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Using a Quadratic Model Based on a Verbal Description to Solve a Problem

A farmer has 100 meters of fencing to enclose a rectangular pen.

Which quadratic equation gives the area (A) of the pen, given its width (w)?

A. [tex]\( A(w) = w^2 - 50w \)[/tex]
B. [tex]\( A(w) = w^2 - 100w \)[/tex]
C. [tex]\( A(w) = 50w - w^2 \)[/tex]
D. [tex]\( A(w) = 100w - w^2 \)[/tex]

The greatest rectangular area that the farmer can enclose with 100 meters of fencing is 625 m².

Find the length and width of the greatest rectangular area that the farmer can enclose with 100 meters of fencing.
- The length is ______ meters.
- The width is ______ meters.

Sagot :

GinnyAnsweridn
To solve this problem, let's break it down step by step:

### Step 1: Understanding the problem
The problem involves finding the correct quadratic equation for the area of a rectangular pen given its width and a total fencing of 100 meters. Additionally, we need to determine the dimensions of the rectangle that maximize this area.

### Step 2: Determine the perimeter relationship
The total perimeter (P) of the rectangle is given by:
[tex]\[ P = 2L + 2W \][/tex]
where [tex]\( L \)[/tex] is the length and [tex]\( W \)[/tex] is the width. The perimeter is given as 100 meters, so:
[tex]\[ 2L + 2W = 100 \][/tex]
Divide both sides by 2:
[tex]\[ L + W = 50 \][/tex]
Solving for [tex]\( L \)[/tex] in terms of [tex]\( W \)[/tex]:
[tex]\[ L = 50 - W \][/tex]

### Step 3: Express the area in terms of width
The area [tex]\( A \)[/tex] of the rectangle is given by:
[tex]\[ A = L \times W \][/tex]
Substituting [tex]\( L \)[/tex] from the previous step:
[tex]\[ A = (50 - W) \times W \][/tex]
[tex]\[ A = 50W - W^2 \][/tex]

### Step 4: Identify the quadratic equation
From the expression [tex]\( A = 50W - W^2 \)[/tex], we see that the correct option from the provided choices is:
[tex]\[ A(w) = 50w - w^2 \][/tex]

### Step 5: Determine the dimensions that maximize the area
The quadratic equation [tex]\( A(w) = 50w - w^2 \)[/tex] represents a parabola that opens downwards (since the coefficient of [tex]\( w^2 \)[/tex] is negative). The width [tex]\( w \)[/tex] that maximizes the area is found at the vertex of the parabola.

For a quadratic equation [tex]\( ax^2 + bx + c \)[/tex], the vertex occurs at:
[tex]\[ w = -\frac{b}{2a} \][/tex]
In our equation, [tex]\( A(w) = -w^2 + 50w \)[/tex], we have [tex]\( a = -1 \)[/tex] and [tex]\( b = 50 \)[/tex].

Thus, the width that maximizes the area is:
[tex]\[ w = -\frac{50}{2(-1)} \][/tex]
[tex]\[ w = \frac{50}{2} \][/tex]
[tex]\[ w = 25 \, \text{meters} \][/tex]

### Step 6: Calculate the corresponding length
Using the relationship [tex]\( L = 50 - W \)[/tex]:
[tex]\[ L = 50 - 25 \][/tex]
[tex]\[ L = 25 \, \text{meters} \][/tex]

### Final answer
The length and width of the greatest rectangular area that the farmer can enclose with 100 meters of fencing are both 25 meters.

Therefore, the length is [tex]\( 25 \, \text{m} \)[/tex] and the width is [tex]\( 25 \, \text{m} \)[/tex].
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