a wire 60cm long is to be cut into two pieces, one of which is to be bent into the shape of a square and the other of which is to be bent into an equilateral triangle. find how the wire should be cut so that the combined area of the square and equilateral is minimum

Question

a wire 60cm long is to be cut into two pieces, one of which is to be bent into the shape of a square and the other of which is to be bent into an equilateral triangle. find how the wire should be cut so that the combined area of the square and equilateral is minimum

Answer 1

If the square has side s and the triangle has side t, then

4s+3t = 60
The area is
a = s^2 + √3/2 t^2
Now, since t = 20 - 4/3 s,
a = s^2 + √3/2 (20 - 4/3 s)^2
da/ds = 2s - 16(15-s) / 3√3
so a is minimum when da/ds = 0

Answer 2

To find how the wire should be cut so that the combined area of the square and equilateral triangle is minimum, we need to consider the relationship between the lengths of sides for each shape.

Let's assume that the wire is cut into two pieces, one of length x and the other of length (60 - x). The piece of length x will be used to form a square, with each side measuring x/4, and the piece of length (60 - x) will be used to form an equilateral triangle, with each side measuring (60 - x)/3.

The area of the square is given by A_square = (x/4)^2 = x^2/16
The area of the equilateral triangle is given by A_triangle = (sqrt(3)/4)(60 - x)^2

Now, we can express the combined area, A_combined, as the sum of the areas of the square and triangle:
A_combined = A_square + A_triangle = x^2/16 + (sqrt(3)/4)(60 - x)^2

To find the value of x that minimizes the combined area, we can take the derivative of A_combined with respect to x, set it equal to zero, and solve for x:

dA_combined/dx = 0

Differentiating A_combined with respect to x:
dA_combined/dx = (2x)/16 - (sqrt(3)/4)(2)(60 - x)

Setting the derivative equal to zero:
2x/16 - (sqrt(3)/4)(120 - 2x) = 0

Simplifying and solving for x:
x/8 - (sqrt(3)/4)(120 - 2x) = 0
x/8 - (sqrt(3)/2)(60 - x) = 0
x/8 - (sqrt(3)/2)*60 + (sqrt(3)/2)*x = 0
(x/8) + ((sqrt(3)/2)*x) = (sqrt(3)/2)*60
x/8 + (sqrt(3)/2)*x = (sqrt(3)/2)*60
(8/8)x + (8sqrt(3)/2)*x = (8sqrt(3)/2)*60
x(1 + 4sqrt(3)) = 4sqrt(3)*60
x = (4sqrt(3)*60)/(1 + 4sqrt(3))

Now we can substitute this value of x back into the equation for A_combined to find the minimum combined area.