A wire 60 cm long is to be cut into two pieces. One of the pieces will be bent into the shape of a square and the other into the shape of an equilateral triangle, as shown in the diagram below:
a diagram showing a 60 cm wire cut into two pieces of length x and 60 minus x respectively. The wire of length x is bent into a square with side length x over 4 cm. The wire of length 60 minus x is bent into an equilateral triangle with sides of length 60 minus x over 3 cm
The wire is to be cut in order that the sum of the areas of the square and the triangle is to be a maximum. An equation that can be used to model the sum of the areas is A at x equals the sum of x squared over 16 and root 3 times quantity 60 minus x squared, over 18. Determine the boundaries and the corresponding areas. You need not solve further.
4. A cylindrical tank is to have a capacity of 1000 m³. It is to fit into a foundry that is 12 m wide with a height of 11 m.
f(x) = x^2 /16 + 3^.5 (60-x)^2 / 18
min or max when derivative = 0
f'(x) = 0 = (1/8) x + sqrt 3 * (60-x) (-1) /18
0 = x/8 - sqrt 3 (60/18) + sqrt 3 ( /18) x
x(sqrt 3 / 18 + 1/8) = (30/9) (sqrt 3)
1000 = h * pi D^2/4
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 triangle is minimum.
A wire 60cm long is to be cut into two pieces, one of which is to be bent into the shape of a circle 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 circle and equilateral triangle is minimum.
could you explain the first problem in a different way
recall that the area of an equilateral triangle of side s is √3/4 s^2. So, s = (60-x)/3
the side of a square of perimeter x is x/4, so the area is x^2/16