QuestionPeter has 2000 yards of fencing. Find the rectangle dimensions that maximize the area and state the maximum area.

Studdy Solution

STEP 1

Assumptions1. Peter has2000 yards of fencing available. . The fencing is to enclose a rectangular area.
3. We need to find the dimensions of the rectangle that maximize the enclosed area.

STEP 2

Let's denote the length of the rectangle as $$ and the width as $W$. Since the fencing is used to enclose the rectangle, the perimeter of the rectangle is equal to the total fencing available.
The formula for the perimeter of a rectangle is $2 +2W$ . We can set this equal to the total fencing available $2 +2W =2000$

STEP 3

We can simplify this equation by dividing each term by2 $+ W =1000$

STEP 4

We can express the width $W$ in terms of the length W =1000 - L$$

STEP 5

The area $A$ of a rectangle is given by the product of its length and width. We can express the area in terms of A = L \times W = L \times (1000 - L)$$

STEP 6

This is a quadratic function, and its maximum value occurs at the vertex. The x-coordinate of the vertex of a parabola given by $y = ax^2 + bx + c$ is $-b/2a$ . In this case, $a = -1$ and $b =1000$ , so the length $that maximizes the area is$ = -\frac{b}{2a} = -\frac{1000}{2(-1)}$$

STEP 7

Calculate the length $that maximizes the area$ = -\frac{1000}{2(-1)} =500 \, yards$$

STEP 8

Substitute $=500$ yards into the equation $W =1000 - L$ to find the width $W$ that maximizes the area $W =1000 -500 =500 \, yards$

STEP 9

Substitute $=500$ yards and $W =500$ yards into the equation $A = L \times W$ to find the maximum area $A =500 \times500$

STEP 10

Calculate the maximum area $A =500 \times500 =250,000 \, square \, yards$ The rectangle that maximizes the enclosed area has a length of500 yards and a width of500 yards. The maximum area is250,000 square yards.

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QuestionPeter has 2000 yards of fencing. Find the rectangle dimensions that maximize the area and state the maximum area.

Studdy Solution

STEP 1

Assumptions1. Peter has2000 yards of fencing available. . The fencing is to enclose a rectangular area.
3. We need to find the dimensions of the rectangle that maximize the enclosed area.

STEP 2

STEP 3

We can simplify this equation by dividing each term by2 $+ W =1000$

STEP 4

We can express the width $W$ in terms of the length W =1000 - L$$

STEP 5

The area $A$ of a rectangle is given by the product of its length and width. We can express the area in terms of A = L \times W = L \times (1000 - L)$$

STEP 6

This is a quadratic function, and its maximum value occurs at the vertex. The x-coordinate of the vertex of a parabola given by $y = ax^2 + bx + c$ is $-b/2a$ . In this case, $a = -1$ and $b =1000$ , so the length $that maximizes the area is$ = -\frac{b}{2a} = -\frac{1000}{2(-1)}$$

STEP 7

Calculate the length $that maximizes the area$ = -\frac{1000}{2(-1)} =500 \, yards$$

STEP 8

Substitute $=500$ yards into the equation $W =1000 - L$ to find the width $W$ that maximizes the area $W =1000 -500 =500 \, yards$

STEP 9

Substitute $=500$ yards and $W =500$ yards into the equation $A = L \times W$ to find the maximum area $A =500 \times500$

STEP 10

Calculate the maximum area $A =500 \times500 =250,000 \, square \, yards$ The rectangle that maximizes the enclosed area has a length of500 yards and a width of500 yards. The maximum area is250,000 square yards.

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QuestionPeter has 2000 yards of fencing. Find the rectangle dimensions that maximize the area and state the maximum area.

Studdy Solution

STEP 1

Assumptions1. Peter has2000 yards of fencing available. . The fencing is to enclose a rectangular area.3. We need to find the dimensions of the rectangle that maximize the enclosed area.

STEP 2

STEP 3

We can simplify this equation by dividing each term by2+W=1000 + W =1000+W=1000

STEP 4

We can express the width WWW in terms of the length W =1000 - L$$

STEP 5

The area AAA of a rectangle is given by the product of its length and width. We can express the area in terms of A = L \times W = L \times (1000 - L)$$

STEP 6

STEP 7

Calculate the length thatmaximizesthearea that maximizes the areathatmaximizesthearea = -\frac{1000}{2(-1)} =500 \, yards$$

STEP 8

Substitute =500 =500=500 yards into the equation W=1000−LW =1000 - LW=1000−L to find the width WWW that maximizes the areaW=1000−500=500 yardsW =1000 -500 =500 \, yardsW=1000−500=500yards

STEP 9

Substitute =500 =500=500 yards and W=500W =500W=500 yards into the equation A=L×WA = L \times WA=L×W to find the maximum areaA=500×500A =500 \times500A=500×500

STEP 10

Calculate the maximum areaA=500×500=250,000 square yardsA =500 \times500 =250,000 \, square \, yardsA=500×500=250,000squareyardsThe rectangle that maximizes the enclosed area has a length of500 yards and a width of500 yards. The maximum area is250,000 square yards.

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Assumptions1. Peter has2000 yards of fencing available. . The fencing is to enclose a rectangular area.
3. We need to find the dimensions of the rectangle that maximize the enclosed area.

We can simplify this equation by dividing each term by2 $+ W =1000$

We can express the width $W$ in terms of the length W =1000 - L$$

The area $A$ of a rectangle is given by the product of its length and width. We can express the area in terms of A = L \times W = L \times (1000 - L)$$

Calculate the length $that maximizes the area$ = -\frac{1000}{2(-1)} =500 \, yards$$

Substitute $=500$ yards into the equation $W =1000 - L$ to find the width $W$ that maximizes the area $W =1000 -500 =500 \, yards$

Substitute $=500$ yards and $W =500$ yards into the equation $A = L \times W$ to find the maximum area $A =500 \times500$

Calculate the maximum area $A =500 \times500 =250,000 \, square \, yards$ The rectangle that maximizes the enclosed area has a length of500 yards and a width of500 yards. The maximum area is250,000 square yards.