Question

Please help me with my pre calc homework, it is attached as a screenshot below:

Answer 1

The Maximum Area of the fence is 10000 square feet

SOLUTION

Problem Statement

The question tells us the fence is made of 400 feet of material and we are asked to calculate the maximum area that can be enclosed by the fencing.

Method

- Let the length and breadth of the rectangle be x and b. Thus, we can write out an expression for the perimeter of the fence as follows:

`\begin{gathered} 2(x+b)=400 \\ \text{Divide both sides by 2} \\ x+b=(400)/(2)=200 \\ \\ \therefore x+b=200\text{ (Equation 1)} \end{gathered}`

- Also, we can write out the Area enclosed by the fence as follows:

`\begin{gathered} A=x* b \\ A=xb\text{ (Equation 2)} \end{gathered}`

Implementation

- For this stage, we shall work on the two Equations we arrived at during our analysis in the Method stage.

- We shall try to write the function of the Area in terms of x. Once this is done, we shall proceed to differentiate this function of A with respect to x.

- The differentiation describes how the Area of the plot surrounded by the fence changes with its length x.

- At maximum Area, this change will be equal to zero because any other change would be a decrease in the area of the plot.

- We shall complete the solution by finding the value of x for which the Area is maximum, finding the corresponding value of breadth, b, and then finding the Maximum Area of the plot using the formula for the Area given above.

- We can break this process down into the following steps:

1. Find the Area function in terms of x alone

2. Differentiate the Area function.

3. Equate to zero and find x.

4. Find the value of b and the Maximum Area.

Now, we can solve as follows:

1. Find the Area function in terms of x alone

`\begin{gathered} x+b=200\text{ (Equation 1)} \\ A=xb\text{ (Equation 2)} \\ \\ \text{From Equation 1,} \\ b=200-x \\ Substitute\text{ this expression for b into Equation 2} \\ \\ A=x(200-x) \\ A=200x-x^2 \end{gathered}`

2. Differentiate the Area function

`\begin{gathered} A=200x-x^2 \\ \text{Differentiating A with respect to x, we have:} \\ \mathrm{(dA)/(dx)}=\mathrm{(d)/(dx)}(200x-x^2) \\ \\ \mathrm{(dA)/(dx)}=\mathrm{(d)/(dx)}(200x)-\mathrm{(d)/(dx)}(x^2) \\ \\ \mathrm{(dA)/(dx)}=200-2x \end{gathered}`

3. Equate to zero and find x:

`\begin{gathered} \mathrm{(dA)/(dx)}=200-2x \\ At\text{ maximum Area, }\mathrm{(dA)/(dx)}=0 \\ \\ 200-2x=0 \\ \therefore2x=200 \\ \text{Divide both sides by }2 \\ (2x)/(2)=(200)/(2) \\ x=100 \end{gathered}`

4. Find the value of b and the Maximum Area:

`\begin{gathered} \text{ We find the value of b by substituting the value of }x\text{ into the Perimeter equation, Equation 1.} \\ x+b=200 \\ 100+b=200 \\ \text{subtract 100 from both sides} \\ b=200-100 \\ \therefore b=100 \\ \\ \text{Thus, the maximum area is:} \\ A=xb=100*100 \\ \therefore A=10000 \end{gathered}`

Final Answer

The Maximum Area of the fence is 10000 square feet