Question

A school wishes to enclose its rectangular playground using 480 meters of fencing.

Answer 1

Answer:

Part a) $A(x)=(-x^2+240x)\ m^2$

Part b) The side length x that give the maximum area is 120 meters

Part c) The maximum area is 14,400 square meters

Step-by-step explanation:

The picture of the question in the attached figure

Part a) Find a function that gives the area A(x) of the playground (in square meters) in terms of x

we know that

The perimeter of the rectangular playground is given by

$P=2(L+W)$

we have

$P=480\ m\\L=x\ m$

substitute

$480=2(x+W)$

solve for W

$240=x+W\\W=(240-x)\ m$

Find the area of the rectangular playground

The area is given by

$A=LW$

we have

$L=x\ m\\W=(240-x)\ m$

substitute

$A=x(240-x)\\A=-x^2+240x$

Convert to function notation

$A(x)=(-x^2+240x)\ m^2$

Part b) What side length x gives the maximum area that the playground can have?

we have

$A(x)=-x^2+240x$

This function represent a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the length that give the maximum area that the playground can have

Convert the quadratic equation into vertex form

$A(x)=-x^2+240x$

Factor -1

$A(x)=-(x^2-240x)$

Complete the square

$A(x)=-(x^2-240x+120^2)+120^2$

$A(x)=-(x^2-240x+14,400)+14,400$

$A(x)=-(x-120)^2+14,400$

The vertex is the point (120,14,400)

therefore

The side length x that give the maximum area is 120 meters

Part c) What is the maximum area that the playground can have?

we know that

The y-coordinate of the vertex represent the maximum area

The vertex is the point (120,14,400) -----> see part b)

therefore

The maximum area is 14,400 square meters

Verify

$x=120\ m$

$W=(240-120)=120\ m$

The playground is a square

$A=120^2=14,400\ m^2$