Maximize and Minimize Area of 2 rectangles

  • Thread starter Nitro171
  • Start date
  • Tags
    Area
In summary, a man has 340 yds of fencing for enclosing two separate fields, one of which is to be rectangular twice as long as it is wide and the other a square. The square field must contain at least 100 sq. yds and the rectangular field must contain at least 800 sq. yds. To find the maximum and minimum values of the width x of the rectangular field, we can set up an equation representing the total length of fencing used in the two fields and an equation representing the total area. By solving for one variable and substituting it into the area equation, we get a parabola that represents the total area as a function of a single variable. The constraints on the minimum sizes of the fields
  • #1
Nitro171
1
0
A man has 340 yds of fencing for enclosing two separate fields, one of which is to be rectangular twice as long as it is wide and the other a square. The square field must contain at least 100 sq. yds. And the rectangular one must contain at least 800 sq. yds. Find the maximum and the minimum values of the width x of the rectangular field. What is the greatest number of square yards that can be enclosed in the two fields?

What I have so far: (not too much)

A= bh

One rectangle is bh (area 100 sq. yards or more), the other us 2b x 2h (area 800 sq. yards or more).

bh+ (2b)(2h) = 340 (?)

I am really having trouble coming up with an equation and solving the problem in general. Any help would be greatly appreciated.
 
Physics news on Phys.org
  • #2
Nitro171 said:
A man has 340 yds of fencing for enclosing two separate fields, one of which is to be rectangular twice as long as it is wide and the other a square. The square field must contain at least 100 sq. yds. And the rectangular one must contain at least 800 sq. yds. Find the maximum and the minimum values of the width x of the rectangular field. What is the greatest number of square yards that can be enclosed in the two fields?

What I have so far: (not too much)

A= bh

One rectangle is bh (area 100 sq. yards or more), the other us 2b x 2h (area 800 sq. yards or more).
You are confusing the two fields. The square field is the one that has to contain at least 100 sq. yds. The rectangular one has to contain at least 800 sq. yds. Also, your dimensions are wrong for the rectangular field.
Nitro171 said:
bh+ (2b)(2h) = 340 (?)

I am really having trouble coming up with an equation and solving the problem in general. Any help would be greatly appreciated.

Draw a sketch of the two fields, and label the sides. The sketch doesn't have to be very accurate, since all you need are variables to represent the dimensions.

Identify your variables with descriptions of what they represent, like this:
Let w = the width of the rectangular field.
Let 2w = the length of the rectangular field.
Let s = the length of any side of the square field.

Now, write an equation that represents the total length of fencing used in the two fields.
Then, write an equation that represents the total area. This equation will involve w and s.

Solve the first equation for one variable, say w.
Substitute for w in the second equation to get the total area as a function of a single variable. The new equation will be that of a parabola. Use the constraints on the minimum sizes of the two fields to find the maximum area of both fields and the minimum area.
 

FAQ: Maximize and Minimize Area of 2 rectangles

What is the formula for finding the area of a rectangle?

The formula for finding the area of a rectangle is length x width.

How do you maximize the area of a rectangle?

In order to maximize the area of a rectangle, you must make sure that the length and width are equal. This will result in a square, which has the largest area for a given perimeter.

How do you minimize the area of a rectangle?

In order to minimize the area of a rectangle, you must make sure that one side is significantly longer than the other. This will result in a long and skinny rectangle, which has the smallest area for a given perimeter.

Can you have two rectangles with the same area?

Yes, it is possible to have two rectangles with the same area but different dimensions. For example, a square with sides of 4 units will have the same area as a rectangle with sides of 2 and 8 units.

What are some real-world applications of maximizing and minimizing the area of rectangles?

Maximizing and minimizing the area of rectangles is useful in many fields, such as architecture, engineering, and manufacturing. For example, architects may want to maximize the area of a building within a given plot of land, while manufacturers may want to minimize the amount of material needed to create a product.

Back
Top