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

Answer 1

#38.4 or 21.6 cm# for a minimum area of #80.4 cm^2#
The limit as #x -> 0# is #225 cm^2#

The total length is 60. The first rectangle has dimensions of #x xx 5x#, and the perimeter is thus #2(x + 5x) = 12x#. The remaining length for the square is then #60 – 12x#. This is equal to the square perimeter, #4y#, so we can express the square side ‘y’ in terms of ‘x’ as #4y = 60 – 12x# ; #y = 15 – 3x#. The square area in terms of ‘x’ is thus #(15 – 3x)^2# = #9x^2 – 90x +225#.
Combined with the area of the rectangle we obtain the total area as: #A_t = 5x^2 + 9x^2 – 90x +225# = #14x^2 – 90x + 225# This quadratic indicates that it will only have one extrema – the minimum for a given length of wire. That point is where the first derivative of the area quadratic equation is zero. #f’(x) = 28x – 90# Minimum at #x = 3.2#
The derived dimensions follow: Rectangle short side: #x = 3.2cm# Rectangle long side: #5x = 16cm# Rectangle area: #51.2 cm^2# Square side: #y = 15 – 3x = 5.4# Square area: #29.2 cm^2# Total Area: #80.4 cm^2# Total lengths: #2(3.2 + 16) + 4(5.4) = 38.4 + 21.6 = 60# Correct. Cut the wire at a length of #60 – 12x = 60 – 38.4 = 21.6cm or 38.4cm#
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Answer 2

Max Area: #"225 cm"^3#
Min Area: #"80.36 cm"^3#

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

    #A' = 28x-90#

    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 #"225 cm"^3# and the minimum value we have is #"80.36 cm"^3#.

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Answer 3

To minimize the total area, the wire should be cut into two equal parts, each measuring 30 cm.

To maximize the total area, the wire should be cut into two unequal parts, with one part measuring 45 cm and the other measuring 15 cm.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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