Mins and maxes

**calc_help123** · April 12th, 2009, 06:22 PM

A man has 340 yards of fencing for enclosing two seprate fields, one of which is to be a rectangle twice as long as it is wide and the other is a square. The square field must contain at least 100 square yards and the rectanglar one must contain at least 800 square yards.
a) if x is the width of the rectangular field, what are the maximum and minimum possible values of x?
b)What is the greatest number of square yards that can be enclosed in the two fields? Explain.

**mollymcf2009** · April 12th, 2009, 06:59 PM

Originally Posted by calc_help123

A man has 340 yards of fencing for enclosing two seprate fields, one of which is to be a rectangle twice as long as it is wide and the other is a square. The square field must contain at least 100 square yards and the rectanglar one must contain at least 800 square yards.
a) if x is the width of the rectangular field, what are the maximum and minimum possible values of x?
b)What is the greatest number of square yards that can be enclosed in the two fields? Explain.

Let x = width of rectangle, so length of rectangle = 2x

Let y = length of one side of square

340 yards of fencing is available to enclose the two fields, so:

perimeter of rectangle + perimeter of square = 340 yds

$<br /> 6x + 4y = 340$

Then find the areas of the two fields:

Area of rectangle = $2x^2 = 800$
Area of square = $y^2 = 100$

Now use these equations to answer the questions. Give it a try and come back if you are still stuck!

**OnMyWayToBeAMathProffesor** · April 12th, 2009, 07:09 PM

I too had reached up to this point. For x of the rectangle, it is greater than or at least 20 yards and greater than or at least 10 yards for the square. but that is where i get stuck. could u or someone please explain the rest of the problem.

**Soroban** · April 12th, 2009, 07:33 PM

Hello, calc_help123!

A man has 340 yards of fencing for enclosing two separate fields,
one of which is to be a rectangle twice as long as it is wide and the other is a square.

The square field must contain at least 100 square yards
and the rectanglar one must contain at least 800 square yards.

a) If x is the width of the rectangular field, what are the maximum
and minimum possible values of x?

The rectangular field has width $x$ and length $2x.$
The area of the rectangular field is: . $A_1 \:=\:2x\cdot x \:=\:2x^2$ yd².

Its perimeter is $6x$ yards, leaving $340-6x$ yards for the square field.

The side of the square field is: . $\frac{340-6x}{4}\,=\,\frac{170-3x}{2}$ yards.

The area of the square field is: . $A_2 \:=\:\left(\frac{170-3x}{2}\right)^2$ yd².

We are told that: . $A_1 \geq 800$
. . $2x^2 \:\geq \:800 \quad\Rightarrow\quad x^2 \:\geq \:400 \quad\Rightarrow\quad x \:\geq \:20$

We are told that: . $A_2 \geq 100$
. . $\left(\frac{170-3x}{2}\right)^2 \:\geq\:100 \quad\Rightarrow\quad \frac{170-3x}{2} \:\geq \:10 \quad\Rightarrow \quad -3x \:\geq\:-150 \quad\Rightarrow\quad x \:\leq\:50$

Therefore: . $20 \:\leq \:x \:\leq\:50$

b) What is the greatest number of square yards that can be enclosed
in the two fields? Explain.

The total area is: . $A \:=\:2x^2 + \left(\frac{170-3x}{2}\right)^2 \quad\Rightarrow\quad A \;=\;\tfrac{1}{4}\left(17x^2 - 1020x + 28,\!900\right)$

The graph is an up-opening parabola.
Its vertex (lowest point) is: . $(30,3400)$

Code:

       |
       |                       *
       |
       |                      *:
       |                     * :
       |     *             *   :
       |     :  *       *      :
       |     :      *          :
       |     :      :          :
       |     :      :          :
     --+-----+------+----------+---
       |    20     30         50
       |

Hence, maximum area occurs at the endpoints of the interval.

Looking at the graph, we suspect it happens at $x = 50$ ,
. . but we can verify this.

At $x=20\!:\;A \;=\;\tfrac{1}{4}\left(17\!\cdot\!20^2 - 1020\!\cdot\!20 + 28,\!900\right) \:=\:3825\text{ yd}^2$

At $x=50\!:\;A \;=\;\tfrac{1}{4}\left(17\!\cdot\!50^2 - 1020\!\cdot\!50 + 28,\!900\right) \:=\:5100\text{ yd}^2$ . . . There!

The rectangle will be 50 × 100 yards.
The square has sides of 10 yards.

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