Math  /  Geometry

QuestionA rancher plans to use 1200 feet of fencing and a side of his barn to form a rectangular boundary for cattle. What dimensions of the rectangle would give the maximum area? What is that area?
The maximum area occurs when each side of fencing that touches the barn is \square feet long and the side opposite the barn is \square feet long.
The maximum area is \square square feet. (Type an integer or a fraction.)

Studdy Solution

STEP 1

What is this asking? We need to figure out the *perfect* rectangle dimensions using 1200 feet of fencing alongside a barn to maximize the area for those lucky cattle! Watch out! Don't forget the barn acts as one side of the rectangle, so we only need fencing for three sides!

STEP 2

1. Set up variables and the constraint equation.
2. Express area in terms of one variable.
3. Find the maximum area.
4. Calculate dimensions for maximum area.

STEP 3

Let ww be the width of the rectangle (the sides touching the barn) and ll be the length (the side opposite the barn).

STEP 4

We have 1200 feet of fencing.
Since the barn is one side, the total fencing used is 2w+l=12002w + l = 1200.
This is our **constraint equation**!

STEP 5

From 2w+l=12002w + l = 1200, we can express the length in terms of the width: l=12002wl = 1200 - 2w.
This helps us work with just *one* variable!

STEP 6

The area of a rectangle is A=lwA = l \cdot w.
Substituting ll, we get A(w)=(12002w)w=1200w2w2A(w) = (1200 - 2w) \cdot w = 1200w - 2w^2.
Now we have the area as a function of *just* the width!

STEP 7

The graph of our area function A(w)A(w) is a parabola opening downwards.
The maximum area occurs at the **vertex** of this parabola.
The x-coordinate (in our case, *w*-coordinate) of the vertex is given by w=b2aw = -\frac{b}{2a}, where aa and bb are coefficients of the quadratic equation A(w)=aw2+bw+cA(w) = aw^2 + bw + c.

STEP 8

In our area function A(w)=1200w2w2A(w) = 1200w - 2w^2, we have a=2a = -2 and b=1200b = 1200.
So, w=12002(2)=12004=300w = -\frac{1200}{2 \cdot (-2)} = -\frac{1200}{-4} = 300.
So, the width that gives the maximum area is **300 feet**!

STEP 9

Using our expression from earlier, l=12002wl = 1200 - 2w, we find l=12002300=1200600=600l = 1200 - 2 \cdot 300 = 1200 - 600 = 600.
The length is **600 feet**!

STEP 10

Now, plug the width back into the area formula: A(300)=12003002(300)2=360000290000=360000180000=180000A(300) = 1200 \cdot 300 - 2 \cdot (300)^2 = 360000 - 2 \cdot 90000 = 360000 - 180000 = 180000.
The **maximum area** is 180,000 square feet!

STEP 11

We found that the width w=300w = 300 feet and the length l=600l = 600 feet give the maximum area.

STEP 12

The maximum area occurs when each side of fencing that touches the barn is 300300 feet long and the side opposite the barn is 600600 feet long.
The maximum area is 180000180000 square feet.

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