Sarah is building a 10-foot-long pen for her pet rabbits. She wants to make gravel bases for two circular food containers to go inside the pen along its width, as shown in the diagram. Let y represent the width of the pen and x represent the
radii of the gravel bases and food containers.
10 feet
Fy>o
She knows that the gravel bases need to fit entirely inside the pen, so the sum of the diameters of the gravel bases
cannot exceed the width, y.
Sarah has at most $150 to spend on materials. The cost will be $4.00 per linear foot of fencing and $2.00 per square foot
of the gravel bases (the cost of the gravel at 1 inch of depth).
(Recall: The area of a circle is A = #r?, where r is the radius.)
Part A
Write the system of inequalities that can be used to determine the length and width of the pen and the radii of the gravel
bases for the food containers. Use the diagram of the pen and containers above to help. Base your inequality statements

y ≥ 4x

y ≤ 8.75 –1/2πx^2

Step-by-step explanation:

Because the diameters of the gravel bases added together cannot exceed the width of the pen, we get the inequality 2x + 2x ≤ y . Rewriting, we get y ≥ 4x as the first inequality in the system.

Next, write an inequality for cost.

To write the expression for the cost of the fencing, find the perimeter of the rectangle, and multiply the perimeter by the cost per foot of fencing. The pen is a rectangle, so the perimeter is 2(10) + 2(y), or 20 + 2y. Multiply the cost of the fencing material ($4.00 per foot) by the perimeter of the fence to get 4(20 + 2y).

Now, write an expression for the gravel bases for the circular food containers. Because A = r2 and the cost of the gravel is $2.00 per square foot, multiply the cost of the material by the sum of these areas to get 2(x2) + 2(x2).

The total cost must be less than or equal to $150. So, we can say that 4(20 + 2y) + 2(x2) + 2(x2) ≤ 150. After simplifying and solving for y: y ≤ 8.75 – x2.

So, this is the system:

y ≥ 4x

y ≤ 8.75 –1/2πx^2

1. perpendicular

2. parallel

3. perpendicular