Peter is building a fenced in area for his dog. He has 250 feet (ft.) of fencing to create an area for his dog. He is building it next to his house so he will only need fencing for 3 sides of the rectangular area. The diagram shows the width, d, of the area Peter is building for his dog.

Question

The function A(d)=d(250−2d) represents the area that will be fenced in for Peter's dog, in square feet.

Complete the statements.

(2 points)
A reasonable scale for the horizontal axis of the function A(d)=d(250−2d) is (options are 1ft, 2ft, 10ft, or 75ft)
, A reasonable scale for the vertical axis of the function A(d)=d(250−2d) is (options are 75 sq ft, 100 sq ft, 500 sq ft, or 2000 sq ft)

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Answer 1

To complete the statements regarding the scales for the axes of the function $A(d)=d(250-2d)$, let's analyze the function a bit more:

1. For the horizontal axis (d):

   • The variable $d$ represents the width, and it must be a positive value. Since Peter has 250 feet of fencing, the maximum width $d$ can be is 125 feet (as $250-2d$ must still be positive). A reasonable scale for the horizontal axis would allow us to visualize a range from 0 to 125 feet effectively.
   • Considering the options: 1 ft, 2 ft, 10 ft, or 75 ft, a scale of 10 ft would be reasonable, allowing for clear visualization of the width increments without making the graph too crowded.

2. For the vertical axis (A(d)):

   • The area $A(d)$ can be calculated as follows: $$A(d)=d(250-2d)$$
   • To find the maximum area, we can analyze the function. The area is a quadratic function, and its maximum value occurs at the vertex. The vertex $d$ value can be calculated as: $$d=-\frac{b}{2a}=-\frac{250}{2(-2)}=62.5$$
   • Substituting back into the area function: $$A(62.5)=62.5(250-2(62.5))=62.5(125)=7812.5\text{ square feet}$$
   • The area goes to 0 when $d=0$ or $d=125$. Thus, the range for the area is from 0 to approximately 7812.5 square feet.
   • Considering the options for the vertical axis: 75 sq ft, 100 sq ft, 500 sq ft, or 2000 sq ft, a scale of 500 sq ft would be appropriate and manageable based on the maximum area.

Combining these observations, we can complete the statements:

A reasonable scale for the horizontal axis of the function $A(d)=d(250-2d)$ is 10 ft, and a reasonable scale for the vertical axis of the function $A(d)=d(250-2d)$ is 500 sq ft.