3. (6 points) a rectangular field of 32 m² area is to be fenced off and then divided by a stone fence into two sections. the outer fence costs $10 per meter, while the special dividing stone fence costs $20 per meter. let x = height and y = length.
a. (1 point) identify the constraints using x and y.
b. (1 points) identify the cost function (in x and y) to be optimized.
c. (1 point) by eliminating y, write the function in terms of x.
d. (2 points) minimize the cost.
fully justify your answer and write the name of the theorem/test that you use.
e. (1 point) find the dimensions (x and y) of the field with the minimum cost.
include the minimum cost.
x=
y=
c=

Question

3. (6 points) a rectangular field of 32 m² area is to be fenced off and then divided by a stone fence into two sections. the outer fence costs $10 per meter, while the special dividing stone fence costs $20 per meter. let x = height and y = length.
 a. (1 point) identify the constraints using x and y.
 b. (1 points) identify the cost function (in x and y) to be optimized.
 c. (1 point) by eliminating y, write the function in terms of x.
 d. (2 points) minimize the cost.
 fully justify your answer and write the name of the theorem/test that you use.
 e. (1 point) find the dimensions (x and y) of the field with the minimum cost.
 include the minimum cost.
 x=
 y=
 c=

Accepted Answer

### a. Constraints
# Explanation:
## Step1: Area Constraint
The area of a rectangle is $A = xy$. Given $A = 32$, so the constraint is $xy=32$.
## Step2: Variable Domain
Since $x$ and $y$ represent dimensions (length and height), $x>0$ and $y > 0$.

# Answer:
The constraints are $xy = 32$, $x>0$, and $y>0$.

### b. Cost Function
# Explanation:
## Step1: Outer - Fence Cost
The outer - fence perimeter of a rectangle is $P_{outer}=2x + 2y$. The cost of the outer fence is $10(2x + 2y)=20x+20y$.
## Step2: Dividing - Fence Cost
The dividing fence has a length of $x$ (assuming it is parallel to the height), and its cost is $20x$.
## Step3: Total Cost
The total cost function $C(x,y)$ is the sum of the outer - fence cost and the dividing - fence cost. So $C(x,y)=20x + 20y+20x=40x + 20y$.

# Answer:
The cost function is $C(x,y)=40x + 20y$.

### c. Function in terms of $x$
# Explanation:
## Step1: Solve for $y$ from the constraint
From $xy = 32$, we get $y=\frac{32}{x}$.
## Step2: Substitute $y$ into the cost function
Substitute $y=\frac{32}{x}$ into $C(x,y)$. Then $C(x)=40x+20\times\frac{32}{x}=40x+\frac{640}{x}$, where $x > 0$.

# Answer:
The function in terms of $x$ is $C(x)=40x+\frac{640}{x}$, $x>0$.

### d. Minimize the cost
# Explanation:
## Step1: Find the first derivative
Differentiate $C(x)=40x+\frac{640}{x}$ with respect to $x$. Using the power rule $(x^n)^\prime=nx^{n - 1}$, $C^\prime(x)=40-\frac{640}{x^{2}}$.
## Step2: Set the first derivative equal to zero
Set $C^\prime(x)=0$, so $40-\frac{640}{x^{2}}=0$.
$$
\begin{align*}
40&=\frac{640}{x^{2}}\
x^{2}& = 16\
x&=4\quad(\text{since }x>0)
\end{align*}
$$
## Step3: Find the second derivative
Differentiate $C^\prime(x)$ with respect to $x$. $C^{\prime\prime}(x)=\frac{1280}{x^{3}}$.
## Step4: Use the second - derivative test
When $x = 4$, $C^{\prime\prime}(4)=\frac{1280}{4^{3}}=\frac{1280}{64}=20>0$. By the second - derivative test, the function $C(x)$ has a local minimum at $x = 4$.

# Answer:
The cost is minimized at $x = 4$. The theorem/test used is the second - derivative test.

### e. Dimensions and minimum cost
# Explanation:
## Step1: Find $y$
Since $xy=32$ and $x = 4$, then $y=\frac{32}{4}=8$.
## Step2: Find the minimum cost
Substitute $x = 4$ into $C(x)=40x+\frac{640}{x}$. $C(4)=40\times4+\frac{640}{4}=160 + 160=320$.

# Answer:
$x = 4$m, $y = 8$m, $C = 320$ dollars.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

a. Constraints

Step1: Area Constraint

Step2: Variable Domain

Step1: Outer - Fence Cost

Step2: Dividing - Fence Cost

Step3: Total Cost

Step1: Solve for \(y\) from the constraint

Step2: Substitute \(y\) into the cost function

Answer:

b. Cost Function