Mathematics, 06.07.2019 20:10 itssamuelu

Arancher has 800 feet of fencing to put around a rectangular field and then subdivide the field into 2 identical smaller rectangular plots by placing a fence parallel to one of the field's shorter sides. find the dimensions that maximize the enclosed area. write your answers as fractions reduced to lowest terms.

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Answer from: aksambo4707

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:

P = 2y + 3x

Where: x and y are the dimensions of the ranch

So, we have:

2y + 3x = 800

Make y the subject

$y = \frac{800 - 3x}{2}$

The area of the ranch is:

A = xy

Substitute $y = \frac{800 - 3x}{2}$

$A = x \times \frac{800 - 3x}{2}$

$A = \frac{800x - 3x^2}{2}$

Split fraction

$A = \frac{800x}{2} - \frac{3x^2}{2}$

Differentiate

$A' = \frac{800}{2} - 2 \times \frac{3x}{2}$

A' = 400 - 3x

Set to 0

400 - 3x = 0

Collect like terms

3x = 400

Divide through by 3

$x = \frac{400}3$

Recall that: $y = \frac{800 - 3x}{2}$

$y = \frac{800 - 3 \times 400/3}{2}$

$y = \frac{800 - 400}{2}$

$y = \frac{400}{2}$

y = 200

So, we have:

$x = \frac{400}3$ and y = 200

Hence, the dimension that maximizes the enclosed area is: 400/3 by 200

Read more about areas at:

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Answer from: live4dramaoy0yf9

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

∴ $b=\frac{800}{3}-\frac{2a}{3}$ ⇒ (1)

- Lets find the area of the field

∵ The area of the rectangle = length × width

∴ A = a × b

∴ $A=(a).(\frac{800}{3}-\frac{2a}{3})=\frac{800a}{3}-\frac{2a^{2}}{3}$

- To find the dimensions of maximum area differentiate the area with

respect to a and equate it by 0

∴ $\frac{dA}{da}=\frac{800}{3}-\frac{4a}{3}$

∵ $\frac{dA}{da}=0$

∴ $\frac{800}{3}-\frac{4}{3}a=0$ ⇒ Add 4/3 a to both sides

∴ $\frac{800}{3}=\frac{4}{3}a$ ⇒ multiply both sides by 3

∴ 800 = 4a ⇒ divide both sides by 4

∴ 200 = a

- Substitute the value of a in equation (1)

∴ $b=\frac{800}{3}-\frac{2}{3}(200)=\frac{800}{3}-\frac{400}{3}=\frac{400}{3}$

* The dimensions of enclosed area are 200 and 400/3 feet

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