Math  /  Geometry

QuestionPaul has 400 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?
A rectangle that maximizes the enclosed area has a length of \square yards and a width of \square yards.

Studdy Solution

STEP 1

What is this asking? We need to figure out the length and width of a rectangle that uses 400 yards of fencing and has the biggest area possible. Watch out! Don't mix up perimeter and area!
Perimeter is the total length of the fence, and area is the space inside the fence.

STEP 2

1. Set up the equations
2. Express width in terms of length
3. Express area in terms of length
4. Find the length that maximizes area
5. Calculate the width
6. Calculate the maximum area

STEP 3

We know the total fencing is 400 yards, which is the perimeter of the rectangle.
The formula for the perimeter of a rectangle is P=2L+2W P = 2L + 2W , where PP is the perimeter, LL is the length, and WW is the width.
So, we have 400=2L+2W 400 = 2L + 2W .

STEP 4

We want to maximize the area of the rectangle.
The formula for the area of a rectangle is A=LW A = L \cdot W , where AA is the area.

STEP 5

Let's solve the perimeter equation for WW.
First, we subtract 2L2L from both sides: 4002L=2W 400 - 2L = 2W .
Now, we divide both sides by 2: W=4002L2=200L W = \frac{400 - 2L}{2} = 200 - L .
So, the width is 200L 200 - L .

STEP 6

Now, we can substitute this expression for WW into the area equation: A=L(200L)=200LL2 A = L \cdot (200 - L) = 200L - L^2 .

STEP 7

Our area equation is a quadratic equation, which graphs as a parabola that opens downwards.
The maximum value occurs at the vertex of the parabola.
The x-coordinate (in our case, the LL value) of the vertex of a parabola in the form y=ax2+bx+c y = ax^2 + bx + c is given by x=b2a x = -\frac{b}{2a} .
In our case, a=1 a = -1 and b=200 b = 200 , so L=2002(1)=2002=100 L = -\frac{200}{2 \cdot (-1)} = \frac{-200}{-2} = 100 .
So, the length that maximizes the area is **100** yards.

STEP 8

Now that we know L=100 L = 100 , we can plug it back into our equation for WW: W=200100=100 W = 200 - 100 = 100 .
So, the width is also **100** yards.

STEP 9

Finally, we can calculate the maximum area: A=100100=10000 A = 100 \cdot 100 = 10000 .
The maximum area is **10,000** square yards.

STEP 10

A rectangle that maximizes the enclosed area has a length of **100** yards and a width of **100** yards.
The maximum area is **10,000** square yards.

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