A rectangular pen with one side along a river will be bounded by a fence on the other three sites. If 600m of fence are available, what the dimensions of the rectangle should be in order to enclose maximum area and how much this area is?

x along river and y

600 = x + 2y so x = 600-2y

A = x y
A = (600-2y)y = 600y - 2y^2

I assume you do not know calculus so complete square to find vertex of parabola.

y^2-300 y = -A/2

y^2 -300 y+(300/2)^2 = -A/2 + 150^2

(y-150)^2 = -(1/2) (A-45,000)
vertex at y = 150 and A = 45,000
so
y = 150
x = 300
A = 45,000

To find the dimensions of the rectangle that will enclose the maximum area, we can use calculus. Let's assume the length of the rectangle is x, and the width is y.

We are given that there is one side along the river, which means the length of the fence used for the length of the rectangle is x. The other two sides will make up the width, so the length of fence used for the width is 2y.

The total fence used is x + 2y, and it must be equal to 600m:

x + 2y = 600

We can solve this equation for y:

2y = 600 - x
y = (600 - x)/2

The area of the rectangle is given by A = x * y. Substituting the expression for y we found:

A = x * (600 - x)/2
A = (600x - x^2)/2
A = 300x - (x^2)/2

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

dA/dx = 300 - x/2 = 0

Solving this equation for x:

x/2 = 300
x = 600

Substituting this value of x back into the equation for y:

y = (600 - 600)/2
y = 0

This implies that the width of the rectangle is 0, which does not make sense. Therefore, there is no rectangle that can enclose the maximum area with the given constraints.

In conclusion, it is not possible to enclose a rectangular pen with maximum area using the given amount of fence (600m) when one side is along the river.

To find the dimensions of the rectangle that will enclose the maximum area, we need to convert the problem into a mathematical equation. Let's assume the length of the rectangle is x (which will be along the river), and the width is y.

From the given information, we know that the fence will be on the other three sides, meaning there will be two equal lengths (y) and one length (x) along the river.

The perimeter of the rectangle can be expressed as:
P = x + 2y + y = x + 3y = 600m

We can rearrange this equation to solve for x in terms of y:
x = 600 - 3y

Now, let's express the area of the rectangle (A) in terms of x and y:
A = x * y

Substituting the value of x obtained above into the area equation:
A = (600 - 3y) * y

To maximize the area, we need to find the value of y that will make A as large as possible. For this, we can use calculus. We can take the derivative of A with respect to y, set it equal to zero, and solve for y:

dA/dy = (600 - 3y)y - (600 - 3y) * 1 = 0

Expanding and simplifying:
600y - 3y^2 - 600 + 3y = 0
-3y^2 + 603y - 600 = 0

Now, we can solve this quadratic equation for y. We can either factor it or use the quadratic formula:

Using the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / 2a

where a = -3, b = 603, and c = -600. Solving this equation will give us two possible values of y.

Once we have the value(s) of y, we can substitute it back into the equation for x:
x = 600 - 3y

This will give us the corresponding value(s) of x.

Finally, we can calculate the area by substituting the values of x and y into the area equation:
A = (600 - 3y) * y

The dimensions of the rectangle that will enclose the maximum area, as well as the corresponding maximum area, can be calculated using these steps.

length of side parallel to river --- y

length of the other two equal sides --- x
2x + y = 600 ---> y = 600-2x

area = xy
= x(600-2x)
= -2x^2 + 600x

this is a downwards opening parabola.
Find the vertex by whatever means you were taught to find the vertex, and extract your answer from that.