applied calculus maximization question

Question

The owner of the Rancho Los Feliz has 2940 yd of fencing to enclose a rectangular piece of grazing land along the straight portion of a river and then subdivide it by means of a fence running parallel to the sides. No fencing is required along the river. (See the figure below.) What are the dimensions of the largest area that can be enclosed?

width yds

length yds

What is this area?

  

Answer 1

Call the two sides of the rectangle w and the side parallel to the river l.

Then we know that 2w + l = 2940. That's our contraint

We're trying to maximize the area.

From the constraint, solve for a variable.

l = 2940 - 2w.

The area equals l*w or (2940 - 2w)*w by substitution.

So, we're maximizing (2940 - 2w)*w

Take the derivative of this and set it equal to 0.

Solve for w.

Use the constraint to solve for l.

Answer 2

Let L = length of the side parallel to the river and W = length of each side perpendicular to the river.

If the area is subdivided by another length of fence perpendicular to the river, then:

L + 3W = 2940, then L = 2940 - 3W

A = area = LW = (2940 - 3W)(W) = -3W² + 2940W

The graph is a parabola opening downward and has maximum at the vertex (which is where dA/dW = 0).

dA/dW = -6W + 2940

dA/dW = 0 when W = 490 ft

The area is maximized when W = 490 ft and L = 2940 - 3(490) = 1470 ft.