A piece of wire 14 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?
14 m this one I got right
(b) How much wire should be used for the square in order to minimize the total area?
I found this to be 6.09 m which is marked wrong and I have only one try left. Please help someone.
thank you!
I stand by mine and Damon's answer for a minimum area.
I even ran a simple 1980's style "BASIC" program to show it
10 a = 5000
20 for x = 0 to 14 step 1.000000E-05
30 y = (14-4*x)/3
40 na = x*x+sqr(3)/4*y*y
50 if na < a then a = na : side = x
60 next x
70 print side
80 end
>run
1.52238
my "a" is an arbitrarily large area , na is the new area
I used the area formula both you and Damon used.
So the part needed for the square is 4x
or 4(1.52238) = 6.08952
I also got 6.09 (actually 6.0895..)
I defined the side of the triangle slightly different from Damon to avoid as many fractions as I could.
let each side of the square be x m
let each side of the equilateral triangle be 2y m
that way I can say that the height of the triangle is √3y , using the ratios of the 1-√3-2 equilateral triangle.
4x + 6y = 14
2x + 3y = 7
x = (7-3y)/2
area = x^2 + (1/2)(2y)(√3y) = x^2 + √3y^2
= (7-3y)^2 / 4 + √3y^2
= (1/4)(7-3y)^2 + √3y^2
d(area)/dy =(-3/2)(7-3y) + 2√3y
= 0 for a max/min
2√3y = (3/2)(7-3y)
4√3y = 21 - 9y
4√3y + 9y = 21
y(4√3+9) = 21
y = 21/(4√3+9) = appr 1.3184
so 6y or 7.91 m should be used for the triangle and
14-7.91 or 6.09 m should be used for the square.
check my arithmetic
How about this then?
if the side of the square is x and the side of the triangle is y, then
4x + 3y = 14
and the total area is x^2 + y^2 √3/4
= x^2 + √3/4 (14-4x)/3)^2
= 1/18 (18x^2 - 2√3 x + 7√3)
just a parabola, with vertex at x = 1/(6√3)
So, the square uses 4/6√3 = 0.385m
Damon and Reiny are correct. I made a syntax error and did not check to see how wolframalpha interpreted my expression.
So, if the answer is marked wrong, there's a problem with the key.
A piece of wire is 36 cm long. What will be the length of each side of the wire used to form.1Q-An equilateral triangle.
Thank you both! But your answers were wrong.
square side = x
triangle side = y
perimeter = 4 x + 3 y = 14
area = x^2 + sqrt3 (y^2)/4
x = (14 - 3 y)/4
area
= (196 - 84 y +9y^2)/16 + 4 (3^.5)y^2/16
= [1/16] (196 - 84 y + 15.93 y^2)
dA/dy = 0 for max or min
0= -84 y + 31.86 y^2
so y = 0 or y = 2.63
so
x = 14/4 or 1.528
so
4 x = 14 or 6.11
we are pretty close to each other here.