Answer:
Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.
Step-by-step explanation:
Given that, a homeowner plans to enclose a 200 square foot rectangle playground.
Let the width of the playground be y and the length of the playground be x which is the side along the boundary.
The perimeter of the playground is = 2(length +width)
=2(x+y) foot
The material costs $1 per foot.
Therefore total cost to give boundary of the play ground
=$[ 2(x+y)×1]
=$[2(x+y)]
But the neighbor will play one third of the side x foot.
So the neighbor will play ![=\$(\frac13x)](https://tex.z-dn.net/?f=%3D%5C%24%28%5Cfrac13x%29)
Now homeowner's total cost for the material is
![=\$[ 2(x+y)-\frac13x]](https://tex.z-dn.net/?f=%3D%5C%24%5B%202%28x%2By%29-%5Cfrac13x%5D)
![=\$[2x+2y-\frac13x]](https://tex.z-dn.net/?f=%3D%5C%24%5B2x%2B2y-%5Cfrac13x%5D)
![=\$[2y+x+x-\frac13x]](https://tex.z-dn.net/?f=%3D%5C%24%5B2y%2Bx%2Bx-%5Cfrac13x%5D)
![=\$[2y+x+\frac{3x-x}{3}]](https://tex.z-dn.net/?f=%3D%5C%24%5B2y%2Bx%2B%5Cfrac%7B3x-x%7D%7B3%7D%5D)
![=\$[2y+x+\frac{2}{3}x]](https://tex.z-dn.net/?f=%3D%5C%24%5B2y%2Bx%2B%5Cfrac%7B2%7D%7B3%7Dx%5D)
![=\$[2y+\frac53x]](https://tex.z-dn.net/?f=%3D%5C%24%5B2y%2B%5Cfrac53x%5D)
where C(x) is total cost of material in $.
Given that the area of the playground is 200.
We know that,
The area of a rectangle is =length×width
=xy square foot
∴xy=200
![\Rightarrow y=\frac{200}{x}](https://tex.z-dn.net/?f=%5CRightarrow%20y%3D%5Cfrac%7B200%7D%7Bx%7D)
Putting the value of y in C(x)
![\therefore C(x)=2(\frac{200}{x})+\frac53x](https://tex.z-dn.net/?f=%5Ctherefore%20C%28x%29%3D2%28%5Cfrac%7B200%7D%7Bx%7D%29%2B%5Cfrac53x)
The domain of C is
.
![\therefore C(x)=2(\frac{200}{x})+\frac53x](https://tex.z-dn.net/?f=%5Ctherefore%20C%28x%29%3D2%28%5Cfrac%7B200%7D%7Bx%7D%29%2B%5Cfrac53x)
Differentiating with respect to x
![C'(x)= - \frac{400}{x^2}+\frac53](https://tex.z-dn.net/?f=C%27%28x%29%3D%20-%20%5Cfrac%7B400%7D%7Bx%5E2%7D%2B%5Cfrac53)
Again differentiating with respect to x
![C''(x) = \frac{800}{x^3}](https://tex.z-dn.net/?f=C%27%27%28x%29%20%3D%20%5Cfrac%7B800%7D%7Bx%5E3%7D)
To find the critical point set C'(x)=0
![\therefore 0= - \frac{400}{x^2}+\frac53](https://tex.z-dn.net/?f=%5Ctherefore%200%3D%20-%20%5Cfrac%7B400%7D%7Bx%5E2%7D%2B%5Cfrac53)
![\Rightarrow \frac{400}{x^2}=\frac{5}{3}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7B400%7D%7Bx%5E2%7D%3D%5Cfrac%7B5%7D%7B3%7D)
![\Rightarrow x^2 =\frac{400\times 3}{5}](https://tex.z-dn.net/?f=%5CRightarrow%20x%5E2%20%3D%5Cfrac%7B400%5Ctimes%203%7D%7B5%7D)
![\Rightarrow x=\sqrt{240}](https://tex.z-dn.net/?f=%5CRightarrow%20x%3D%5Csqrt%7B240%7D)
![\Rightarrow x=15.49 \approx15.5](https://tex.z-dn.net/?f=%5CRightarrow%20x%3D15.49%20%5Capprox15.5)
Therefore
![\left C''(x) \right|_{x=15.5}=\frac{800}{15.5^3}>0](https://tex.z-dn.net/?f=%5Cleft%20C%27%27%28x%29%20%5Cright%7C_%7Bx%3D15.5%7D%3D%5Cfrac%7B800%7D%7B15.5%5E3%7D%3E0)
Therefore at x= 15.5 , C(x) is minimum.
Putting the value of x in
we get
![\therefore y=\frac{200}{15.5}](https://tex.z-dn.net/?f=%5Ctherefore%20y%3D%5Cfrac%7B200%7D%7B15.5%7D)
=12.9
Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.