A rectangular field will be fenced on all four sides. Fencing for the north and south sides costs $5 per foot and fencing for the other two sides costs $9 per foot. What is the maximum area that can be enclosed for $6000?

6000=2(5x)+2(9x)

6000=10x+18x
6000=28x
214.3=x

214.3*214.3 = 45924.5 sq ft

A square gives the maximum area. So therefore the sides are equal(x).

Kerbie:

HOW THE SIDES ARE EQUAL SINCE IT IS A RECTANGULAR GARDEN?

Well, if I were you, I'd be tempted to build a fence around a field of infinite length. After all, you want to maximize that area, right? Just imagine all the endless possibilities – fields galore, all enclosed within your budget!

Unfortunately, the real world doesn't work that way. We have to deal with limitations, like limited funds and pesky physics. So, let's find a more reasonable solution.

Let's assume the length of the field is 'x' feet, and the width is 'y' feet.

The cost of fencing the north and south sides would be 2 * x * $5 = $10x.
The cost of fencing the other two sides would be 2 * y * $9 = $18y.

Given that the total cost should not exceed $6000, we can form the equation:

$10x + $18y ≤ $6000

Now, we need to find the maximum area. The area of a rectangle is simply A = x * y.

To make things easier, let's solve the first equation for 'y':

$18y ≤ $6000 - $10x
y ≤ ($6000 - $10x) / $18

Now, substitute this value for 'y' in the area equation:

A = x * (($6000 - $10x) / $18)

To keep things simple, let's multiply both sides of the equation by 18:

18A = x * ($6000 - $10x)

Alright, now we have a quadratic equation to work with:

18A = -10x² + $6000x

Let's find the maximum area by finding the vertex of this quadratic equation. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a = -10 and b = $6000.

x = -$6000 / (2 * -10)
x = $6000 / 20
x = $300

Plug this value back into the equation for 'y':

y = ($6000 - $10($300)) / $18
y = ($6000 - $3000) / $18
y = $3000 / $18
y ≈ 166.67

So, the maximum area that can be enclosed for $6000 is approximately 300 feet by 166.67 feet.

To find the maximum area that can be enclosed for $6000, we need to consider the cost of fencing for each side. Let's assume the length of the rectangular field is L and the width is W.

The north and south sides of the field each require two lengths of fencing, so the cost for the north and south sides would be 2L * $5/foot = $10L.

Similarly, the cost for the other two sides (east and west) would be 2W * $9/foot = $18W.

To calculate the total cost of fencing, we add the cost for the north and south sides with the cost for the east and west sides: $10L + $18W.

We are given that the total cost of fencing is $6000, so we can express this relationship as an equation: $10L + $18W = $6000.

To find the maximum area, we need to maximize the product of the length (L) and width (W) of the rectangular field.

To simplify the equation, let's divide both sides by 2: $5L + $9W = $3000.

Now, let's express one variable in terms of the other. Solving for L in terms of W, we get L = ($3000 - $9W) / $5.

Substituting this expression for L into the area formula, the area (A) can be expressed as A = L * W = (($3000 - $9W) / $5) * W.

To maximize the area, we can differentiate A with respect to W and set it equal to zero.

dA/dW = ($3000 - $9W) / $5 - ($9W) / $5 = ($3000 - $18W) / $5 = 0.

Simplifying the equation, we get $3000 - $18W = 0.

Solving for W, we find W = $3000 / $18 = 166.67 feet.

Now, we can substitute this value of W back into the area formula to find the maximum area:

A = (($3000 - $9W) / $5) * W = (($3000 - $9 * 166.67) / $5) * 166.67 = 4999.98 square feet.

Therefore, the maximum area that can be enclosed for $6000 is approximately 4999.98 square feet.