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

Answer from: Quest

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right

Draw lines to match each expression on the left with the correct product on the right.

Answer from: Quest

answer: my brother can figure it out, imma go get him so i can answer : ) edit: my brother is sleeping, sorry man

step-by-step explanation:

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