Asked by Nane

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!

Answers

Answered by Damon
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.
Answered by Reiny
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
Answered by Nane
Thank you both! But your answers were wrong.
Answered by Steve
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
Answered by Reiny
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
Answered by Steve
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.
Answered by Artika
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.
There are no AI answers yet. The ability to request AI answers is coming soon!

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