A rectangular field is to be enclosed by a fence. Two fences parallel to one side of the field divide the field into three rectangular fields. If 2400m of fence are available find the dimensions giving the max area.

Because there are 2 dividers, the perimeter = 2l + 4w, where l is the length of the field, and w is the width of the field (and the dividers)

2l + 4w = 2400

let x = l
2x + 4w = 2400
4w = 2400 - 2x
w = 600 - 0.5x

Now you can find the area.
A = l*w
A = x(600 - 0.5x)

Now you need to optimize area. A calculator is the easiest way, although if you are in Calculus you would likely have to show more work.

If you can use a graphing calculator, enter the equation in and find the maximum value on the graph.

I will demonstrate the calculus way below:
A = 600x - 0.5x^2
dA/dt = 600 - x

Find where dA/dt = 0

600 - x = 0
x = 600

dA/dt changes signs from + to - at x = 600, so there is a relative maximum at x = 600

Because x = length, the length is 600
w = 600 - 0.5x
w = 600 - 300
w = 300

The dimensions are 300m * 600m

let the width of the whole rectangle be x m (there will be 4 of these)

let the length be y m
then 4x + 2y = 2400
2x + y = 1200
y = 1200 - 2x

Area = xy
= x(1200-2x)
= -2x^2 + 1200x

Now, I don't know if you are studying Calculus.
If you do, then
d(Area)/dx = -4x + 1200
= 0 for a max area
x = 300

then y = 600
and the max area is (300(600) = 180000

If you don't know Calculus, complete the square on the above quadratic
you should end up with
Area = -2(x-300)^2 + 180000

Hey, we think alike, lol

Interesting...I did not think to complete the square to find the maximum area though. Based on steven's other questions I was not sure whether he was in Calculus.

a farmer has a rectangle field with an area of 3/4 square like the field is 1/2 mule wide what is the length of the field

Why did the scarecrow win an award? Because he was outstanding in his field!

To find the dimensions that give the maximum area, we need to use the concept of derivatives and optimization. Here's how you can solve this problem step by step:

1. Let's assume the length of the rectangular field is 'L' and the width is 'W'.

2. According to the problem statement, two fences parallel to one side of the field divide the field into three rectangular fields. This means that there are two smaller rectangular fields on either side of the main rectangular field.

3. The dimensions of the two smaller rectangular fields are the same, so let's call their width 'w'. Therefore, the length of each smaller rectangular field would be (L - 2w).

4. Now, we can calculate the total length of the fence required by adding up the lengths of each side of the rectangular field. Since there are four sides, the total length of the fence required is: 2L + 3w.

5. According to the problem statement, we have 2400m of fence available. Therefore, we can set up the equation: 2L + 3w = 2400.

6. Our goal is to maximize the area of the rectangular field. The area of a rectangle is given by the formula A = length * width. In this case, the area is A = L * W.

7. We can use the equation from step 5 to find an expression for L in terms of w. Rearranging the equation, we get: L = (2400 - 3w) / 2.

8. Substituting this expression for L into the area equation, we get: A = [(2400 - 3w) / 2] * W.

9. Simplifying the expression, we have: A = (2400W - 3wW) / 2.

10. To maximize the area, we need to take the derivative of the area equation with respect to w and set it equal to zero. This will give us a critical point where the maximum area occurs.

11. Differentiating the area equation with respect to w, we get: dA/dw = (2400W - 6w) / 2.

12. Setting the derivative equal to zero, we have: (2400W - 6w) / 2 = 0.

13. Solving for w, we get: 2400W - 6w = 0. Rearranging, we have: w = 400W.

14. Now, we substitute this value for w back into the equation we obtained in step 5 to find the corresponding value of L. Plugging in w = 400W, we get: 2L + 3(400W) = 2400.

15. Solving for L, we have: 2L + 1200W = 2400. Rearranging, we get: L = (2400 - 1200W) / 2.

16. Finally, we have expressions for both L and W in terms of a single variable, W. To find the values of L and W that give the maximum area, we can substitute these expressions into the area formula from step 6 and calculate the value of A.

17. Find the maximum value of A by taking the second derivative of the area equation with respect to W. If the second derivative is negative, it is a maximum.

18. Calculate the dimensions L and W that maximize the area.

By following these steps, you can find the dimensions that give the maximum area for the rectangular field given the available fence.