm}" />).

## Where Is It Used?

### Business & Economics

**Example:**

The production cost of <InlineMath math="x" /> units of goods (in thousands of rupiah) is <InlineMath math="C(x) = 3x^2 - 60x + 500" />. How many units should be produced to minimize the cost?

- Cost function: <InlineMath math="C(x) = 3x^2 - 60x + 500" /> (<InlineMath math="a=3, b=-60" />). <InlineMath math="a>0" />, so there is a minimum.
- Number of units for minimum cost: <InlineMath math="x_p = -b / (2a) = -(-60) / (2 \times 3) = 60 / 6 = 10" /> units.
- Minimum cost: <InlineMath math="C(10) = 3(10)^2 - 60(10) + 500 = 3(100) - 600 + 500 = 300 - 600 + 500 = 200" /> thousand rupiah (or Rp 200,000).

## Exercise

The sum of two positive numbers is 16. Determine these two numbers so that the sum of their squares is minimum, and calculate the minimum sum of squares!

### Answer Key

1.  Let the first number be <InlineMath math="x" />, the second number be <InlineMath math="y" />. (<InlineMath math="x > 0, y > 0" />).
2.  Relationship: <InlineMath math="x + y = 16" />. So <InlineMath math="y = 16 - x" />.
3.  Sum of Squares: <InlineMath math="S = x^2 + y^2" />.

    <div className="flex flex-col gap-4">
      <BlockMath math="S(x) = x^2 + (16 - x)^2" />
      <BlockMath math="S(x) = x^2 + (256 - 32x + x^2)" />
      <BlockMath math="S(x) = 2x^2 - 32x + 256" />
    </div>

4.  Sum of Squares Function: <InlineMath math="S(x) = 2x^2 - 32x + 256" />.

    (<InlineMath math="a = 2, b = -32" />). Since <InlineMath math="a > 0" />, there is a minimum value.

5.  Value of <InlineMath math="x" /> for minimum sum of squares:

    <BlockMath math="x_p = -b / (2a) = -(-32) / (2 \times 2) = 32 / 4 = 8" />.

6.  Value of <InlineMath math="y" />:

    <BlockMath math="y = 16 - x = 16 - 8 = 8" />.

7.  Minimum Sum of Squares:

    <BlockMath math="S(8) = 2(8)^2 - 32(8) + 256 = 2(64) - 256 + 256 = 128" />.

Therefore, the two numbers are 8 and 8 for the sum of their squares to be minimum (which is 128).