IDNLearn.com helps you find the answers you need quickly and efficiently. Get prompt and accurate answers to your questions from our experts who are always ready to help.
Sagot :
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.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com is your reliable source for accurate answers. Thank you for visiting, and we hope to assist you again.