Answer :
To determine the dimensions that maximize the area of the enclosure Matt is planning to build, we will follow a series of logical and mathematical steps.
### Part (a):
We are given:
- Matt has 850 feet of fencing.
- One side of the enclosure is against the barn, meaning he needs no fencing on that side.
We need to express the area \(A\) of the rectangular enclosure in terms of the width \(w\).
1. Understand the fencing used:
- There are three sides to be fenced: two widths and one length.
- If \(w\) is the width, then the total length of the fencing needed for the widths is \(2w\).
- Let \(l\) be the length, then the remaining fencing is used for the length of the enclosure.
2. Express the total fencing:
[tex]\[ 2w + l = 850 \][/tex]
3. Solve for \(l\) in terms of \(w\):
[tex]\[ l = 850 - 2w \][/tex]
4. Express the area \(A\) in terms of \(w\):
- The area of the rectangle is given by the product of the width and the length.
[tex]\[ A(w) = w \cdot l = w \cdot (850 - 2w)\][/tex]
- Simplify the area function:
[tex]\[\boxed{A(w) = 850w - 2w^2} \][/tex]
### Part (b):
To maximize the area, we need to find the critical points where the derivative of \(A(w)\) with respect to \(w\) equals zero. Let's find the width \(w\) that maximizes \(A(w)\).
1. Differentiate the area function \(A(w)\):
[tex]\[ \frac{dA}{dw} = 850 - 4w \][/tex]
2. Set the derivative equal to zero and solve for \(w\):
[tex]\[ 850 - 4w = 0 \][/tex]
[tex]\[ 4w = 850 \][/tex]
[tex]\[ w = \frac{850}{4} \][/tex]
[tex]\[ w = 212.5 \text{ feet} \][/tex]
Thus, the width \(w\) that maximizes the area is:
[tex]\[\boxed{212.5 \text{ feet}}\][/tex]
### Part (a):
We are given:
- Matt has 850 feet of fencing.
- One side of the enclosure is against the barn, meaning he needs no fencing on that side.
We need to express the area \(A\) of the rectangular enclosure in terms of the width \(w\).
1. Understand the fencing used:
- There are three sides to be fenced: two widths and one length.
- If \(w\) is the width, then the total length of the fencing needed for the widths is \(2w\).
- Let \(l\) be the length, then the remaining fencing is used for the length of the enclosure.
2. Express the total fencing:
[tex]\[ 2w + l = 850 \][/tex]
3. Solve for \(l\) in terms of \(w\):
[tex]\[ l = 850 - 2w \][/tex]
4. Express the area \(A\) in terms of \(w\):
- The area of the rectangle is given by the product of the width and the length.
[tex]\[ A(w) = w \cdot l = w \cdot (850 - 2w)\][/tex]
- Simplify the area function:
[tex]\[\boxed{A(w) = 850w - 2w^2} \][/tex]
### Part (b):
To maximize the area, we need to find the critical points where the derivative of \(A(w)\) with respect to \(w\) equals zero. Let's find the width \(w\) that maximizes \(A(w)\).
1. Differentiate the area function \(A(w)\):
[tex]\[ \frac{dA}{dw} = 850 - 4w \][/tex]
2. Set the derivative equal to zero and solve for \(w\):
[tex]\[ 850 - 4w = 0 \][/tex]
[tex]\[ 4w = 850 \][/tex]
[tex]\[ w = \frac{850}{4} \][/tex]
[tex]\[ w = 212.5 \text{ feet} \][/tex]
Thus, the width \(w\) that maximizes the area is:
[tex]\[\boxed{212.5 \text{ feet}}\][/tex]
