A pieces of wire 10cm long is cut in to two pieces, one pc. is bent into a square and the other is bent into an equalateral triangle.How shoul the wire be cut so that the total enclose area is:a)...
A pieces of wire 10cm long is cut in to two pieces, one pc. is bent into a square and the other is bent into an equalateral triangle.
How shoul the wire be cut so that the total enclose area is:
You need to cut the wire into two pieces whose lengths are not given. You need to consider these lengths as `3x` and `4y` .
The problem provides the information that the total length of wire is of 10 cm such that:
`3x+4y = 10`
Consider that you need to use the length of `3x` to form the equilateral triangle and the length of `4y` to form the square.
You need to evaluate the area of equilateral triangle using the formula:
`A_(Delta) = l^2sqrt3/4`
Since the length of side of triangle is of `3x/3 = x` yields:
You need to evaluate the area of the square using the formula:
`A_(square) = l^2`
Since the length of side of square is of `4y/4 =y` yields:
`A_(square) = y^2`
The area that needs to be optimized is:
`A = A_(Delta) + A_(square)`
`A = x^2sqrt3/4 + y^2`
You should write this equation in term of one variable, hence, you need to use the relation `x+y=10` to write y in terms of x such that:
`y = 10-x`
Substituting `10-x` for y in equation of area yields:
`A = x^2sqrt3/4 + (10-x)^2`
`A = x^2sqrt3/4 + 100 - 20x + x^2`
You should remember that the value of area is maximum or minimum at `A'(x) = 0` , hence, you need to differentiate the equation `A = x^2sqrt3/4 + 100 - 20x + x^2` with respect to x such that:
`A'(x) = xsqrt3/2 - 20 + 2x`
You need to solve the equation `A'(x) = 0` such that:
`xsqrt3/2 - 20 + 2x = 0 => 4x + xsqrt3 = 40`
Factoring out x yields:
`x(4 + sqrt3) = 40 => x = 40/(4 + sqrt3)`
Hence, you may find y substituting `40/(4 + sqrt3)` for x in equation `y = 10- x` such that:
`y = 10 - 40/(4 + sqrt3) => y = (40 + 10sqrt3 - 40)/(4 + sqrt3)`
`y = 10sqrt3/(4 + sqrt3)`
Since, evaluating the second derivative, yields positive, then the found values `x = 40/(4 + sqrt3)` and `y = 10sqrt3/(4 + sqrt3)` minimize the area.
The problem with the above answer (by sciecesolve) is that with the above values of x and y we need a wire length other than 10cm.
because length of wire = 3x+4y
Substituting the calculated values of x and y i.e. 40/(4+sqrt(3)) and 10*sqrt(3)/(4+sqrt(3)) we get:
Length of wire = 3*40/(4+sqrt(3))+4*10*sqrt(3)/(4+sqrt(3))
=> (120+40*sqrt(3))/(4+sqrt(3)) cm > 10cm
Thus the length of wire required according to the calculated values of x and y axceed the given length of wire 10cm.
In fact the areas of regular shapes increase for the same length of wire in sequence of an equilateral triangle, square, regular pentagon, regular hexagon, regular heptagon, regular octagon and so on till we get a circle.
The area of equilateral triangle is smallest and that of a circle is largest for the same perimeter.
Maximum or minimum area of a combination of two shapes cannot be achieved unless one of the shapes has a zero area with the same perimeter length.
The area is maximum when the wire is cut in length 10cm for square and Zero cm for triangle.
The maximum area is (10/4)^2+0 = 25/4cm^2 or 6.25cm^2
The area is minimum when the wire is cut in length 0cm for square and 10cm for triangle.
The minimum area is 0+sqrt(3)*(10/3)^2/4 = sqrt(3)*25/9cm^2
Note: area of an equilateral triangle
thank you so much..
I can not give any reference off the cuff but it is also true in case of volume which increases for a fixed surface area in sequence of tetrahedron, cube and sphere. I shall let you know the reference if I found one.
Thank you for you help. If you will mind, can I ask some reference in solving this kind of problem?? thanks..