The length (longer side) is 107.8 yd.
The width (shorter side) is 16.2 yd.
Let l be the length and w be the width of the pen.
We know that 2l + 2w = 248 and lw = 3619.
Solving for l and w, we get:
l = (248 - 2w)/2
lw = 3619
Substituting the first equation into the second equation, we get:
w(248 - 2w) = 3619
Expanding the equation, we get:
248w - 2w^2 = 3619
Rearranging the equation, we get:
2w^2 - 248w + 3619 = 0
Dividing both sides by 2, we get:
w^2 - 124w + 1809.5 = 0
Using the quadratic formula, we get:
w = (124 +/- sqrt(124^2 - 4 * 1809.5)) / 2
w = (124 +/- sqrt(15376 - 7238)) / 2
w = (124 +/- sqrt(8138)) / 2
w = (124 +/- 89.6) / 2
w1 = 106.8 and w2 = 16.2
Since the width cannot be greater than the length, we choose w = 16.2. Substituting w back into the equation l = (248 - 2w)/2,
we get: l = (248 - 2 * 16.2) / 2 l = (248 - 32.4) / 2 l = 107.8
Therefore, the dimensions of the pen are 107.8 yd by 16.2 yd.
The length (longer side) is 107.8 yd.
The width (shorter side) is 16.2 yd.
Question
Clark's Country Pet Resort is fencing a new play area for dogs. The manager has purchased 248 yd of fence to enclose a rectangular pen. The area of the pen must be 3619yd^2. What are the dimensions of the pen?
The length (longer side) is □ yd.
The width (shorter side) is □ yd.
(Type exact answers, using radicals as needed. Simplify your answers.)