Determining Maximum Area

What is Maximum Area?

Ever seen a garden fence? Sometimes, we have a limited length of fence, but we want to make the garden as big as possible. Well, quadratic functions can help us find the garden dimensions that give the largest area! Magic, right?

Types of Quadratic Functions

A quadratic function looks like a smile ($ax^2 + bx + c$ if $a > 0$) or a frown ($ax^2 + bx + c$ if $a < 0$).

If we want to find the biggest value (maximum), we use the frown shape, so the value of $a$ is negative ($a < 0$).

The general form is:

Here, $a$, $b$, and $c$ are numbers, and $a$ cannot be zero (because if it's zero, it becomes linear).

How to Find the Highest Point (Vertex)

The largest value is at the very top point of the frowning graph. This point is called the vertex.

To find the position of the vertex (the $x$ value), we use the formula:

After getting $x_p$, we plug it back into the quadratic function to get the largest value (the $y$ or $f(x)$ value):

Or you can use this shortcut formula:

where $D = b^2 - 4ac$ (D is called the discriminant).

Example to Understand

Let's say a farmer has 20 meters of fence. He wants to build a rectangular chicken coop. What are the dimensions of the coop so that the area is maximized?

1. Give Names: Let the length be $p$ meters and the width be $l$ meters.

2. Fence Length Relation: The fence is the perimeter.

   $$2p + 2l = 20$$

   Simplify (divide everything by 2):

   $$p + l = 10$$

   Meaning:

   $$p = 10 - l$$

3. Area Formula: The area of the coop is $A = p \times l$. Substitute $p$:

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

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

4. Quadratic Function Form: Arrange it neatly:

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

   This is a quadratic function with $a = -1$, $b = 10$, and $c = 0$. Since $a$ is negative, the graph is a frown, so there is a maximum value.

5. Find Maximum Width (l): Use the formula $x_p = -b / (2a)$ (but replace $x$ with $l$):

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

   So, the width must be 5 meters for the maximum area.

6. Find Maximum Area (A): Plug $l = 5$ into the area formula $A(l) = 10l - l^2$:

   $$A(5) = 10(5) - (5)^2 = 50 - 25 = 25$$

   The largest area is 25 square meters.

7. Conclusion: For the maximum coop area (25 m²), the width $l=5$ meters. The length is $p = 10 - l = 10 - 5 = 5$ meters. The dimensions must be 5 meters x 5 meters (it turns out to be a square!).

Where is This Used?

Business

Finding the biggest profit.

Example:

A toy store sells dolls. If they sell $x$ dolls, the profit (in thousands of rupiah) is $P(x) = 100x - 2x^2$. How many dolls should be sold to maximize profit?

• Profit function: $P(x) = -2x^2 + 100x$ ($a=-2, b=100$).
• Number of dolls for max profit: $x_p = -b / (2a) = -100 / (2 \times -2) = -100 / -4 = 25$ dolls.
• Maximum profit: $P(25) = 100(25) - 2(25^2) = 2500 - 2(625) = 2500 - 1250 = 1250$ thousand rupiah (or IDR 1,250,000).

Physics

Calculating the highest point of a thrown object's jump.

Example:

A toy rocket flies! Its height after $t$ seconds is $h(t) = 40t - 5t^2$ meters. When is the rocket highest and what is its maximum height?

• Quadratic function: $h(t) = -5t^2 + 40t$. $a = -5, b = 40$.
• Time to reach max height: $t_p = -b / (2a) = -40 / (2 \times -5) = -40 / -10 = 4$ seconds.
• Maximum height: $h(4) = 40(4) - 5(4^2) = 160 - 5(16) = 160 - 80 = 80$ meters.

Exercise

A rectangle has a perimeter of 60 cm. Determine its length and width to maximize its area, and calculate the maximum area!

Answer Key

1. Let length be $p$, width be $l$.

2. Perimeter:

   $$2p + 2l = 60$$

   $$p + l = 30$$

   $$p = 30 - l$$

3. Area:

   $$L(l) = p \times l = (30 - l)l$$

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

4. Area Function:

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

   ($a = -1, b = 30$).

5. Width for max area ($l_p$):

   $$l_p = -b / (2a) = -30 / (2 \times -1) = -30 / -2 = 15 \text{ cm}$$

6. Length at max area ($p$):

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

7. Maximum area ($L_{max}$):

   $$L(15) = 30(15) - (15)^2 = 450 - 225 = 225 \text{ cm}^2$$

So, for the maximum area ($L_{max} = 225 \text{ cm}^2$), the dimensions must be 15 cm x 15 cm (a square again!).