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To find the maximum possible area of the rectangular field given by the quadratic equation [tex]\(A(x) = -2x^2 + 10x + 32\)[/tex], follow these steps:
1. Understand the nature of the quadratic equation:
The equation is in the form [tex]\(A(x) = ax^2 + bx + c\)[/tex], where [tex]\(a = -2\)[/tex], [tex]\(b = 10\)[/tex], and [tex]\(c = 32\)[/tex]. Since the coefficient of [tex]\(x^2\)[/tex] (which is [tex]\(a\)[/tex]) is negative, the parabola opens downwards, indicating that the quadratic function has a maximum value.
2. Find the vertex of the parabola:
The vertex of the parabola represented by the quadratic equation [tex]\(ax^2 + bx + c\)[/tex] gives the maximum (or minimum) point. For a quadratic equation [tex]\(ax^2 + bx + c\)[/tex], the [tex]\(x\)[/tex]-coordinate of the vertex can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
In this case:
[tex]\[ a = -2, \quad b = 10 \][/tex]
Substitute these values into the vertex formula:
[tex]\[ x = -\frac{10}{2(-2)} = -\frac{10}{-4} = 2.5 \][/tex]
Hence, the width [tex]\(x\)[/tex] at which the area is maximized is [tex]\(2.5\)[/tex] meters.
3. Calculate the maximum area:
To find the maximum area, plug the [tex]\(x\)[/tex]-coordinate of the vertex back into the quadratic equation [tex]\(A(x)\)[/tex]:
[tex]\[ A(2.5) = -2(2.5)^2 + 10(2.5) + 32 \][/tex]
Calculate each term step-by-step:
[tex]\[ (2.5)^2 = 6.25 \][/tex]
[tex]\[ -2(6.25) = -12.5 \][/tex]
[tex]\[ 10(2.5) = 25 \][/tex]
Now add all the terms together:
[tex]\[ A(2.5) = -12.5 + 25 + 32 = 44.5 \][/tex]
Thus, the maximum possible area of the field is [tex]\(44.5\)[/tex] square meters.
Through these detailed steps, we determined that the width [tex]\(x\)[/tex] at which the area is maximized is [tex]\(2.5\)[/tex] meters, and the maximum possible area of the field is [tex]\(44.5\)[/tex] square meters.
1. Understand the nature of the quadratic equation:
The equation is in the form [tex]\(A(x) = ax^2 + bx + c\)[/tex], where [tex]\(a = -2\)[/tex], [tex]\(b = 10\)[/tex], and [tex]\(c = 32\)[/tex]. Since the coefficient of [tex]\(x^2\)[/tex] (which is [tex]\(a\)[/tex]) is negative, the parabola opens downwards, indicating that the quadratic function has a maximum value.
2. Find the vertex of the parabola:
The vertex of the parabola represented by the quadratic equation [tex]\(ax^2 + bx + c\)[/tex] gives the maximum (or minimum) point. For a quadratic equation [tex]\(ax^2 + bx + c\)[/tex], the [tex]\(x\)[/tex]-coordinate of the vertex can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
In this case:
[tex]\[ a = -2, \quad b = 10 \][/tex]
Substitute these values into the vertex formula:
[tex]\[ x = -\frac{10}{2(-2)} = -\frac{10}{-4} = 2.5 \][/tex]
Hence, the width [tex]\(x\)[/tex] at which the area is maximized is [tex]\(2.5\)[/tex] meters.
3. Calculate the maximum area:
To find the maximum area, plug the [tex]\(x\)[/tex]-coordinate of the vertex back into the quadratic equation [tex]\(A(x)\)[/tex]:
[tex]\[ A(2.5) = -2(2.5)^2 + 10(2.5) + 32 \][/tex]
Calculate each term step-by-step:
[tex]\[ (2.5)^2 = 6.25 \][/tex]
[tex]\[ -2(6.25) = -12.5 \][/tex]
[tex]\[ 10(2.5) = 25 \][/tex]
Now add all the terms together:
[tex]\[ A(2.5) = -12.5 + 25 + 32 = 44.5 \][/tex]
Thus, the maximum possible area of the field is [tex]\(44.5\)[/tex] square meters.
Through these detailed steps, we determined that the width [tex]\(x\)[/tex] at which the area is maximized is [tex]\(2.5\)[/tex] meters, and the maximum possible area of the field is [tex]\(44.5\)[/tex] square meters.
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