Answer:
Part 1) The expression is ![A(x)=150x-2x^2](https://tex.z-dn.net/?f=A%28x%29%3D150x-2x%5E2)
Part 2) The area of the schoolyard when x=40 m is A=2,800 m^2
Part 3) The domain is all real numbers greater than zero and less than 75 meters
Step-by-step explanation:
Part 1) Write an expression for A(x)
Let
x -----> the length of the rectangular schoolyard
y ---> the width of the rectangular schoolyard
we know that
The perimeter of the fencing (using the wall of the school for one side) is
![P=2x+y](https://tex.z-dn.net/?f=P%3D2x%2By)
![P=150\ m](https://tex.z-dn.net/?f=P%3D150%5C%20m)
so
![150=2x+y](https://tex.z-dn.net/?f=150%3D2x%2By)
-----> equation A
The area of the rectangular schoolyard is
---> equation B
substitute equation A in equation B
![A=x(150-2x)](https://tex.z-dn.net/?f=A%3Dx%28150-2x%29)
![A=150x-2x^2](https://tex.z-dn.net/?f=A%3D150x-2x%5E2)
Convert to function notation
![A(x)=150x-2x^2](https://tex.z-dn.net/?f=A%28x%29%3D150x-2x%5E2)
Part 2) What is the area of the schoolyard when x=40?
For x=40 m
substitute in the expression of Part 1) and solve for A
![A(40)=150(40)-2(40^2)](https://tex.z-dn.net/?f=A%2840%29%3D150%2840%29-2%2840%5E2%29)
![A(40)=2,800\ m^2)](https://tex.z-dn.net/?f=A%2840%29%3D2%2C800%5C%20m%5E2%29)
Part 3) What is a reasonable domain for A(x) in this context
we know that
A represent the area of the rectangular schoolyard
x represent the length of of the rectangular schoolyard
we have
![A(x)=150x-2x^2](https://tex.z-dn.net/?f=A%28x%29%3D150x-2x%5E2)
This is a vertical parabola open downward
The vertex is a maximum
The x-coordinate of the vertex represent the length for the maximum area
The y-coordinate of the vertex represent the maximum area
The vertex is the point (37.5, 2812.5)
using a graphing tool, see the attached figure
therefore
The maximum area is 2,812.5 m^2
The x-intercepts are x=0 m and x=75 m
The domain for A is the interval -----> (0, 75)
All real numbers greater than zero and less than 75 meters