Respuesta :
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 [tex]=\$(\frac13x)[/tex]
Now homeowner's total cost for the material is
[tex]=\$[ 2(x+y)-\frac13x][/tex]
[tex]=\$[2x+2y-\frac13x][/tex]
[tex]=\$[2y+x+x-\frac13x][/tex]
[tex]=\$[2y+x+\frac{3x-x}{3}][/tex]
[tex]=\$[2y+x+\frac{2}{3}x][/tex]
[tex]=\$[2y+\frac53x][/tex]
[tex]\therefore C(x)=2y+\frac53x[/tex]
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
[tex]\Rightarrow y=\frac{200}{x}[/tex]
Putting the value of y in C(x)
[tex]\therefore C(x)=2(\frac{200}{x})+\frac53x[/tex]
The domain of C is[tex](0,\infty )[/tex].
[tex]\therefore C(x)=2(\frac{200}{x})+\frac53x[/tex]
Differentiating with respect to x
[tex]C'(x)= - \frac{400}{x^2}+\frac53[/tex]
Again differentiating with respect to x
[tex]C''(x) = \frac{800}{x^3}[/tex]
To find the critical point set C'(x)=0
[tex]\therefore 0= - \frac{400}{x^2}+\frac53[/tex]
[tex]\Rightarrow \frac{400}{x^2}=\frac{5}{3}[/tex]
[tex]\Rightarrow x^2 =\frac{400\times 3}{5}[/tex]
[tex]\Rightarrow x=\sqrt{240}[/tex]
[tex]\Rightarrow x=15.49 \approx15.5[/tex]
Therefore
[tex]\left C''(x) \right|_{x=15.5}=\frac{800}{15.5^3}>0[/tex]
Therefore at x= 15.5 , C(x) is minimum.
Putting the value of x in [tex]y=\frac{200}{x}[/tex] we get
[tex]\therefore y=\frac{200}{15.5}[/tex]
=12.9
Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.