A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (Round your answers to two decimal places.)

Answers

The maximum and the minimum total area inclosed are 6.25m² and 2.72m²

What is an area?

⇒ The area is the region bounded by the shape of an object.

Calculation:

Let x is the length of the piece of wire for the equilateral triangle, whose side will then be (x/3) m long.

⇒ (10 – x) = length of the remaining piece for the square, whose side will then be (10 – x)/4 m long.

h = height of the equilateral triangle = h = a×√3/2 = (x/3)×(√3/2)

(Where a is the side of the triangle)

A(x) = Total area of the square formed + Total area ofthe triangle formed

Then A(x) = [(10 – x)/4]²+ (1/2)(x/3)(x/6)√3

= [(100 – 20x + x²)/16] + (√3/36)x² .

The maximum total area enclosed :

⇒ The entire 10m length should be allocated to the square because a square produces more area per unit of its perimeter than does a triangle.

Thus if all square, then x=0 and A(0) = [2.5]² = 6.25 m² = Maximumarea

If all triangle, then x = 10 and A(10) = (1.732/36)(100) = 4.811 m² .

So, the maximum area occurs when it’s all used to make a square of side 2.5 m.

The minimum total area enclosed is :

⇒ We want a relatively small square and a small triangle.

We are going to Find x by setting the derivative of A(x) to zero.

d[ A(x)]/dx = [(-20 + 2x)/16] + (2√3/36)x = 0

= -5/4 + (1/8)x + (√3/18)x = 0

x = 5/[4 (1/8 + √3/18)] = 5/[ 4(0.125 + 0.09622)] = 5/(0.88489) ⇒5.65 m perimeter of the triangle

and (10 – x) = 4.35 = perimeter of square

And

A(5.65) = [4.35/4]² + (√3/36)(5.65)²

= 1.1827 from the square+1.5358 from the triangle

= 2.719m² total area = Minimum area

⇒ On rounding off it will be 2.72m²

Hence the maximum and the minimum areas are 6.25m² and 2.72m²

Learn more about the area here:

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Questions: Apiece of wire 10m long is cut into two pieces. one piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is a. a maximum? b. minimum?

maximumand theminimumtotal area inclosed are 6.25m² and 2.72m²What is an area?areais the regionboundedby the shape of an object.Calculation:lengthof the piece of wire for theequilateraltriangle, whose side will then be (x/3) m long.square, whose side will then be (10 – x)/4 m long.A(x)= Total area of thesquareformed + Total area oftriangleformedThe maximum total area enclosed :squarebecause a squareproducesmore area per unit of its perimeter than does a triangle.maximumarea occurs when it’s all used to make a square of side 2.5 m.The minimum total area enclosed is :square+1.5358 from thetriangleareahere:Disclaimer:The question was given incomplete on the portal. Here is the complete question.Questions:A