A farmer wants to fence a small rectangular yard next to a barn. Fence for side parallel to the barn will cost 50 per foot and the fence for the other two sides will cost20 per foot. The farmer has a total of 2000 dollars to spend on the project. Find the dimensions for the yard that will have the largest possible area

A = x y

50 x + 20 (2y) = 2000
or
5 x + 4 y = 200

5 x + 4 A/x = 200

5 x^2 - 200 x - 4A = 0
where is vertex of parabola?
-b/2a = 200/10 =20
so x = 20
5(20) + 4 y = 200
4 y = 100
y = 25

To find the dimensions that will result in the largest possible area, we can set up an equation and maximize it using calculus. Let's denote the width of the yard as "x" and the length as "y."

Since we are given that the fence parallel to the barn costs $50 per foot and the other two sides cost $20 per foot, we can calculate the total cost of the fencing.

The cost of the fence parallel to the barn (the width) will be $50 multiplied by the length "x." The cost of the other two sides (the length) will be $20 multiplied by the sum of the width "x" and twice the length "y" because there are two of these sides. Hence, the total cost of the fencing can be expressed as:

C(x, y) = 50x + 20(2x + 2y)

We are given that the total budget for the project is $2000, so we can set up an equation:

50x + 20(2x + 2y) = 2000

Simplifying the equation further, we get:

50x + 40x + 40y = 2000
90x + 40y = 2000
9x + 4y = 200

Now, we want to find the dimensions that give the largest possible area. The area of a rectangle is given by length multiplied by width. In this case, area A(x, y) = xy.

To maximize the area, we can solve for "y" in terms of "x" using the equation 9x + 4y = 200:

4y = 200 - 9x
y = (200 - 9x)/4

Now, substitute this value of "y" into the area equation:

A(x) = x[(200 - 9x)/4]

Expanding the equation and simplifying, we get:

A(x) = (200x - 9x^2)/4

To maximize this function, we can take its derivative with respect to "x" and set it equal to zero:

A'(x) = (200 - 18x)/4 = 0

Solving for "x," we find:

200 - 18x = 0
18x = 200
x = 200/18
x ≈ 11.11

Substituting this value of "x" back into our equation for "y," we get:

y = (200 - 9(11.11))/4
y ≈ 10.42

Therefore, the dimensions that will result in the largest possible area for the given budget are approximately x ≈ 11.11 feet and y ≈ 10.42 feet.

To find the dimensions for the yard that will have the largest possible area, we can start by assigning variables to represent the unknown dimensions.

Let's say the length of the yard, parallel to the barn, is x feet, and the width, perpendicular to the barn, is y feet.

Since the farmer wants to fence three sides of the yard (two sides of length x and one side of length y), we can calculate the total cost of the fence.

The fence parallel to the barn will cost 50 per foot, so the cost of fencing those two sides will be 2x * 50 = 100x dollars.

The other two sides of the yard, perpendicular to the barn, will cost 20 per foot, so the cost of fencing those two sides will be 2y * 20 = 40y dollars.

The total cost of the fence is given as 2000 dollars, so we can write the equation:

100x + 40y = 2000

Now, we need to express the area of the yard in terms of x and y, and aim to maximize it.

The area of a rectangle is given by the formula A = length * width.

In this case, the length is x feet, and the width is y feet.

Therefore, the area of the yard (A) is given by:

A = x * y

Now, we can rewrite this equation in terms of one variable by solving the equation for either x or y.

Let's solve it for y:

100x + 40y = 2000

40y = 2000 - 100x

y = (2000 - 100x) / 40

Now, we can substitute this expression for y into the area formula:

A = x * ((2000 - 100x) / 40)

Simplifying this equation, we get:

A = (1/40) * (2000x - 100x^2)

To find the dimensions that will maximize the area, we need to find the maximum value of A by finding the critical points of the equation. We can do this by taking the derivative of A with respect to x, setting it equal to zero, and solving for x.

dA/dx = (1/40) * (2000 - 200x) = 0

2000 - 200x = 0

200x = 2000

x = 2000 / 200

x = 10

Now, we can substitute this value of x back into the equation for y to find the corresponding value:

y = (2000 - 100x) / 40

y = (2000 - 100*10) / 40

y = (2000 - 1000) / 40

y = 1000 / 40

y = 25

Therefore, the dimensions of the yard that will have the largest possible area are 10 feet by 25 feet.