Answer:
[tex]Length=10.5\ ft[/tex]
[tex]Width=10.5\ ft[/tex]
[tex]Area=110.25\ ft^{2}[/tex]
Step-by-step explanation:
Let
x----> the length of the rectangular garden
y---> the width of the rectangular garden
we know that
The perimeter of the rectangle is equal to
[tex]P=2(x+y)[/tex]
we have
[tex]P=42\ ft[/tex]
so
[tex]42=2(x+y)[/tex]
simplify
[tex]21=(x+y)[/tex]
[tex]y=21-x[/tex]------> equation A
Remember that the area of rectangle is equal to
[tex]A=xy[/tex] ----> equation B
substitute equation A in equation B
[tex]A=x(21-x)[/tex]
[tex]A=21x-x^{2}[/tex]----> this is a vertical parabola open downward
The vertex is a maximum
The y-coordinate of the vertex is the maximum area
The x-coordinate of the vertex is the length side of the rectangle that maximize the area
using a graphing tool
The vertex is the point [tex](10.5,110.25)[/tex]
see the attached figure
so
[tex]x=10.5\ ft[/tex]
Find the value of y
[tex]y=21-10.5=10.5\ ft[/tex]
The garden is a square
the area is equal to
[tex]A=(10.5)(10.5)=110.25\ ft^{2}[/tex] ----> is equal to the y-coordinate of the vertex is correct
