Math / Geometry

Questiona. A rectangular pen is built with one side against a barn. If 400 m of fencing are used for the other three sides of the pen, what dimensions maximize the area of the pen? b. A rancher plans to make four identical and adjacent rectangular pens against a barn, each with an area of $225 \mathrm{~m}^{2}$ . What are the dimensions of each pen that minimize the amount of fence that must be used? \begin{tabular}{|l|l|l|l|l|} \hline \multicolumn{4}{|c|}{ Bam } \\ \hline 225 & 225 & 225 & 225 \\ \hline \end{tabular} a. To maximize the area of the pen, the sides perpendicular to the barn should be 100 m long and the side parallel to the barn should be 200 m long. (Type exact answers, using radicals as needed.) b. To minimize the amount of fence that must be used, each of the sides perpendicular to the barn should be 33 m long and each of the sides parallel to the barn should be $3 \sqrt{5} \mathrm{~m}$ long. (Type exact answers, using radicals as needed.)

Studdy Solution

STEP 1

What is this asking? We need to find the best dimensions for rectangular pens next to a barn to maximize area with limited fencing, and minimize fencing with a target area. Watch out! Don't forget that one side of the pen is the barn, so we only need fencing for three sides!
Also, in the second part, there are four pens.

STEP 2

1. Maximize area with fixed fencing
2. Minimize fencing with fixed area

STEP 3

Let's define what we know!
We have $\text{400 m}$ of fencing for three sides.
Let $x$ be the length of each side perpendicular to the barn, and $y$ be the length of the side parallel to the barn.

STEP 4

We know that $2x + y = 400$ .
We want to maximize the area, which is $A = x \cdot y$ .

STEP 5

Let's rewrite the fencing equation as $y = 400 - 2x$ .
Now, we can substitute this into the area equation: $A(x) = x \cdot (400 - 2x) = 400x - 2x^2$ .

STEP 6

To maximize $A(x)$ , we can find the vertex of this parabola.
The x-coordinate of the vertex is given by $x = -\frac{b}{2a}$ , where $a = -2$ and $b = 400$ .
So, $x = -\frac{400}{2(-2)} = \frac{400}{4} = 100$ .

STEP 7

Now, let's find $y$ ! $y = 400 - 2(100) = 400 - 200 = 200$ .

STEP 8

Each pen has an area of $\text{225 } \text{m}^2$ .
Since the four pens are identical and adjacent, let $x$ be the width of each pen (perpendicular to the barn), and $y$ be the total length of the pens parallel to the barn.
The area of one pen is $x \cdot \frac{y}{4} = 225$ .

STEP 9

The total fencing needed is $F = 4x + y$ .
We want to minimize this.

STEP 10

From the area equation, we have $y = \frac{4 \cdot 225}{x} = \frac{900}{x}$ .
Let's substitute this into the fencing equation: $F(x) = 4x + \frac{900}{x}$ .

STEP 11

To minimize $F(x)$ , we take the derivative and set it equal to zero: $F'(x) = 4 - \frac{900}{x^2} = 0$ .

STEP 12

Solving for $x$ , we get $4x^2 = 900$ , so $x^2 = 225$ , and $x = 15$ (since length must be positive).

STEP 13

Now, let's find $y$ ! $y = \frac{900}{15} = 60$ .
The length of each pen parallel to the barn is $\frac{y}{4} = \frac{60}{4} = 15$ .

STEP 14

a. Dimensions to maximize area: $\text{100 m}$ perpendicular to the barn and $\text{200 m}$ parallel to the barn. b. Dimensions of each pen to minimize fencing: $\text{15 m}$ perpendicular to the barn and $\text{15 m}$ parallel to the barn.

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Studdy Solution

STEP 1

STEP 2

1. Maximize area with fixed fencing
2. Minimize fencing with fixed area

STEP 3

Let's define what we know!
We have $\text{400 m}$ of fencing for three sides.
Let $x$ be the length of each side perpendicular to the barn, and $y$ be the length of the side parallel to the barn.

STEP 4

We know that $2x + y = 400$ .
We want to maximize the area, which is $A = x \cdot y$ .

STEP 5

Let's rewrite the fencing equation as $y = 400 - 2x$ .
Now, we can substitute this into the area equation: $A(x) = x \cdot (400 - 2x) = 400x - 2x^2$ .

STEP 6

To maximize $A(x)$ , we can find the vertex of this parabola.
The x-coordinate of the vertex is given by $x = -\frac{b}{2a}$ , where $a = -2$ and $b = 400$ .
So, $x = -\frac{400}{2(-2)} = \frac{400}{4} = 100$ .

STEP 7

Now, let's find $y$ ! $y = 400 - 2(100) = 400 - 200 = 200$ .

STEP 8

Each pen has an area of $\text{225 } \text{m}^2$ .
Since the four pens are identical and adjacent, let $x$ be the width of each pen (perpendicular to the barn), and $y$ be the total length of the pens parallel to the barn.
The area of one pen is $x \cdot \frac{y}{4} = 225$ .

STEP 9

The total fencing needed is $F = 4x + y$ .
We want to minimize this.

