Question

A rectangular schoolyard is to be fenced in using the wall of the school for one side and 150

Answer 1

Answer:

Part 1) The expression is $A(x)=150x-2x^2$

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

$P=2x+y$

$P=150\ m$

so

$150=2x+y$

$y=150-2x$ -----> equation A

The area of the rectangular schoolyard is

$A=xy$ ----> equation B

substitute equation A in equation B

$A=x(150-2x)$

$A=150x-2x^2$

Convert to function notation

$A(x)=150x-2x^2$

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

$A(40)=150(40)-2(40^2)$

$A(40)=2,800\ m^2$    

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

$A(x)=150x-2x^2$

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)

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