Answer
The length of the holding pen is 25 ft. while the width is 21 ft.
Work Step by Step
First we need to write the equations that would satisfy the given conditions such as having a perimeter of $92$ feet, and an area of $525$ square feet.
Using the formula in getting the perimeter and area of a rectangle, these equations translate to:
$Perimeter: P = 2(L + W) = 92$ (equation 1)
$Area: A = L\:x\:W = 525$ (equation 2)
Next, we are going to use the first equation to solve for the second equation: $$2(L + W) = 92$$ $$L + W = \frac{92}{2}$$ $$L + W = 46$$
Solving for $W$, we have $$W = 46 - L$$
Substituting to equation 2: $$L\:x\:W = 525$$ $$L (46-L) = 525$$ $$46L - L^{2} = 525$$
To be able to find the roots, we need to rewrite this equation into the quadratic form $ax^{2} +bx + c = 0$, and use the quadratic formula to solve for the value/s $L$.
Thus, we have: $$-L^{2} + 46L -525 = 0$$ with $a = -1$, $b = 46$, and $c = -525$
Using the quadratic formula: $$L = \frac{-b ± \sqrt{(b^{2}-4ac)}}{2a}$$ $$L = \frac{-46 ± \sqrt{(46^{2}-4(-1)(-525)}}{2(-1)}$$ $$L = \frac{-46 ± \sqrt{16}}{-2}$$ $$L = 21$$ $$L= 25$$
When $L=21$, equation 1 where $P = 2(L+W) = 92$ will be: $$2(21 + W) = 92$$ $$21 + W = 46$$ $$W = 25$$
When $L=25$, equation 1 where $P = 2(L+W) = 92$ will be: $$2(25 + W) = 92$$ $$25 + W = 46$$ $$W = 21$$
Since it is a holding pen, then the width will be less than the length, hence, the $L = 25 ft.$ while the $W = 21 ft$.
