Answer:
c) 330 ft by 220 ft with the divider 220 ft long.
Step-by-step explanation:
We are given that
Let length of playground=y
Breadth of playground=x
Fencing used=1320 ft
We have to find the dimensions of playground that will enclose the greatest total area.
According to question
Fencing used=x+x+x+y+y=3x+2y
[tex]3x+2y=1320[/tex]
[tex]y=1320-3x[/tex]
[tex]y=\frac{1320-3x}{2}[/tex]
Area of rectangular playground=[tex]l\times b[/tex]
Area of rectangular playground=[tex]x\times y[/tex]
Substitute the values then ,we get
[tex]A(x)=x\times \frac{1320-3x}{2}[/tex]
[tex]A(x)=\frac{1}{2}(1320x-3x^2)[/tex]
Differentiate w.r.t x
[tex]A'(x)=\frac{1}{2}(1320-6x)[/tex]
Using formula: [tex]\frac{dx^n}{dx}=nx^{n-1}[/tex]
Substitute A'(x)=0
[tex]\frac{1}{2}(1320-6x)=0[/tex]
[tex]1320-6x=0[/tex]
[tex]6x=1320[/tex]
[tex]x=\frac{1320}{6}=220[/tex]
Again differentiate w.r.t x
[tex]A''(x)=\frac{1}{2}(-6)=-3 <0[/tex]
Hence, the area of rectangular playground is maximum at x=220
Substitute x=220
Then, [tex]y=\frac{1320-3(220)}{2}=330[/tex]
Length of rectangular playground=330 ft
Breadth of rectangular playground=220 ft
Option c is true.
c) 330 ft by 220 ft with the divider 220 ft long.
