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 $\displaystyle x$ and length $\displaystyle 2x.$
The area of the rectangular field is: .$\displaystyle A_1 \:=\:2x\cdot x \:=\:2x^2$ yd˛.
Its perimeter is $\displaystyle 6x$ yards, leaving $\displaystyle 3406x$ yards for the square field.
The side of the square field is: .$\displaystyle \frac{3406x}{4}\,=\,\frac{1703x}{2}$ yards.
The area of the square field is: .$\displaystyle A_2 \:=\:\left(\frac{1703x}{2}\right)^2$ yd˛.
We are told that: .$\displaystyle A_1 \geq 800$
. . $\displaystyle 2x^2 \:\geq \:800 \quad\Rightarrow\quad x^2 \:\geq \:400 \quad\Rightarrow\quad x \:\geq \:20$
We are told that: .$\displaystyle A_2 \geq 100$
. . $\displaystyle \left(\frac{1703x}{2}\right)^2 \:\geq\:100 \quad\Rightarrow\quad \frac{1703x}{2} \:\geq \:10 \quad\Rightarrow \quad 3x \:\geq\:150 \quad\Rightarrow\quad x \:\leq\:50$
Therefore: .$\displaystyle 20 \:\leq \:x \:\leq\:50$
b) What is the greatest number of square yards that can be enclosed
in the two fields? Explain.
The total area is: .$\displaystyle A \:=\:2x^2 + \left(\frac{1703x}{2}\right)^2 \quad\Rightarrow\quad A \;=\;\tfrac{1}{4}\left(17x^2  1020x + 28,\!900\right)$
The graph is an upopening parabola.
Its vertex (lowest point) is: .$\displaystyle (30,3400)$ Code:

 *

 *:
 * :
 * * :
 : * * :
 : * :
 : : :
 : : :
++++
 20 30 50

Hence, maximum area occurs at the endpoints of the interval.
Looking at the graph, we suspect it happens at $\displaystyle x = 50$,
. . but we can verify this.
At $\displaystyle x=20\!:\;A \;=\;\tfrac{1}{4}\left(17\!\cdot\!20^2  1020\!\cdot\!20 + 28,\!900\right) \:=\:3825\text{ yd}^2$
At $\displaystyle x=50\!:\;A \;=\;\tfrac{1}{4}\left(17\!\cdot\!50^2  1020\!\cdot\!50 + 28,\!900\right) \:=\:5100\text{ yd}^2$ . . . There!
The rectangle will be 50 × 100 yards.
The square has sides of 10 yards.