How much wire should be used for the square in order to minimize the total area?
A piece of wire 28 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.
(b) How much wire should be used for the square in order to minimize the total area?
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You should come up with the following notations: x expresses the length of side of square and y expresses the length of side of equilateral triangle.
You need to evaluate the perimeters of square and equilateral triangle such that:
`P square = 4x`
P triangle = `3y`
The problem provides the information that the total length of piece of wire is of 28 m such that:
`28 = 4x + 3y =gt y = (28-4x)/3:`
You need to evaluate the total area of square and triangle such that:
`A = A square + ` A triangle
`A = x^2 + y^2*sqrt3/4`
You need to write the function of area in terms of one variable, hence you may substitute `(28-4x)/3` for y in equation of total area such that:
`A(x) = x^2 + (28-4x)^2*sqrt3/36`
`A(x) = x^2 + 16(7-x)^2*sqrt3/36`
`A(x) = x^2 + 4(7-x)^2*sqrt3/9`
You need to differentiate the function A(x) with respect to x and then you need to solve the equation A'(x)=0 to find how much wire needs to be used for square to minimize the total area.
`A'(x) = 2x- 8sqrt3/9*(7-x)`
You need to solve the equation A'(x)=0 such that:
`2x - 8sqrt3/9*(7-x)= 0`
`2x- 56sqrt3/9 + 8sqrt3/9*x = 0`
`x(2 + 8sqrt3/9) = 56sqrt3/9 =gt x = (56sqrt3/9)/((18+8sqrt3)/9)`
`x = 56sqrt3/(18+8sqrt3) =gt x = 96.994/31.856`
`x = 3.044 m`
`P square = 4*3.044 =gt P square = 12.176 m`
Hence, evaluating the length of wire to be used for square to minimize the total area yields 12.176 meters.
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