Question

Find the smallest perimeter and the dimensions for a rectangle with an area of 36 in squared. The smallest perimeter for a rectangle with an area of 36 in squared is nothing in. ​(Simplify your​ answer.) The dimensions of this rectangle are nothing in. ​(Simplify your answers. Use a comma to separate​ answers.)

Answer 1

The smallest perimeter for a rectangle with an area of 36 square inches is 24 inches, and the dimensions that achieve this are those of a square, specifically 6 inches by 6 inches.

Step-by-step explanation:

The smallest perimeter of a rectangle with an area of 36 square inches is achieved when the rectangle is a square because the sides are equal, and a square minimizes the perimeter for a given area. The formula for the area of a square is area = side², so if we have an area of 36 square inches, the side length of the square is √36, which equals 6 inches. Using this side length, we can calculate the perimeter of the square, which is perimeter = 4 × side, giving us a perimeter of 6 inches × 4, which is 24 inches.

The smallest perimeter we can have is 24 inches, and the dimensions of the rectangle (in this case, a square) that give us this smallest perimeter are 6 inches by 6 inches.

Answer 2

The dimensions of the rectangle are:
`6√(2),3√(2)`

The smallest perimeter therefore will be:
`18√(2)`

Step-by-step explanation:

Let the dimension of the rectangle be x and y

Area of the Rectangle = xy

Area=36 Inch Squared

Therefore: xy=36

`x=(36)/(y)`

Perimeter of the Rectangle, P(x,y)=2(x+y)

Substituting
`x=(36)/(y)` into P(x,y)

`P(y)=2((36)/(y)+y)=(72+y^2)/(y)`

The minimum value of the perimeter occurs at the point where the derivative =0.

`P'(y)=(y^2-72)/(y^2)`

`Setting\:P'(y)=0\\(y^2-72)/(y^2)=0\\y^2-72=0\\y^2=72\\y=√(72)=6 √(2)`

Recall:

`x=(36)/(y)\\=(36)/(6√(2) )\\x=3√(2)`

The dimensions of the rectangle are:
`6√(2),3√(2)`

The smallest perimeter therefore will be:

`=2(6√(2)+3√(2)) \\=2(9√(2))\\=18√(2)`