Fill in the blanks using the available answer choices. Dana wants to build a rectangular pen for her goats. The length of the pen should be at least 50 feet, and the perimeter of t... Fill in the blanks using the available answer choices. Dana wants to build a rectangular pen for her goats. The length of the pen should be at least 50 feet, and the perimeter of the pen should be no more than 190 feet. Dana's goat pen could be ___ wide and ___ long. How many different sets of dimensions are possible for this situation? Read more

Steps to Solve

1. Perimeter Formula

Recall the formula for the perimeter $P$ of a rectangle with length $l$ and width $w$:

$$P = 2l + 2w$$

2. Use the given perimeter constraint

We are given that the perimeter should be no more than 190 feet, so we have:

$$2l + 2w \le 190$$

3. Simplify inequality

Divide both sides of the inequality by 2:

$$l + w \le 95$$

4. Analyze the length constraint

We are given that the length should be at least 50 feet:

$$l \ge 50$$

5. Test the given values for width (Blank 1)

We have two options for the width: 29 feet and 35 feet. Let's test each:

Case 1: Width = 29 feet

$$w = 29$$ $$l + 29 \le 95$$ $$l \le 95 - 29$$ $$l \le 66$$

Now, we need to check if any of the length options (40, 65, 76) satisfy both $l \ge 50$ and $l \le 66$.

• 40 is not greater or equal to 50.
• 65 satisfies $50 \le 65 \le 66$.
• 76 is not less or equal to 66.

So, if the width is 29 feet, the length could be 65 feet.

Case 2: Width = 35 feet

$$w = 35$$ $$l + 35 \le 95$$ $$l \le 95 - 35$$ $$l \le 60$$

Now, we need to check if any of the length options (40, 65, 76) satisfy both $l \ge 50$ and $l \le 60$.

• 40 is not greater or equal to 50.
• 65 is not less or equal to 60.
• 76 is not less or equal to 60.

So, if the width is 35 feet, there are no possible values for the length among the answer choices.

6. Answer to Blanks 1 and 2

From the analysis above, the width can be 29 feet and the length can be 65 feet.

7. Determine the number of dimension sets possible (Blank 3)

Since $l$ and $w$ must be integers, we want to find the number of integer solutions to:

$$l + w \le 95$$ $$l \ge 50$$

We can rewrite the first inequality as $w \le 95 - l$. Since $l \ge 50$, the maximum possible value for $l$ such that $w$ is positive is $l = 94$ because if $l=95$, $w$ can be zero. We want to find how many positive integer values can $l$ take, knowing that it must at least be 50. The constraint is that $l \le 95 - w$ and $w \ge 1$. So we can rewrite the inequality as $l+w \le 95$, where $l \ge 50$ and $w \ge 1$.

If $l = 50$, then $w \le 95 - 50 = 45$, so $1 \le w \le 45$, 45 possibilities. If $l = 51$, then $w \le 95 - 51 = 44$, so $1 \le w \le 44$, 44 possibilities. ... If $l = 94$, then $w \le 95 - 94 = 1$, so $w = 1$, 1 possibility.

The number of possibilities is the sum of integers from 1 to 45, which is:

$$\sum_{i=1}^{45} i = \frac{45(45+1)}{2} = \frac{45 \cdot 46}{2} = 45 \cdot 23 = 1035$$

However, the wording of the question says "sets of dimensions", which implies that the different sets of dimensions should be obtained only from the values provided as options, so we need to re-think our approach. The question asks "how many different sets of dimensions are possible for this situation?". We already found one possible set of dimensions from the provided answer choices: width=29 and length = 65. Let us find all other possibilities keeping in mind that the values for width must be either 29 or 35, and the values for length must be either 40, 65, or 76.

Since the pen must be at least 50 feet long, a length of 40 is impossible, so the length must be either 65 or 76.

We already did the calculations for width = 29.

Consider again the case of width = 35; then $l \le 95-35 = 60$. Since the length has to be at least 50 and no larger than 60, the only length among the available options that satisfies this condition is None.

Consider the case where the length is 76. Based on the constraint $l+w \le 95$ becomes $76+w \le 95$, so $w \le 19$. The closest option to the width available is 29 and 35 none them satisfy the constraint $w \le 19$. It means that having the length being 76 never will be fulfilled with the constraints.

Then there is only one possible combination: width = 29 and length = 65.