lacy has $600.00 set aside to build a rectangular excercise kennel for her dogs. She will buy fencing material for $15/ft. Because the side of an existing barn will be used for one of the sides of the kennel, only three sides need to be fenced. Lacy needs the area of the kennel to be 150^2. Determine the dimensions of a kennel that uses all possible fencing material.
6 years ago
Answered By Mandie Z
We need to set up a few equations for this question. First we need to find out how much fencing can be purchased for $600:
$600\div15=40$600÷15=40
Then we define the perimeter of the pen and set it equal to the maximum amount of fencing available. Since there is only 3 sides the perimeter is:
P = 2x + y
40 = 2x + y
The next equation would be for the area of the pen:
A = xy
150 = xy
Then we can solve the system of equations through substitution, finding both x and y.
6 years ago
Answered By Mandie Z
We need to set up a few equations for this question. First we need to find out how much fencing can be purchased for $600:
$600\div15=40$600÷15=40
Then we define the perimeter of the pen and set it equal to the maximum amount of fencing available. Since there is only 3 sides the perimeter is:
P = 2x + y
40 = 2x + y
The next equation would be for the area of the pen:
A = xy
150 = xy
Then we can solve the system of equations through substitution, finding both x and y.
4 years ago
Answered By Swapan S
$Area=x\cdot y$Area=x·y
Since areas shall be greater than 150 sqft
$x\cdot y>150$x·y>150 ------(1)
$FencingMaterial=\text{600/15}=40feet$FencingMaterial=600/15=40ƒ eet
Since one side is walled 20 feet is required to cover only 3 sides
So $2x+y=15feet$2x+y=15ƒ eet ------(2)
solving for 1 and 2
$2x^2-40x>150$2x2−40x>150
which gives x = 0.38 to 19.62 y= 39.62 to 20.38 (becuase 2x+y=40)
Since Square will give the most area so we can take x=y= 20
4 years ago
Answered By Swapan S
See the graph
Attached Graph:
4 years ago
Answered By Swapan S
Area=x·y
Since areas shall be greater than 150 sqft
$x\cdot y>150$x·y>150 ------(1)
$FencingMaterial=\text{600/15}=40feet$FencingMaterial=600/15=40ƒ eet
Since one side is walled 20 feet is required to cover only 3 sides
So $2x+y=15feet$2x+y=15ƒ eet ------(2)
solving for 1 and 2
$2x^2-40x>150$2x2−40x>150
which gives x = 0.38 to 19.62 y= 39.62 to 20.38 (becuase 2x+y=40)
Since Square will give the most area so we can take x=y= 13.33