Question

My friend has $1044 to spend on a fence for her rectangular garden. She wants to use cedar fencing which costs $20/ meter on one side, and cheaper metal fencing which costs $9/ meter for the other three sides. What are the dimensions of the garden with the largest area she can enclose? length for cedar side = meter width for other side = What is the largest area that can be enclosed? area = meter-squared If necessary, round your answers accurate to two decimal

Answer 1

To maximize the area of the rectangular garden, we can set up equations based on the cost of the fencing and the area of the rectangle. By taking the derivative of the area equation, we can find the values of L and W that maximize the area. The largest area that can be enclosed is approximately 340.84 square meters.

Step-by-step explanation:

To maximize the area of the rectangular garden, we need to find the dimensions that maximize the area. Let's assume the length of the garden is L meters and the width is W meters. We know that one side of the garden will use cedar fencing which costs $20 per meter, so the cost of that side will be 20L dollars. The other three sides will use metal fencing which costs $9 per meter, so the cost of those sides will be 3(9W) dollars. The total cost of the fence is given as $1044, so we can set up the equation:

20L + 3(9W) = 1044

Next, we need to express the area of the rectangle in terms of L and W. The area of a rectangle is given by A = LW. We want to maximize the area, so we can use the equation:

A = LW

Now we have two equations with two variables. We can solve the first equation for L and substitute it into the second equation:

20L + 27W = 1044

L = (1044 - 27W) / 20

A = (1044 - 27W) / 20 * W

To find the dimensions that maximize the area, we can take the derivative of A with respect to W and set it equal to zero:

dA/dW = (1044 - 27W) / 20 - 27W / 20 = 0

1044 - 27W - 27W = 0

54W = 1044

W = 19.33 (rounded to two decimal places)

ubstituting this value of W back into the equation for L:

L = (1044 - 27(19.33)) / 20 = 18.46 (rounded to two decimal places)

So, the dimensions of the garden with the largest area she can enclose are approximately 18.46 meters by 19.33 meters.

To find the largest area, we can substitute the values of L and W into the equation for A:

A = (1044 - 27(19.33)) / 20 * 19.33 = 340.84 (rounded to two decimal places)

Therefore, the largest area that can be enclosed is approximately 340.84 square meters.

Answer 2

Your friend can enclose the largest area with a garden measuring:

Length (cedar side): 16.5 meters

Width (metal side): 31.5 meters

This results in a maximum area of:

Area: 521.25 square meters

Step-by-step explanation:

Define Variables:

Let x be the length of the garden (cedar side).

Let y be the width of the garden (metal side).

Cost Constraint:

The total cost of the fence should be $1044.

Express the cost as an equation: 20x + 9(2y) = 1044

Simplify: 20x + 18y = 1044

Maximize Area:

The area of the garden is A = xy.

We want to maximize A subject to the cost constraint.

Solve for y:

From the cost equation, we can express y in terms of x: y = (1044 - 20x) / 18

Substitute this into the area equation: A(x) = x * ((1044 - 20x) / 18)

Find Maximum Area:

To find the maximum area, we need to find the maximum value of A(x).

This can be done by taking the derivative of A(x) with respect to x and setting it equal to 0: d(A)/dx = 0

Solving the resulting equation for x gives the optimal length x ≈ 16.5 meters.

Substitute this back into the equation for y to find the optimal width y ≈ 31.5 meters.

Calculate Maximum Area:

Plug the optimal dimensions into the area equation: A ≈ 16.5 * 31.5 ≈ 521.25 square meters.

Therefore, your friend can enclose the largest area of 521.25 square meters with a garden measuring 16.5 meters by 31.5 meters, using cedar fencing for one side and metal fencing for the other three sides.