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Sagot :
To solve this problem, let's start with the given information:
1. The farmer has 100 meters of fencing to enclose a rectangular pen.
2. We need to find the quadratic equation that gives the area [tex]\( A \)[/tex] of the pen in terms of its width [tex]\( w \)[/tex].
3. Finally, we need to determine the greatest rectangular area that can be enclosed with the given fencing.
First, let's understand the relation between the width [tex]\( w \)[/tex] and the length [tex]\( l \)[/tex] of the pen.
Since the total perimeter is 100 meters, we set up the equation for the perimeter of a rectangle:
[tex]\[ 2w + 2l = 100 \][/tex]
Dividing the entire equation by 2:
[tex]\[ w + l = 50 \][/tex]
From this equation, we can express the length [tex]\( l \)[/tex] in terms of the width [tex]\( w \)[/tex]:
[tex]\[ l = 50 - w \][/tex]
Next, we need to express the area [tex]\( A \)[/tex] of the rectangle in terms of [tex]\( w \)[/tex]:
[tex]\[ A = w \cdot l \][/tex]
Using the expression for [tex]\( l \)[/tex] in terms of [tex]\( w \)[/tex]:
[tex]\[ A = w \cdot (50 - w) \][/tex]
[tex]\[ A = 50w - w^2 \][/tex]
Thus, the quadratic equation for the area [tex]\( A \)[/tex] in terms of the width [tex]\( w \)[/tex] is:
[tex]\[ A(w) = 50w - w^2 \][/tex]
Now, to find the greatest rectangular area that the farmer can enclose, we need to maximize the area function [tex]\( A(w) \)[/tex]. The quadratic equation [tex]\( A(w) = 50w - w^2 \)[/tex] is a downward-opening parabola (since the coefficient of [tex]\( w^2 \)[/tex] is negative), and its maximum value occurs at the vertex.
The vertex form of a quadratic equation [tex]\( ax^2 + bx + c \)[/tex] has its maximum (or minimum) at [tex]\( x = -\frac{b}{2a} \)[/tex]. Here, [tex]\( a = -1 \)[/tex] and [tex]\( b = 50 \)[/tex].
Therefore, the width [tex]\( w \)[/tex] that maximizes the area is:
[tex]\[ w = -\frac{b}{2a} \][/tex]
[tex]\[ w = -\frac{50}{2 \times -1} \][/tex]
[tex]\[ w = 25 \][/tex]
Using this width, we can calculate the maximum area:
[tex]\[ A = 50w - w^2 \][/tex]
[tex]\[ A = 50 \cdot 25 - 25^2 \][/tex]
[tex]\[ A = 1250 - 625 \][/tex]
[tex]\[ A = 625 \][/tex]
Therefore, the greatest rectangular area that the farmer can enclose with 100 meters of fencing is:
[tex]\[ \boxed{625 \, \text{m}^2} \][/tex]
Summarizing, the quadratic equation that gives the area of the pen in terms of its width [tex]\( w \)[/tex] is:
[tex]\[ A(w) = 50w - w^2 \][/tex]
And the greatest rectangular area that the farmer can enclose with 100 meters of fencing is:
[tex]\[ \boxed{625 \, \text{m}^2} \][/tex]
So the correct quadratic equation is:
[tex]\[ A(w) = 50w - w^2 \][/tex]
1. The farmer has 100 meters of fencing to enclose a rectangular pen.
2. We need to find the quadratic equation that gives the area [tex]\( A \)[/tex] of the pen in terms of its width [tex]\( w \)[/tex].
3. Finally, we need to determine the greatest rectangular area that can be enclosed with the given fencing.
First, let's understand the relation between the width [tex]\( w \)[/tex] and the length [tex]\( l \)[/tex] of the pen.
Since the total perimeter is 100 meters, we set up the equation for the perimeter of a rectangle:
[tex]\[ 2w + 2l = 100 \][/tex]
Dividing the entire equation by 2:
[tex]\[ w + l = 50 \][/tex]
From this equation, we can express the length [tex]\( l \)[/tex] in terms of the width [tex]\( w \)[/tex]:
[tex]\[ l = 50 - w \][/tex]
Next, we need to express the area [tex]\( A \)[/tex] of the rectangle in terms of [tex]\( w \)[/tex]:
[tex]\[ A = w \cdot l \][/tex]
Using the expression for [tex]\( l \)[/tex] in terms of [tex]\( w \)[/tex]:
[tex]\[ A = w \cdot (50 - w) \][/tex]
[tex]\[ A = 50w - w^2 \][/tex]
Thus, the quadratic equation for the area [tex]\( A \)[/tex] in terms of the width [tex]\( w \)[/tex] is:
[tex]\[ A(w) = 50w - w^2 \][/tex]
Now, to find the greatest rectangular area that the farmer can enclose, we need to maximize the area function [tex]\( A(w) \)[/tex]. The quadratic equation [tex]\( A(w) = 50w - w^2 \)[/tex] is a downward-opening parabola (since the coefficient of [tex]\( w^2 \)[/tex] is negative), and its maximum value occurs at the vertex.
The vertex form of a quadratic equation [tex]\( ax^2 + bx + c \)[/tex] has its maximum (or minimum) at [tex]\( x = -\frac{b}{2a} \)[/tex]. Here, [tex]\( a = -1 \)[/tex] and [tex]\( b = 50 \)[/tex].
Therefore, the width [tex]\( w \)[/tex] that maximizes the area is:
[tex]\[ w = -\frac{b}{2a} \][/tex]
[tex]\[ w = -\frac{50}{2 \times -1} \][/tex]
[tex]\[ w = 25 \][/tex]
Using this width, we can calculate the maximum area:
[tex]\[ A = 50w - w^2 \][/tex]
[tex]\[ A = 50 \cdot 25 - 25^2 \][/tex]
[tex]\[ A = 1250 - 625 \][/tex]
[tex]\[ A = 625 \][/tex]
Therefore, the greatest rectangular area that the farmer can enclose with 100 meters of fencing is:
[tex]\[ \boxed{625 \, \text{m}^2} \][/tex]
Summarizing, the quadratic equation that gives the area of the pen in terms of its width [tex]\( w \)[/tex] is:
[tex]\[ A(w) = 50w - w^2 \][/tex]
And the greatest rectangular area that the farmer can enclose with 100 meters of fencing is:
[tex]\[ \boxed{625 \, \text{m}^2} \][/tex]
So the correct quadratic equation is:
[tex]\[ A(w) = 50w - w^2 \][/tex]
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