Math / Geometry

QuestionA farmer wants to fence in a rectangular plot of land adjacent to the north wall of his barn. No fencing is needed along the barn, and the fencing along the west side of the plot is shared with a neighbor who will split the cost of that portion of the fence. If the fencing costs $\$ 10$ per linear foot to install and the farmer is not willing to spend more than $\$ 7000$ , find the dimensions for the plot that would enclose the most area. (Enter the dimensions as a comma separated list.)

Studdy Solution

STEP 1

What is this asking? We need to figure out the length and width of a rectangular fence that gives the biggest area, but there's a limited budget, and one side doesn't need a fence, and the neighbor is helping pay for another side! Watch out! Don't forget about the shared side and the free wall when calculating the total fence cost!

STEP 2

1. Set up the cost equation.
2. Express width in terms of length.
3. Define the area function.
4. Find the length that maximizes the area.
5. Calculate the width.

STEP 3

Let's call the length of the fence parallel to the barn $l$ and the width of the fence perpendicular to the barn $w$ .
Since the barn acts as one side of the rectangle, we only need fencing for three sides.

STEP 4

The west side costs $\$$ 10 per foot and is shared with the neighbor, so the farmer pays $\$10 \cdot \frac{1}{2} = \$5$ per foot for this side ( $w$ feet).
The cost for this side is $5w$ .

STEP 5

The east side costs $\$10$ per foot and has length $w$ feet, so the cost is $10w$ .

STEP 6

The south side costs $\$10$ per foot and has length $l$ feet, so the cost is $10l$ .

STEP 7

The total cost is the sum of the costs of the three sides: $5w + 10w + 10l$ , and this must be equal to $\$7000$ .
So, $15w + 10l = 7000$ .

STEP 8

We can simplify the cost equation from $15w + 10l = 7000$ by dividing both sides by 5 to get $3w + 2l = 1400$ .

STEP 9

Now, let's solve for $w$ in terms of $l$ : $3w = 1400 - 2l$ , so $w = \frac{1400 - 2l}{3}$ .

STEP 10

The area of the rectangular plot is given by $A = l \cdot w$ .

STEP 11

Substituting the expression for $w$ from the previous step, we get $A(l) = l \cdot \frac{1400 - 2l}{3} = \frac{1400l - 2l^2}{3}$ .

STEP 12

To maximize the area, we can find the vertex of the parabola represented by the area function.
The x-coordinate (in our case, $l$ ) of the vertex of a parabola in the form $f(x) = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$ .

STEP 13

In our area function, $a = -\frac{2}{3}$ and $b = \frac{1400}{3}$ .
So, $l = -\frac{\frac{1400}{3}}{2 \cdot -\frac{2}{3}} = -\frac{1400}{3} \cdot -\frac{3}{4} = \frac{1400}{4} = 350$ .

STEP 14

Therefore, the length that maximizes the area is 350 feet.

STEP 15

Now we can plug the optimal length $l = 350$ back into the equation for $w$ : $w = \frac{1400 - 2 \cdot 350}{3} = \frac{1400 - 700}{3} = \frac{700}{3}$ .

STEP 16

So, the width that maximizes the area is $\frac{700}{3}$ feet.

STEP 17

The dimensions that maximize the area are 350 feet by 700/3 feet.

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Set up the cost equation.
2. Express width in terms of length.
3. Define the area function.
4. Find the length that maximizes the area.
5. Calculate the width.

STEP 3

Let's call the length of the fence parallel to the barn $l$ and the width of the fence perpendicular to the barn $w$ .
Since the barn acts as one side of the rectangle, we only need fencing for three sides.

STEP 4

The west side costs $\$$ 10 per foot and is shared with the neighbor, so the farmer pays $\$10 \cdot \frac{1}{2} = \$5$ per foot for this side ( $w$ feet).
The cost for this side is $5w$ .

STEP 5

The east side costs $\$10$ per foot and has length $w$ feet, so the cost is $10w$ .

STEP 6

The south side costs $\$10$ per foot and has length $l$ feet, so the cost is $10l$ .

STEP 7

The total cost is the sum of the costs of the three sides: $5w + 10w + 10l$ , and this must be equal to $\$7000$ .
So, $15w + 10l = 7000$ .

STEP 8

We can simplify the cost equation from $15w + 10l = 7000$ by dividing both sides by 5 to get $3w + 2l = 1400$ .

STEP 9

Now, let's solve for $w$ in terms of $l$ : $3w = 1400 - 2l$ , so $w = \frac{1400 - 2l}{3}$ .

STEP 10

The area of the rectangular plot is given by $A = l \cdot w$ .

STEP 11

Substituting the expression for $w$ from the previous step, we get $A(l) = l \cdot \frac{1400 - 2l}{3} = \frac{1400l - 2l^2}{3}$ .

STEP 12

To maximize the area, we can find the vertex of the parabola represented by the area function.
The x-coordinate (in our case, $l$ ) of the vertex of a parabola in the form $f(x) = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$ .

STEP 13

In our area function, $a = -\frac{2}{3}$ and $b = \frac{1400}{3}$ .
So, $l = -\frac{\frac{1400}{3}}{2 \cdot -\frac{2}{3}} = -\frac{1400}{3} \cdot -\frac{3}{4} = \frac{1400}{4} = 350$ .

