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Quadratic Functions

Determining Minimum Area

Nabil Akbarazzima Fatih

Mathematics

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 (ax2+bx+cax^2 + bx + c when a>0a > 0) or a frown (ax2+bx+cax^2 + bx + c when a<0a < 0).

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

The general form remains the same:

f(x)=ax2+bx+cf(x) = ax^2 + bx + c

(aa, bb, and cc are numbers, a0a \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 xx value):

xp=b2ax_p = -\frac{b}{2a}

To find the smallest value (the yy or f(x)f(x) value):

yp=f(xp)=a(xp)2+b(xp)+cy_p = f(x_p) = a(x_p)^2 + b(x_p) + c

Or use the discriminant shortcut formula:

yp=D4ay_p = -\frac{D}{4a}

where D=b24acD = 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 xx cm, and the side of the second square is yy cm.

  2. Wire Length Relationship:

    • Wire for the first square: Perimeter is 4x4x cm.

    • Wire for the second square: Perimeter is 4y4y cm.

    • Total wire length:

      4x+4y=404x + 4y = 40

      Simplify (divide by 4):

      x+y=10x + y = 10

      This means:

      y=10xy = 10 - x
  3. Total Area Formula: The total area of both squares is A=Area1+Area2A = \text{Area}_1 + \text{Area}_2.

    A(x)=x2+y2A(x) = x^2 + y^2

    Replace yy with 10x10 - x:

    A(x)=x2+(10x)2A(x) = x^2 + (10 - x)^2
    A(x)=x2+(10020x+x2)A(x) = x^2 + (100 - 20x + x^2)
    A(x)=2x220x+100A(x) = 2x^2 - 20x + 100
  4. Quadratic Function Form: We now have A(x)=2x220x+100A(x) = 2x^2 - 20x + 100.

    This is a quadratic function with a=2a = 2, b=20b = -20, c=100c = 100. Since aa is positive, the graph is a smile, so there is a minimum value.

  5. Find Side x for Minimum Area: Use the formula xp=b/(2a)x_p = -b / (2a):

    xp=202×2=204=5x_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: y=10x=105=5y = 10 - x = 10 - 5 = 5 cm.

    • Minimum Total Area: Plug x=5x = 5 into A(x)A(x):

      A(5)=2(5)220(5)+100=2(25)100+100=50A(5) = 2(5)^2 - 20(5) + 100 = 2(25) - 100 + 100 = 50

      The minimum total area is 50 cm250 \text{ cm}^2.

  7. Conclusion: For the total area of both squares to be minimal (50 cm250 \text{ cm}^2), the wire must be cut so that both squares have the same side length, which is 5 cm5 \text{ cm}. (This means the wire is cut into two equal parts, 20 cm20 \text{ cm} and 20 cm20 \text{ cm}).

Where Is It Used?

Business & Economics

Example:

The production cost of xx units of goods (in thousands of rupiah) is C(x)=3x260x+500C(x) = 3x^2 - 60x + 500. How many units should be produced to minimize the cost?

  • Cost function: C(x)=3x260x+500C(x) = 3x^2 - 60x + 500 (a=3,b=60a=3, b=-60). a>0a>0, so there is a minimum.
  • Number of units for minimum cost: xp=b/(2a)=(60)/(2×3)=60/6=10x_p = -b / (2a) = -(-60) / (2 \times 3) = 60 / 6 = 10 units.
  • Minimum cost: C(10)=3(10)260(10)+500=3(100)600+500=300600+500=200C(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 xx, the second number be yy. (x>0,y>0x > 0, y > 0).

  2. Relationship: x+y=16x + y = 16. So y=16xy = 16 - x.

  3. Sum of Squares: S=x2+y2S = x^2 + y^2.

    S(x)=x2+(16x)2S(x) = x^2 + (16 - x)^2
    S(x)=x2+(25632x+x2)S(x) = x^2 + (256 - 32x + x^2)
    S(x)=2x232x+256S(x) = 2x^2 - 32x + 256
  4. Sum of Squares Function: S(x)=2x232x+256S(x) = 2x^2 - 32x + 256.

    (a=2,b=32a = 2, b = -32). Since a>0a > 0, there is a minimum value.

  5. Value of xx for minimum sum of squares:

    xp=b/(2a)=(32)/(2×2)=32/4=8x_p = -b / (2a) = -(-32) / (2 \times 2) = 32 / 4 = 8
    .

  6. Value of yy:

    y=16x=168=8y = 16 - x = 16 - 8 = 8
    .

  7. Minimum Sum of Squares:

    S(8)=2(8)232(8)+256=2(64)256+256=128S(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).