Maximize Area of Rectangle w/ 2400ft Fence

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In summary, the problem is to find the dimensions of a rectangular field that has the largest area given a fixed amount of fencing and a straight river border. To solve this, we use the fact that the area of a rectangle is A = xy, and express A in terms of one variable by eliminating y. This leads to the function A(x) = 2400x - 2x^2, which we then maximize using the first derivative test to find the critical number x = 600. To confirm that this is a maximum, we use the second derivative test, which gives A''(x) = -4. Since the second derivative is negative, this confirms that x = 600 is a maximum, and we can solve
  • #1
smart_worker
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Homework Statement



A farmer has 2400 feet of fencing and want to fence of a
rectangular field that borders a straight river. He needs no fence along the river.
What are the dimensions of the field that has the largest area ?

Homework Equations



We wish to maximize the area A of the rectangle. Let x and y be the width and length of the rectangle (in feet). Then we express A in terms of x and y as A = xy.
We want to express A as a function of just one variable, so we eliminate y by expressing it in terms of x. To do this we use the given information that the total length of the fencing is 2400 ft. Therefore 2x + y = 2400
Hence y = 2400 − 2x and the area is A= x (2400 – 2x) = 2400 x − 2x2
Note that x ≥ 0 and x ≤ 1200 (otherwise A < 0). So the function that we
wish to maximize is
A (x) = 2400 x − 2x2, 0 ≤ x ≤ 1200.

The Attempt at a Solution



A′(x) = 2400 − 4x, so to find the critical numbers we solve the equation 2400 − 4x = 0 which gives x = 600. The maximum of A must occur either at this critical number or at an end point of the interval. Since A(0) = 0, A(600) = 7,20,000 and A(1200) = 0, thus the maximum value is A (600) = 720,000. When x = 600, y = 2400 − 1200 = 1200.but my teacher insists me to solve the problem using second derivative test

so,A''(x) = − 4

after this what should we do?
since the second derivative of x is negative so it is a local maximum
similarly the second derivative of y is also negative
so how to find the x and y values?
 
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  • #2
smart_worker said:
A′(x) = 2400 − 4x, so to find the critical numbers we solve the equation 2400 − 4x = 0 which gives x = 600. The maximum of A must occur either at this critical number or at an end point of the interval.
Note that ##A'(x)## has a single root, therefore it is either the maximum you are looking for, or there is no solution (except for ##x \rightarrow \pm\infty##). You therefore only need to show that ##x=600## is a maximum. Your teacher wants you to do that using ##A''(x)##. You obtain the final answer the same way you already have.
 

1. What is the formula for finding the maximum area of a rectangle with a given perimeter?

The formula for finding the maximum area of a rectangle with a given perimeter is A = (P/4)^2, where A is the area and P is the perimeter.

2. How do you solve for the dimensions of the rectangle with a given perimeter?

To solve for the dimensions, you can use the formula A = lw, where A is the area and l and w are the length and width of the rectangle. Plug in the perimeter for the value of P, and then solve for either l or w. Using that value, you can then solve for the other dimension.

3. What is the maximum area that can be achieved with a 2400ft fence?

The maximum area that can be achieved with a 2400ft fence is 600,000 square feet. This can be achieved by creating a square with sides of 600ft (600ft x 600ft = 600,000 sq ft).

4. How does the shape of the rectangle affect the maximum area that can be achieved with a 2400ft fence?

The shape of the rectangle can greatly affect the maximum area that can be achieved. A square will have the largest area, while a long and narrow rectangle will have a smaller area. It is important to find the balance between length and width to maximize the area.

5. Are there any real-life applications for maximizing the area of a rectangle with a given perimeter?

Yes, this concept can be applied in landscaping and gardening to create a garden or plant bed with the largest possible area within a given perimeter. It can also be used in construction to determine the largest possible floor area for a given amount of fencing material.

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