Consider the following problem: A farmer with 950 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?
(a) Draw several diagrams illustrating the situation, some with shallow, wide pens and some with deep, narrow pens. Find the total areas of these configurations. Does it appear that there is a maximum area? If so, estimate it.
(b) Draw a diagram illustrating the general situation. Let x denote the length of each of two sides and three dividers. Let y denote the length of the other two sides.
(c) Write an expression for the total area A in terms of both x and y.
A =
(d) Use the given information to write an equation that relates the variables.
(e) Use part (d) to write the total area as a function of one variable.
A(x) =
(f) Finish solving the problem by finding the largest area. ft2

Answer:

Step-by-step explanation:

(a)

Suppose we came up with an ideology whereby we pick a value for the length including the length dividing the inside into 4 parts(5 parallel sides), then we can get the value for breath by using the following process.

Let assume the length of the rectangle is 50;

Then, the breath can be calculated as follows:

= 50 × 5 = 250   ( since the breath is divided into 5 parallel sides)

The fencing is said to be 950 ft

So, 950 - 250 = 700

Then divided by 2, we get:

= 700/2

= 350

So for the first diagram; the length = 50 and the breath = 350

The area = 50 × 350 = 17500 ft²

Now, let's go up a little bit.

If the length increase to 100;

Then 100 × 5 = 500

⇒ 950 - 500 = 450

⇒ 450/2 = 225

The area = 225 × 100 = 22500 ft²

Suppose the length increases to 150

Then 150 × 5 = 750

⇒ 950 - 750 = 200

⇒ 200/2 = 100

The area = 150 × 100 = 15000 ft²

The diagrams for each of the outline above can be seen in the image attached below.

(b) The diagram illustrating the general solution can be seen in the second image provided below.

(c) The expression for  the total area A in terms of both x and y is:

Area A = x×y

(d) Recall that:

The fencing is said to be 950 ft.

And the length is divided inside into 5 parallel sides;

Then:

5x + 2y = 950  (from the illustration in the second image below)

2y  = 950 - 5x

[tex]y = \dfrac{950}{2} - \dfrac{5}{2}x[/tex]

[tex]y = 475- \dfrac{5}{2}x[/tex]

(e)

From (c); replace the value of y in (d) into (c)

Then:

Area A = x×y

[tex]f(x)= x\times ( 475 -\dfrac{5}{2}x)[/tex]

Open brackets

[tex]f(x)= ( 475 x-\dfrac{5}{2}x^2)[/tex]

(f)

By differentiating what we have in (e)

[tex]f(x)= ( 475 x-\dfrac{5}{2}x^2)[/tex]

[tex]f'(x)= ( 475 (1)-\dfrac{5}{2}(2x))[/tex]

[tex]f'(x)= 475 -5x[/tex]

[tex]\implies 475 = 5x[/tex]

x = 475/5

x = 95

From (d):

[tex]y = 475- \dfrac{5}{2}x[/tex]

[tex]y = 475- \dfrac{5}{2}(95)[/tex]

[tex]y =237.5[/tex]

∴

Area A = x × y

Area A = 95 × 237.5

Area A = 22562.5 ft²

What is the measure of ABC

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