STEP 10

From the area equation, we have $y = \frac{4 \cdot 225}{x} = \frac{900}{x}$ .
Let's substitute this into the fencing equation: $F(x) = 4x + \frac{900}{x}$ .

STEP 11

To minimize $F(x)$ , we take the derivative and set it equal to zero: $F'(x) = 4 - \frac{900}{x^2} = 0$ .

STEP 12

Solving for $x$ , we get $4x^2 = 900$ , so $x^2 = 225$ , and $x = 15$ (since length must be positive).

STEP 13

Now, let's find $y$ ! $y = \frac{900}{15} = 60$ .
The length of each pen parallel to the barn is $\frac{y}{4} = \frac{60}{4} = 15$ .

STEP 14

a. Dimensions to maximize area: $\text{100 m}$ perpendicular to the barn and $\text{200 m}$ parallel to the barn. b. Dimensions of each pen to minimize fencing: $\text{15 m}$ perpendicular to the barn and $\text{15 m}$ parallel to the barn.

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Studdy Solution

STEP 1

STEP 2

1. Maximize area with fixed fencing2. Minimize fencing with fixed area

STEP 3

Let's **define** what we know!We have 400 m\text{400 m}400 m of fencing for *three* sides.Let xxx be the length of each side perpendicular to the barn, and yyy be the length of the side parallel to the barn.

STEP 4

We know that 2x+y=4002x + y = 4002x+y=400.We want to **maximize** the area, which is A=x⋅yA = x \cdot yA=x⋅y.

STEP 5

Let's **rewrite** the fencing equation as y=400−2xy = 400 - 2xy=400−2x.Now, we can **substitute** this into the area equation: A(x)=x⋅(400−2x)=400x−2x2A(x) = x \cdot (400 - 2x) = 400x - 2x^2A(x)=x⋅(400−2x)=400x−2x2.

STEP 6

STEP 7

Now, let's **find** yyy! y=400−2(100)=400−200=200y = 400 - 2(100) = 400 - 200 = 200y=400−2(100)=400−200=200.

STEP 8

STEP 9

The **total fencing** needed is F=4x+yF = 4x + yF=4x+y.We want to **minimize** this.

STEP 10

From the area equation, we have y=4⋅225x=900xy = \frac{4 \cdot 225}{x} = \frac{900}{x}y=x4⋅225​=x900​.Let's **substitute** this into the fencing equation: F(x)=4x+900xF(x) = 4x + \frac{900}{x}F(x)=4x+x900​.

STEP 11

To **minimize** F(x)F(x)F(x), we take the derivative and set it equal to zero: F′(x)=4−900x2=0F'(x) = 4 - \frac{900}{x^2} = 0F′(x)=4−x2900​=0.

STEP 12

Solving for xxx, we get 4x2=9004x^2 = 9004x2=900, so x2=225x^2 = 225x2=225, and x=15x = 15x=15 (since length must be positive).

STEP 13

Now, let's **find** yyy! y=90015=60y = \frac{900}{15} = 60y=15900​=60.The length of *each* pen parallel to the barn is y4=604=15\frac{y}{4} = \frac{60}{4} = 154y​=460​=15.

STEP 14

a. Dimensions to maximize area: 100 m\text{100 m}100 m perpendicular to the barn and 200 m\text{200 m}200 m parallel to the barn. b. Dimensions of each pen to minimize fencing: 15 m\text{15 m}15 m perpendicular to the barn and 15 m\text{15 m}15 m parallel to the barn.

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1. Maximize area with fixed fencing
2. Minimize fencing with fixed area

Let's define what we know!
We have $\text{400 m}$ of fencing for three sides.
Let $x$ be the length of each side perpendicular to the barn, and $y$ be the length of the side parallel to the barn.

We know that $2x + y = 400$ .
We want to maximize the area, which is $A = x \cdot y$ .

Let's rewrite the fencing equation as $y = 400 - 2x$ .
Now, we can substitute this into the area equation: $A(x) = x \cdot (400 - 2x) = 400x - 2x^2$ .

Now, let's find $y$ ! $y = 400 - 2(100) = 400 - 200 = 200$ .

The total fencing needed is $F = 4x + y$ .
We want to minimize this.

From the area equation, we have $y = \frac{4 \cdot 225}{x} = \frac{900}{x}$ .
Let's substitute this into the fencing equation: $F(x) = 4x + \frac{900}{x}$ .

To minimize $F(x)$ , we take the derivative and set it equal to zero: $F'(x) = 4 - \frac{900}{x^2} = 0$ .

Solving for $x$ , we get $4x^2 = 900$ , so $x^2 = 225$ , and $x = 15$ (since length must be positive).

Now, let's find $y$ ! $y = \frac{900}{15} = 60$ .
The length of each pen parallel to the barn is $\frac{y}{4} = \frac{60}{4} = 15$ .

a. Dimensions to maximize area: $\text{100 m}$ perpendicular to the barn and $\text{200 m}$ parallel to the barn. b. Dimensions of each pen to minimize fencing: $\text{15 m}$ perpendicular to the barn and $\text{15 m}$ parallel to the barn.