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A rectangular playground is to be fenced off and divided into two parts by a fence parallel to one side of the playground. A total of 1320 feet of fencing is used. Find the dimensions of the playground that will enclose the greatest total area.

a) 340 by 230 feet with the divider 340 feet long
b) 330 by 330 feet with the divider 330 feet long
c) 330 by 220 feet with the divider 220 feet long
d) 350 by 230 feet with the divider 230 feet long
e) 325 by 225 feet with the divider 326 feet long

Answer :

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.

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Rewritten by : Jeany