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Determining Minimum Area

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 ( $ax^2 + bx + c$ when ) or a frown ( when ).

Determining Minimum AreaMinimize areas and costs using quadratic functions with practical examples, efficient solutions, and business applications. Learn cost optimization techniques.

On this page

What is Minimum Area?
Getting to Know Quadratic Functions Again
How to Find the Lowest Point (Valley)
Example to Understand
Where Is It Used?
- Business & Economics
Exercise
- Answer Key

Comments

a > 0

ax^2 + bx + c

a < 0

Now, if we want to find the smallest value (minimum), we use the smile-shaped function, so the value of $a$ is positive ( $a > 0$ ).

The general form remains the same:

f(x) = ax^2 + bx + c

( $a$ , $b$ , and $c$ are numbers, $a \neq 0$ ).

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 $x$ value):

x_p = -\frac{b}{2a}

To find the smallest value (the $y$ or $f(x)$ value):

a < 0

Or use the discriminant shortcut formula:

y_p = -\frac{D}{4a}

where $D = b^2 - 4ac$ .

Example to Understand

Let's say we have a $40 \text{ cm}$ 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)?

Name Things: Let's say the side of the first square is $x \text{ cm}$ , and the side of the second square is $y \text{ cm}$ .
Wire Length Relationship:
- Wire for the first square: Perimeter is $4x \text{ cm}$ .
- Wire for the second square: Perimeter is $4y \text{ cm}$ .
- Total wire length:
  
  $4x + 4y = 40$
  
  Simplify (divide by $4$ ):
  
  $x + y = 10$
Total Area Formula: The total area of both squares is $A = \text{Area}_1 + \text{Area}_2$ .

$A(x) = x^2 + y^2$
Quadratic Function Form: We now have $A(x) = 2x^2 - 20x + 100$ .

This is a quadratic function with $a = 2$ , $b = -20$ , . Since is positive, the graph is a smile, so there is a minimum value.
Find Side x for Minimum Area: Use the formula $x_p = -b / (2a)$ :

$x_p = -\frac{-20}{2 \times 2} = \frac{20}{4} = 5$
Find Side y and Minimum Area:
- Side of the second square: $y = 10 - x = 10 - 5 = 5 \text{ cm}$ .
- Minimum Total Area: Plug $x = 5$ into :
Conclusion: For the total area of both squares to be minimal ( $50 \text{ cm}^2$ ), the wire must be cut so that both squares have the same side length, which is $5 \text{ cm}$ . (This means the wire is cut into two equal parts, $20 \text{ cm}$ and $20 \text{ cm}$ ).

Where Is It Used?

Business & Economics

Example:

The production cost of $x \text{ units}$ of goods (in thousands of rupiah) is $C(x) = 3x^2 - 60x + 500$ . How many units should be produced to minimize the cost?

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

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

Let the first number be $x$ , the second number be $y$ . ( $x > 0, y > 0$ ).
Relationship: $x + y = 16$ . So $y = 16 - x$ .
Sum of Squares: $S = x^2 + y^2$ .
$S(x) = x^2 + (16 - x)^2$
Sum of Squares Function: $S(x) = 2x^2 - 32x + 256$ .

( $a = 2, b = -32$ ). Since , there is a minimum value.
Value of $x$ for minimum sum of squares:

$x_p = -b / (2a) = -(-32) / (2 \times 2) = 32 / 4 = 8$
Value of $y$ :

$y = 16 - x = 16 - 8 = 8$
Minimum Sum of Squares:

$S(8) = 2(8)^2 - 32(8) + 256 = 2(64) - 256 + 256 = 128$

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

Determining Maximum Area

This means:

y = 10 - x

Replace $y$ with $10 - x$ :

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

c = 100

a

So, the side of the first square must be $5 \text{ cm}$ for the sum of areas to be minimal.

A(x)

A(5) = 2(5)^2 - 20(5) + 100 = 2(25) - 100 + 100 = 50

The minimum total area is $50 \text{ cm}^2$ .

\text{Rp }200{,}000

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

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

a > 0