Question

Jon is building a dog run up against his house. He has 40 feet of fencing to use on the other three sides. What would be the maximum area possible for his dog run?* *This question is weighted four times as heavily as the other questions. In order to receive full or partial credit for this question, use the space provided to justify your answer or show your work.

175 feet²
100 feet²
225 feet²
200 feet²​

Answer 1

Let's call the length of the two parallel sides of the dog run "x" and the length of the side perpendicular to the house "y". We know that the total length of the three sides is 40 feet, so:

2x + y = 40

Solving for y, we get:

y = 40 - 2x

The area of the dog run is given by:
A = xy

Substituting y in terms of x, we get:

A = x(40 - 2x)

Simplifying, we get: A = 40x - 2x^2

To find the maximum area, we need to find the value of x that maximizes this function. We can do this by taking the derivative of A with respect to x and setting it equal to zero:

dA/dx = 40 - 4x = 0

Solving for x, we get:

x = 10

Substituting x = 10 back into the equation for y, we get:

y = 40 - 2x = 20

Therefore, the dimensions of the dog run that maximize the area are 10 feet by 20 feet. The maximum area is:

A= xy= 10 • 20. = 200 ft^2

* Last option is correct