Respuesta :
1) It would reach a maximum after 1.5 seconds, and would take 5.94 seconds to hit the ground.
2) 7.5 by 7.5 would maximize the area.
Explanation
To find the maximum, we find the axis of symmetry:
x = -b/2a = -48/2(-16) = -48/-32 = 1.5
This is the x-value, which is time.
We solve the related equation
0=280 + 48t - 16t²
to answer this. Using the quadratic formula, we have:
[tex]t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-48\pm \sqrt{48^2-4(-16)(280)}}{2(-16)} \\ \\=\frac{-48\pm \sqrt{2304--17920}}{-32}=\frac{-48\pm \sqrt{2304+17920}}{-32} \\ \\=\frac{-48\pm \sqrt{20224}}{-32}=\frac{-48\pm 142.2}{-32} \\ \\=\frac{-48+142.2}{-32}\text{ or }\frac{-48-142.2}{-32} \\ \\=\frac{94.2}{-32}\text{ or }\frac{-190.2}{-32}=-2.94\text{ or }5.94[/tex]
Since a negative number for time makes no sense, our answer is 5.94.
2) To maximize area and minimize perimeter, we make the dimensions as close to equal as possible. Since she has 15 feet of fence to work with and 2 sides to fence in, we can make each side 15/2 = 7.5 feet to maximize the area.
2) 7.5 by 7.5 would maximize the area.
Explanation
To find the maximum, we find the axis of symmetry:
x = -b/2a = -48/2(-16) = -48/-32 = 1.5
This is the x-value, which is time.
We solve the related equation
0=280 + 48t - 16t²
to answer this. Using the quadratic formula, we have:
[tex]t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-48\pm \sqrt{48^2-4(-16)(280)}}{2(-16)} \\ \\=\frac{-48\pm \sqrt{2304--17920}}{-32}=\frac{-48\pm \sqrt{2304+17920}}{-32} \\ \\=\frac{-48\pm \sqrt{20224}}{-32}=\frac{-48\pm 142.2}{-32} \\ \\=\frac{-48+142.2}{-32}\text{ or }\frac{-48-142.2}{-32} \\ \\=\frac{94.2}{-32}\text{ or }\frac{-190.2}{-32}=-2.94\text{ or }5.94[/tex]
Since a negative number for time makes no sense, our answer is 5.94.
2) To maximize area and minimize perimeter, we make the dimensions as close to equal as possible. Since she has 15 feet of fence to work with and 2 sides to fence in, we can make each side 15/2 = 7.5 feet to maximize the area.
