# A piece of wire 60 cm in length is cut into two pieces. The first piece forms a rectangle 5 times as wide as it is long. The second piece forms a square. Where should the wire be cut to? 1)minimize the total area 2)maximize the total area

##### 2 Answers

Max Area:

Min Area:

#### Explanation:

Steps:

**1) Draw out a rectangle and a square**

**2) Label all your variables based on the information from the question**

- Since the length of the rectangle is five times as the width. Therefore, we know that
#5x# is the length and#x# is the width. - Since it is a square and we do not know the values of the sides, we can label the sides as
#y# .

**3) Find the helper equation for the equation which you are trying to optimize**

- The wire
#"60 cm"# is the circumference for both the rectangle and the square #12x+4y=60 #

**4) Find the equation for #y# of the helper equation in terms of #x#**

#y = 15-3x#

**5) Find the equation that you are trying to optimize**

#"Area" = AR + AS = 5x^2 + y^2#

**6) Plug the equation of #y# into the equation you are trying to optimize**

#5x^2 + y^2# #5x^2 +(15-3x)^2 = 14x^2-90x+225#

**7) Take the derivative of the equation and find the critical point(s) while determine whether they are max or min**

**8) Check the endpoints!**

Endpoints indicate the maximum and minimum value you can have for the rectangle

#12x = 60, x = 5# <-- indicating all material used to create the rectangle#x = 0 # <-- indicating all material used to create the square

**9) Find the answers by plugging in endpoints and critical number into the original area equation**

#A(3.2143) = "80.36 cm"^3# #A(0) = "225 cm"^3# #A(5) = "125 cm"^3#

Therefore, the maximum of the total area we have is

The limit as

#### Explanation:

The total length is 60. The first rectangle has dimensions of

Combined with the area of the rectangle we obtain the total area as:

The derived dimensions follow:

Rectangle short side:

Rectangle long side:

Rectangle area:

Square side:

Square area:

Total Area:

Total lengths: