+23 **I made up this problen and I want to see if you can solve it...please help**
Suppose you want to enclose a rectangular garden plot against a house using fencing on three sides. Assume you have 50 ft of fencing material and want to create a garden with an area of 150ft².
+2592 Ummm how about Length(l) = 4/3w
Then it is just width =15 and then length 20. With the length parallel to the house.
+2592 Ummm how about Length(l) = 4/3w
Then it is just width =15 and then length 20. With the length parallel to the house.
+128664 We have that a perimeter that is composed of a width and two congruent lengths of fencing
So we have
50 = w + 2l subtract w from both sides
50 - w = 2l divide both sides by 2
[ 50 - w] / 2 = l
And area = l * w so we have
150 = [50 - w] / 2 * w multiply both sides by 2
300 = [ 50 - w ] * w simplify
300 = 50w - w^2 rearrange
w^2 - 50w + 300 = 0 and using the quadratic formula....
w = 25 + 5√13 ft [about 43.03 ft]
And the length = [50 - (25 + 5√13) ] / 2 ft =
[ 25 - 5√13] / 2 ft = [about 3.487 ft]
Check the area = l * w = 150 ft^2 ???
[ 25 + 5√13] * [ 25 - 5√13] / 2 = 150 ft^2
