Ernie wants to fence in a rectangular play yard for his two dogs. He has 12 pieces of fence that are one unit each. What are the dimensions of the largest play yard that Ernie could make? Show your work
2 L + 2 w = 12 so L = 6-w
area = A = L w
L w = (6-w)w = A
w^2 - 6 w = -A
find vertex by completing square
w^2 - 6 w + 9 = -A+9
(w-3) = -1(A-9)
max at w = L =3 and area = 9
By the way, you could have guessed that our rectangle would turn out to be a square.
To find the dimensions of the largest play yard that Ernie could make, we need to determine the length and width of the rectangle.
Let's assume the length of the play yard is L, and the width is W.
Since there are 12 pieces of fence, each with a length of 1 unit, the total amount of fence pieces used to enclose the yard is equal to the perimeter of the play yard.
Perimeter of a rectangle = 2 * (Length + Width)
Given that the perimeter is 12 units, we can substitute the values and equation as:
2 * (L + W) = 12
Divide both sides of the equation by 2 to get:
L + W = 6
To maximize the area of the rectangle, we want to find the largest possible values for both the length and width.
One way to do this is by using the concept of the arithmetic mean and the geometric mean.
By applying the Arithmetic Mean-Geometric Mean inequality, we know that for any two positive numbers a and b:
(a + b) / 2 ≥ √(ab)
In our case, a = L and b = W. Rearranging the equation, we have:
L + W ≥ 2√(LW)
Substituting L + W = 6, we now have:
6 ≥ 2√(LW)
Dividing both sides of the inequality by 2:
3 ≥ √(LW)
Now, let's square both sides of the inequality to remove the square root:
3^2 ≥ (√(LW))^2
9 ≥ LW
So, the largest possible area of the play yard is 9 square units.
To find the dimensions, we need to determine the possible pairs of whole numbers whose product is 9. These pairs are:
1 and 9
3 and 3
Therefore, the dimensions of the largest play yard Ernie can make with 12 pieces of fence are:
- Length (L) = 3 units
- Width (W) = 3 units