Marcus wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn, so he needs no fence on that side. The other three sides will be enclosed with wire fencing. If Marcus has 600 feet of fencing, you can find the dimensions that maximize the area of the enclosure.

a) Let  
w
  be the width of the enclosure (perpendicular to the barn) and let  
l
  be the length of the enclosure (parallel to the barn). Write an function for the area  
A
  of the enclosure in terms of  
w
 . (HINT first write two equations with  
w
  and  
l
  and  
A
 . Solve for  
l
  in one equation and substitute for  
l
  in the other).
A
(
w
)
=
  

b) What width  
w
  would maximize the area?
w
  = 
 ft
c) What is the maximum area?
A
  = 
 square feet

To solve this problem, let's break it down step by step.

### Part (a): Write the function for the area \( A(w) \) in terms of the width \( w \).

Let:
- \( w \) be the width of the enclosure (perpendicular to the barn).
- \( l \) be the length of the enclosure (parallel to the barn).

Since one side of the enclosure is against the barn, Marcus only needs fencing for the other three sides, which include two widths and one length. The total amount of fencing is given as 600 feet.

**Step 1: Express the total fencing used.**

The total length of the fencing can be written as:
\[
2w + l = 600
\]

**Step 2: Solve for \( l \) in terms of \( w \).**

Rearrange the equation to solve for \( l \):
\[
l = 600 - 2w
\]

**Step 3: Write the area \( A \) in terms of \( w \).**

The area \( A \) of the rectangular enclosure is given by:
\[
A = l \times w
\]

Substitute \( l = 600 - 2w \) into the area formula:
\[
A(w) = w(600 - 2w)
\]

Simplify the expression:
\[
A(w) = 600w - 2w^2
\]

So, the function for the area in terms of the width \( w \) is:
\[
A(w) = 600w - 2w^2
\]

### Part (b): Find the width \( w \) that maximizes the area.

To find the value of \( w \) that maximizes the area, we can use the method of finding the vertex of the parabola represented by the quadratic function \( A(w) = -2w^2 + 600w \).

The width \( w \) that maximizes the area can be found using the formula for the vertex of a parabola:
\[
w = -\frac{b}{2a}
\]
where \( A(w) = aw^2 + bw + c \).

Here, \( a = -2 \) and \( b = 600 \). Substituting these values:
\[
w = -\frac{600}{2(-2)} = \frac{600}{4} = 150 \text{ feet}
\]

So, the width \( w \) that maximizes the area is **150 feet**.

### Part (c): Find the maximum area.

To find the maximum area, substitute \( w = 150 \) feet back into the area function \( A(w) \):
\[
A(150) = 600(150) - 2(150)^2
\]

Calculate the values:
\[
A(150) = 90000 - 45000 = 45000 \text{ square feet}
\]

So, the maximum area of the enclosure is **45,000 square feet**.

### Summary:
- **(a)** The area function is \( A(w) = 600w - 2w^2 \).
- **(b)** The width \( w \) that maximizes the area is 150 feet.
- **(c)** The maximum area is 45,000 square feet.

Would you like more details on any part of this solution? Here are five related questions:

1. How do you determine whether a quadratic function opens upwards or downwards?
2. What is the significance of the vertex of a parabola in optimization problems?
3. Can the maximum area be determined using calculus methods like derivatives?
4. What happens to the area if the width \( w \) is greater than 150 feet?
5. How would the problem change if Marcus needed to enclose all four sides?

**Tip:** When solving optimization problems involving quadratic functions, always check if the function has a maximum or minimum by examining the sign of the leading coefficient.

Explore how to maximize the area of a rectangular enclosure using a given amount of fencing. Learn to solve quadratic equations and optimize area with step-by-step guidance. Suitable for students in Grades 9-12.

Math Problem Statement

b) What width
w would maximize the area? w

ft c) What is the maximum area? A

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