A farmer wants to build a pen in the following shape to keep his ostriches and emu separate. The pen for the ostriches is a square and the pen for the emu is a rectangle which has the length 10 feet more than its width. The ostriches need 4.5 times as much area as the emus. If the total amount of fencing is 160 feet, find the outside dimensions of the pen.
Note: I had to make up the shape of the pens, and I had to make up the
sentence in red because it can't be worked with only the information you gave.
Area of ostrich pen = side² = x²
Area of emu pen = = length·width = y·(y+10)
Area of ostrich pen = 4.5(area of emu pen)
x² = 4.5y·(y+10)
Fencing = 4 sides of square + top of rectangle + 2 vertical sides of rectangle
160 = 4x + (y+10) + 2y
So we have this system:
x² = 4.5y·(y+10)
160 = 4x + (y+10) + 2y
Simplifying the first:
x² = 4.5y²+45y
Simplifying the second:
160 = 4x + y + 10 + 2y
160 = 4x + 3y + 10
150 = 4x + 3y
Solve for x
150 - 3y = 4x
= x
Substitute in x² = 4.5y²+45y
x² = 4.5y²+45y
x² = 4.5y²+45y
= 4.5y² + 45y
= 4.5y² + 45y
Multiply through by 16
(150-3y)² = 72y² + 720y
[3(50-y)]² = 72y² + 720y
9(50-y)² = 72y² + 720y
Divide both sides by 9
(50-y)² = 8y² + 80y
2500-100y+y² = 8y² + 80y
0 = 7y² + 180y - 2500
Factoring:
0 = (y-10)(7y+250)
Use zero-factor property:
y-10 = 0; 7y+250 = 0
y = 10 7y = -250
y =
Discard the negative value. Substitute y = 10 into
= x
= x
= x
= x
30 = x
Edwin