STEP 14

Therefore, the length that maximizes the area is 350 feet.

STEP 15

Now we can plug the optimal length $l = 350$ back into the equation for $w$ : $w = \frac{1400 - 2 \cdot 350}{3} = \frac{1400 - 700}{3} = \frac{700}{3}$ .

STEP 16

So, the width that maximizes the area is $\frac{700}{3}$ feet.

STEP 17

The dimensions that maximize the area are 350 feet by 700/3 feet.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Set up the cost equation.2. Express width in terms of length.3. Define the area function.4. Find the length that maximizes the area.5. Calculate the width.

STEP 3

Let's call the length of the fence parallel to the barn lll and the width of the fence perpendicular to the barn www.Since the barn acts as one side of the rectangle, we only need fencing for three sides.

STEP 4

The west side costs $\$$10 per foot and is shared with the neighbor, so the farmer pays $10⋅12=$5\$10 \cdot \frac{1}{2} = \$5$10⋅21​=$5 per foot for this side (www feet).The cost for this side is 5w5w5w.

STEP 5

The east side costs $10\$10$10 per foot and has length www feet, so the cost is 10w10w10w.

STEP 6

The south side costs $10\$10$10 per foot and has length lll feet, so the cost is 10l10l10l.

STEP 7

The **total cost** is the sum of the costs of the three sides: 5w+10w+10l5w + 10w + 10l5w+10w+10l, and this must be equal to $7000\$7000$7000.So, 15w+10l=700015w + 10l = 700015w+10l=7000.

STEP 8

We can simplify the **cost equation** from 15w+10l=700015w + 10l = 700015w+10l=7000 by dividing both sides by 5 to get 3w+2l=14003w + 2l = 14003w+2l=1400.

STEP 9

Now, let's solve for www in terms of lll: 3w=1400−2l3w = 1400 - 2l3w=1400−2l, so w=1400−2l3w = \frac{1400 - 2l}{3}w=31400−2l​.

STEP 10

The **area** of the rectangular plot is given by A=l⋅wA = l \cdot wA=l⋅w.

STEP 11

Substituting the expression for www from the previous step, we get A(l)=l⋅1400−2l3=1400l−2l23A(l) = l \cdot \frac{1400 - 2l}{3} = \frac{1400l - 2l^2}{3}A(l)=l⋅31400−2l​=31400l−2l2​.

STEP 12

To **maximize** the area, we can find the vertex of the parabola represented by the area function.The x-coordinate (in our case, lll) of the vertex of a parabola in the form f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c is given by x=−b2ax = -\frac{b}{2a}x=−2ab​.

STEP 13

STEP 14

Therefore, the length that maximizes the area is **350 feet**.

STEP 15

Now we can plug the **optimal length** l=350l = 350l=350 back into the equation for www: w=1400−2⋅3503=1400−7003=7003w = \frac{1400 - 2 \cdot 350}{3} = \frac{1400 - 700}{3} = \frac{700}{3}w=31400−2⋅350​=31400−700​=3700​.

STEP 16

So, the width that maximizes the area is 7003\frac{700}{3}3700​ **feet**.

STEP 17

The dimensions that maximize the area are 350 feet by 700/3 feet.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Set up the cost equation.
2. Express width in terms of length.
3. Define the area function.
4. Find the length that maximizes the area.
5. Calculate the width.

Let's call the length of the fence parallel to the barn $l$ and the width of the fence perpendicular to the barn $w$ .
Since the barn acts as one side of the rectangle, we only need fencing for three sides.

The west side costs $\$$ 10 per foot and is shared with the neighbor, so the farmer pays $\$10 \cdot \frac{1}{2} = \$5$ per foot for this side ( $w$ feet).
The cost for this side is $5w$ .

The east side costs $\$10$ per foot and has length $w$ feet, so the cost is $10w$ .

The south side costs $\$10$ per foot and has length $l$ feet, so the cost is $10l$ .

The total cost is the sum of the costs of the three sides: $5w + 10w + 10l$ , and this must be equal to $\$7000$ .
So, $15w + 10l = 7000$ .

We can simplify the cost equation from $15w + 10l = 7000$ by dividing both sides by 5 to get $3w + 2l = 1400$ .

Now, let's solve for $w$ in terms of $l$ : $3w = 1400 - 2l$ , so $w = \frac{1400 - 2l}{3}$ .

The area of the rectangular plot is given by $A = l \cdot w$ .

Substituting the expression for $w$ from the previous step, we get $A(l) = l \cdot \frac{1400 - 2l}{3} = \frac{1400l - 2l^2}{3}$ .

To maximize the area, we can find the vertex of the parabola represented by the area function.
The x-coordinate (in our case, $l$ ) of the vertex of a parabola in the form $f(x) = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$ .

Therefore, the length that maximizes the area is 350 feet.

Now we can plug the optimal length $l = 350$ back into the equation for $w$ : $w = \frac{1400 - 2 \cdot 350}{3} = \frac{1400 - 700}{3} = \frac{700}{3}$ .

So, the width that maximizes the area is $\frac{700}{3}$ feet.