Math / Algebra

QuestionErik has 40 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?

Studdy Solution

STEP 1

1. Erik has a fixed amount of fencing, 40 yards, to use as the perimeter of a rectangle.
2. The goal is to maximize the area enclosed by the rectangle.
3. The relationship between the perimeter and the dimensions of the rectangle will be used to express the area as a function of one variable.

STEP 2

1. Express the perimeter constraint in terms of the rectangle's dimensions.
2. Express the area as a function of one variable using the perimeter constraint.
3. Find the critical points of the area function to determine the maximum area.
4. Calculate the maximum area and corresponding dimensions.

STEP 3

Let the length and width of the rectangle be $l$ and $w$ respectively. The perimeter $P$ of a rectangle is given by:
$P = 2l + 2w$
Given $P = 40$ , we have:
$2l + 2w = 40$

STEP 4

Solve the perimeter equation for one variable. Let's solve for $w$ :
$2l + 2w = 40$ $l + w = 20$ $w = 20 - l$

STEP 5

Express the area $A$ of the rectangle in terms of $l$ using the expression for $w$ :
$A = l \times w$ $A = l \times (20 - l)$ $A = 20l - l^2$

STEP 6

To find the maximum area, take the derivative of $A$ with respect to $l$ and set it to zero to find critical points:
$A = 20l - l^2$ $\frac{dA}{dl} = 20 - 2l$
Set the derivative equal to zero:
$20 - 2l = 0$

STEP 7

Solve for $l$ :
$20 - 2l = 0$ $2l = 20$ $l = 10$

STEP 8

Substitute $l = 10$ back into the equation for $w$ :
$w = 20 - l$ $w = 20 - 10$ $w = 10$

STEP 9

Calculate the maximum area using the dimensions $l = 10$ and $w = 10$ :
$A = l \times w$ $A = 10 \times 10$ $A = 100$
The dimensions that maximize the enclosed area are $10$ yards by $10$ yards, and the maximum area is $100$ square yards.

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QuestionErik has 40 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?

Studdy Solution

STEP 1

1. Erik has a fixed amount of fencing, 40 yards, to use as the perimeter of a rectangle.
2. The goal is to maximize the area enclosed by the rectangle.
3. The relationship between the perimeter and the dimensions of the rectangle will be used to express the area as a function of one variable.

STEP 2

1. Express the perimeter constraint in terms of the rectangle's dimensions.
2. Express the area as a function of one variable using the perimeter constraint.
3. Find the critical points of the area function to determine the maximum area.
4. Calculate the maximum area and corresponding dimensions.

STEP 3

Let the length and width of the rectangle be $l$ and $w$ respectively. The perimeter $P$ of a rectangle is given by:
$P = 2l + 2w$
Given $P = 40$ , we have:
$2l + 2w = 40$

STEP 4

Solve the perimeter equation for one variable. Let's solve for $w$ :
$2l + 2w = 40$ $l + w = 20$ $w = 20 - l$

STEP 5

Express the area $A$ of the rectangle in terms of $l$ using the expression for $w$ :
$A = l \times w$ $A = l \times (20 - l)$ $A = 20l - l^2$

STEP 6

To find the maximum area, take the derivative of $A$ with respect to $l$ and set it to zero to find critical points:
$A = 20l - l^2$ $\frac{dA}{dl} = 20 - 2l$
Set the derivative equal to zero:
$20 - 2l = 0$

STEP 7

Solve for $l$ :
$20 - 2l = 0$ $2l = 20$ $l = 10$

STEP 8

Substitute $l = 10$ back into the equation for $w$ :
$w = 20 - l$ $w = 20 - 10$ $w = 10$

STEP 9

Calculate the maximum area using the dimensions $l = 10$ and $w = 10$ :
$A = l \times w$ $A = 10 \times 10$ $A = 100$
The dimensions that maximize the enclosed area are $10$ yards by $10$ yards, and the maximum area is $100$ square yards.

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QuestionErik has 40 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?

Studdy Solution

STEP 1

1. Erik has a fixed amount of fencing, 40 yards, to use as the perimeter of a rectangle.2. The goal is to maximize the area enclosed by the rectangle.3. The relationship between the perimeter and the dimensions of the rectangle will be used to express the area as a function of one variable.

STEP 2

1. Express the perimeter constraint in terms of the rectangle's dimensions.2. Express the area as a function of one variable using the perimeter constraint.3. Find the critical points of the area function to determine the maximum area.4. Calculate the maximum area and corresponding dimensions.

STEP 3

Let the length and width of the rectangle be l l l and w w w respectively. The perimeter P P P of a rectangle is given by:P=2l+2w P = 2l + 2w P=2l+2wGiven P=40 P = 40 P=40, we have:2l+2w=40 2l + 2w = 40 2l+2w=40

STEP 4

Solve the perimeter equation for one variable. Let's solve for w w w:2l+2w=40 2l + 2w = 40 2l+2w=40 l+w=20 l + w = 20 l+w=20 w=20−l w = 20 - l w=20−l

STEP 5

Express the area A A A of the rectangle in terms of l l l using the expression for w w w:A=l×w A = l \times w A=l×w A=l×(20−l) A = l \times (20 - l) A=l×(20−l) A=20l−l2 A = 20l - l^2 A=20l−l2

STEP 6

To find the maximum area, take the derivative of A A A with respect to l l l and set it to zero to find critical points:A=20l−l2 A = 20l - l^2 A=20l−l2 dAdl=20−2l \frac{dA}{dl} = 20 - 2l dldA​=20−2lSet the derivative equal to zero:20−2l=0 20 - 2l = 0 20−2l=0

STEP 7

Solve for l l l:20−2l=0 20 - 2l = 0 20−2l=0 2l=20 2l = 20 2l=20 l=10 l = 10 l=10

STEP 8

Substitute l=10 l = 10 l=10 back into the equation for w w w:w=20−l w = 20 - l w=20−l w=20−10 w = 20 - 10 w=20−10 w=10 w = 10 w=10

STEP 9

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1. Erik has a fixed amount of fencing, 40 yards, to use as the perimeter of a rectangle.
2. The goal is to maximize the area enclosed by the rectangle.
3. The relationship between the perimeter and the dimensions of the rectangle will be used to express the area as a function of one variable.

1. Express the perimeter constraint in terms of the rectangle's dimensions.
2. Express the area as a function of one variable using the perimeter constraint.
3. Find the critical points of the area function to determine the maximum area.
4. Calculate the maximum area and corresponding dimensions.

Let the length and width of the rectangle be $l$ and $w$ respectively. The perimeter $P$ of a rectangle is given by:
$P = 2l + 2w$
Given $P = 40$ , we have:
$2l + 2w = 40$

Solve the perimeter equation for one variable. Let's solve for $w$ :
$2l + 2w = 40$ $l + w = 20$ $w = 20 - l$

Express the area $A$ of the rectangle in terms of $l$ using the expression for $w$ :
$A = l \times w$ $A = l \times (20 - l)$ $A = 20l - l^2$

To find the maximum area, take the derivative of $A$ with respect to $l$ and set it to zero to find critical points:
$A = 20l - l^2$ $\frac{dA}{dl} = 20 - 2l$
Set the derivative equal to zero:
$20 - 2l = 0$

Solve for $l$ :
$20 - 2l = 0$ $2l = 20$ $l = 10$

Substitute $l = 10$ back into the equation for $w$ :
$w = 20 - l$ $w = 20 - 10$ $w = 10$