math 151 college algebra

Question

A rancher has 260 meters of fence with which to enclose three sides of a rectangular plot (the fourth side is a cliff wall and will not require fencing). Find the dimensions of the plot with the largest possible area. (For the purpose of this problem, the width will be the smaller dimension (needing two sides); the length with be the longer dimension (needing one side).)

length = meters

width = meters

What is the largest area possible for this plot?

area = meters-squared

Enter your answers as numbers. If necessary, round to the nearest hundredths.

Answer 1

260 = L +2W

L= 260-2W

A= LW = (260-2W)W = 260W-2W^2

A' = 260 -4W =0

4W = 260

W = 260/4 = 65 feet wide

L = 260-130 = 130 feet long

maximum area = 130x65 = 8450 square feet

Answer 2

For optimization problems, you have two equations. One is the constraint. The constraint is the equation they give that has to be kept true at all times. In the example of this problem, the rancher has 260 feet of fence, thus the constraint is the perimeter (minus the cliffside) has to be 260 feet.

You also have the equation that you're optimizing. This is going to be either getting a maximum or minimum of some variable.

Here, we're trying to maximize the area enclosed by the fence

Let's write the two equations down

Constraint: 2x + 2y - y = 260 (we subtract y since one side is the cliff)

Optimizing: A = xy

To solve these problems, make the optimization equation in terms of one variable. We can do that by solving for a variable in the constraints and plugging it in to the optimization equation

2x + 2y - y = 260

2x + y = 260

y = 260 - 2x

Plug this into the optimizing equation:

A = x (260 - 2x) = 260x - 2x²

Great, now to optimize this we just find an absolute maximum.

Set dA/dx equal to zero

dA/dx = 260 - 4x = 0

Solve for x.

Then use your constraint to solve for y.

Lastly, calculate the area with this x and y