Jim is setting up a rectangular dog run in
his backyard. He buys six 3-foot sections
of fencing and a 3-foot wide gate. What
are the dimensions of the dog run with
the greatest area in each situation?
a) Jim uses the yard fence for one side of
the dog run.
b) Jim uses the corner of the yard fence
for two sides of the dog run.
To find the dimensions of the dog run with the greatest area, we need to consider the given materials and constraints in each situation.
a) Jim uses the yard fence for one side of the dog run:
Let's assume that the length of the yard fence is L feet. Therefore, the dimensions of the dog run will be L by W, where W is the width of the dog run.
Given that Jim has six 3-foot sections of fencing, he can use four of them to create three sides (two sides and a gate) of the dog run. So, the remaining two sides will be the width of the dog run, W.
We can write an equation based on the perimeter of the dog run:
2L + W + 3 = 6 * 3
2L + W + 3 = 18
Simplifying the equation, we get:
2L + W = 15
Now, let's focus on the area of the dog run, which is A = L * W. To find the greatest area, we can rewrite the equation for the area in terms of one variable:
A = (15 - 2L) * L
To maximize the area, we need to find the value of L that will result in the largest possible value for A. To do this, we take the derivative of A with respect to L, set it equal to zero, and find the critical point:
dA/dL = 2L - 30 = 0
2L = 30
L = 15
Therefore, if Jim uses the yard fence for one side of the dog run, the dimensions with the greatest area will be 15 feet by 3 feet.
b) Jim uses the corner of the yard fence for two sides of the dog run:
In this situation, Jim will use the yard fence for the length (L) and one of the 3-foot sections for half of the width (W). Again, we can write the equation based on the perimeter of the dog run:
L + W + 3 + 2W = 6 * 3
L + 3W + 3 = 18
L + 3W = 15
To find the area, we use A = L * W. Substituting the value of L from the perimeter equation, we have:
A = (15 - 3W) * W
To maximize the area, we take the derivative of A with respect to W, set it equal to zero, and find the critical point:
dA/dW = 12 - 6W = 0
6W = 12
W = 2
Therefore, if Jim uses the corner of the yard fence for two sides of the dog run, the dimensions with the greatest area will be 3 feet by 2 feet.
The largest area would be closest to a square.
a) two 3-foot sections on each side, with two 3-foot sections and the gate on the center.
b) three 3-foot sections on the two sides with the gate on one side.
I'll let you do the calculations.