Key to "Consider the following problem:A farmer wants to fence..."

High School

nemired9

Answered question

2021-12-21

Consider the following problem:A farmer wants to fence in a rectangular area with 750 feet of fencing, then divide it into four pens by running fencing parallel to one of the sides. What is the largest possible total area of the four pens? Draw several diagrams illustrating the situation, some with shallow, wide pens and some with deep, n $\Rightarrow$ pens. Find the total areas of these configurations. Does it appear that there is a maximum area? If so estimate it

Answer & Explanation

Jenny Sheppard

Beginner2021-12-22Added 35 answers

We can devise a formula that allows us to choose a width value and obtain values for the lengths of the other sides, including the lengths that divide the interior into four pens. If the width is w then start with
$750 - 2 w$
Since you have 750 ft of fencing to work with, subtract 2w from it to get what is left for the other sides. 2w can not be 750 or more for the formula to work. Then divide that by 5, since there are 5 parallel sides.
length of a parallel side $p = \frac{750 - 2 w}{5}$
As an alternative, you could create a formula where you could choose the length of the five parallel sides.
$70 \cdot 200 = 14000 f t^{2}$
$30 \cdot 300 = 9000 f t^{2}$
$110 \cdot 100 = 11000 f t^{2}$
Experimenting with the numbers, this seems to be the maximum area you can get:
$75 \cdot 187.5 = 14062.5 f t^{2}$

Wendy Boykin

Beginner2021-12-23Added 35 answers

STEP 1:
First, we should write down what we know. We know A = xy and the perimeter equals 750ft. Using a diagram, we can form the equation

750 = 5 x + 2 y

Step 2:
Solve for y in terms of x. You should get

y = (\begin{matrix} - \frac{5}{2} \end{matrix}) x + 375

This information is going to help us get our Area equation in terms of one variable only (just in terms of x)
Step 3: Using the relationship you found between x & y in Step 2, put your Area equation in terms of x only. You should get

A = x y = x \cdot [\begin{matrix} (\begin{matrix} - \frac{5}{2} \end{matrix}) x + 375 \end{matrix}] = (\begin{matrix} - \frac{5}{2} \end{matrix}) x^{2} + 375 x

Step 4:
Derive your area equation, and you should get

A^{'} = - 5 a + 375

Step 5:
Find the critical numbers to get your potential minimums & maximums. Set. A'=0 (since A’ exists everywhere, we do not need to worry about A' DNE) and solve for x. You should get x = 75
Step 6:
Find y but plugging 75 in for x in your perimeter equation. You should find y

= 187.5

Step 7:
Plug in your x and y in the Area equation, and you should arrive at your answer of

14, 062.5 f t^{2}

nick1337

Expert2021-12-27Added 777 answers

For any rectangle, the one with the largest area will be the one whose dimensions are as close to 6 square as possible.
However, the dividers change the process to find this maximum somewhat.
Letting x represent two sides of the rectangle anc the 3 parallel dividers, we have 2x+3x = 5x.
Letting y represent the other two sides of the rectangle, we have 2y.
$We know that 2 y + 5 x = 750$
Solving for y, we first subtract 5x from each side:
$2 y + 5 x - 5 x = 750 - 5 x$
$2 y = - 5 x + 750$
Next we divide both sides by 2:
$2 y / 2 = - 5 x / 2 + 750 / 2$
$y = - 25 x + 375$
We know that the ares of a rectangle is given by
A= lw, where | is the length and w is the width. In this rectangle, one dimension is x and the other is y, making the area
$A = x y$
Substituting the expression for y we just found above, we have
$A = x (- 2.5 x + 375)$
$A = - 2.5 x^{2} + 375 x$
This is a quadratic equation, with values a = -2.5, b = 375 and c=0.
To find the maximum, we will find the vertex. First we find the axis of symmetry, using the equation
$x = - b / 2 a$
$x = - 375 / 2 (- 2.5) = - 375 / - 5 = 75$
Substituting this back in place of every x in our area equation, we have
$A = - 2.5 x^{2} + 375 x$
$A = - 2.5 (75)^{2} + 375 (75) = - 2.5 (5625) + 28125 = - 14062.5 + 28125 = 14062.5$

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