Answer:
The dimensions of enclosed area are 200 and 400/3 feet
Step-by-step explanation:
* Lets explain how to solve the problem
- There are 800 feet of fencing
- We will but it around a rectangular field
- We will divided the field into 2 identical smaller rectangular plots
by placing a fence parallel to one of the field's shorter sides
- Assume that the long side of the rectangular field is a and the
shorter side is b
∵ The length of the fence is the perimeter of the field
∵ We will fence 2 longer sides and 3 shorter sides
∴ 2a + 3b = 800
- Lets find b in terms of a
∵ 2a + 3b = 800 ⇒ subtract 2a from both sides
∴ 3b = 800 - 2a ⇒ divide both sides by 3
∴ [tex]b=\frac{800}{3}-\frac{2a}{3}[/tex] ⇒ (1)
- Lets find the area of the field
∵ The area of the rectangle = length × width
∴ A = a × b
∴ [tex]A=(a).(\frac{800}{3}-\frac{2a}{3})=\frac{800a}{3}-\frac{2a^{2}}{3}[/tex]
- To find the dimensions of maximum area differentiate the area with
respect to a and equate it by 0
∴ [tex]\frac{dA}{da}=\frac{800}{3}-\frac{4a}{3}[/tex]
∵ [tex]\frac{dA}{da}=0[/tex]
∴ [tex]\frac{800}{3}-\frac{4}{3}a=0[/tex] ⇒ Add 4/3 a to both sides
∴ [tex]\frac{800}{3}=\frac{4}{3}a[/tex] ⇒ multiply both sides by 3
∴ 800 = 4a ⇒ divide both sides by 4
∴ 200 = a
- Substitute the value of a in equation (1)
∴ [tex]b=\frac{800}{3}-\frac{2}{3}(200)=\frac{800}{3}-\frac{400}{3}=\frac{400}{3}[/tex]
* The dimensions of enclosed area are 200 and 400/3 feet
The area of a ranch is the amount of space on the ranch.
The dimension that maximizes the enclosed area is: 400/3 by 200
The perimeter is given as: 800
This means that:
[tex]P = 2y + 3x[/tex]
Where: x and y are the dimensions of the ranch
So, we have:
[tex]2y + 3x = 800[/tex]
Make y the subject
[tex]y = \frac{800 - 3x}{2}[/tex]
The area of the ranch is:
[tex]A = xy[/tex]
Substitute [tex]y = \frac{800 - 3x}{2}[/tex]
[tex]A = x \times \frac{800 - 3x}{2}[/tex]
[tex]A = \frac{800x - 3x^2}{2}[/tex]
Split fraction
[tex]A = \frac{800x}{2} - \frac{3x^2}{2}[/tex]
Differentiate
[tex]A' = \frac{800}{2} - 2 \times \frac{3x}{2}[/tex]
[tex]A' = 400 - 3x[/tex]
Set to 0
[tex]400 - 3x = 0[/tex]
Collect like terms
[tex]3x = 400[/tex]
Divide through by 3
[tex]x = \frac{400}3[/tex]
Recall that: [tex]y = \frac{800 - 3x}{2}[/tex]
[tex]y = \frac{800 - 3 \times 400/3}{2}[/tex]
[tex]y = \frac{800 - 400}{2}[/tex]
[tex]y = \frac{400}{2}[/tex]
[tex]y = 200[/tex]
So, we have:
[tex]x = \frac{400}3[/tex] and [tex]y = 200[/tex]
Hence, the dimension that maximizes the enclosed area is: 400/3 by 200
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