Thank you for visiting 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. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
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