A piece of wire 6 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. A piece of wire 6 m long is cut into two pieces. One piece is bent...
A piece of wire 6 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.
A piece of wire 6 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.
(a) How much wire should be used for the square in order to maximize the total area?
(b) How much wire should be used for the square in order to minimize the total area?
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a)The problem provides you the information that the piece of wire is cut into two pieces, a square and equilateral triangle.
You should come up with the substitutions for the perimeters of the equilateral triangle and perimeter of square such that:
x = perimeter of square
y = perimeter of equilateral triangle
The total length of 6 meters may be evaluated using the perimeters of equilateral triangle and the square such that:
`6 = x+y`
Notice that the length of the side of square is of `x/4 ` and the length of the side of equilateral triangle is of `y/3` .
You need to express the area of square...
(The entire section contains 346 words.)
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Well the let us assume that length of wire used for square is x and for the triangle is 6-x
then the area of the two shapes are going to be:
Square: (x/4)^2
Triangle: 1/2*a*b*sinC : ((sqrt3)/2)*(6-x)^2
and the total area will equal to the sum of these two equatons.
Then use differentiation to find the turning points of the curve which will be the max/min value of the curve.
Below is a graph using Wolfram showing the results
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