A wire 100 inches long is to be cut into two pieces. One of the pieces will be bent into the shape of a circle and the other into the shape of a equilateral triangle. How should the wire be cut so as to maximize the sum of the area of the areas of the circle and triangle will be maximized?
L=total length,
x=length of piece for triangle
So
L-x=circumference of circle
At(x)=area of triangle
=√3 *x²/4
Ac(x)=area of circle
=π((L-x)/(2π))²
=(L-x)²/(4π)
Total Area, A(x)
= At(x)+Ac(x)
To find the maximum/minimum,
Equate
A'(x)=0 and solve for x.
Find A(x) = max. area.
To maximize the sum of the areas of the circle and equilateral triangle, we need to determine the lengths of the wire that should be allocated to each shape.
Let's denote the length of wire allocated to the circle as x, and the remaining length allocated to the equilateral triangle as 100 - x.
For the circle, the circumference is equal to the length of wire allocated:
2πr = x
Solving for r, we have:
r = x / (2π)
For the equilateral triangle, the sum of the lengths of its sides is equal to the length of wire allocated:
3s = 100 - x
Solving for s, we have:
s = (100 - x) / 3
Now, let's express the area of the circle and equilateral triangle in terms of their respective formulas:
Area of circle = πr^2
Area of triangle = (sqrt(3) / 4) * s^2
Substituting the expressions for r and s obtained earlier:
Area of circle = π(x^2) / (4π^2)
Area of triangle = (sqrt(3) / 4) * ((100 - x)^2) / 9
To maximize the sum of these areas, we need to find the value of x that yields the maximum value for:
Area of circle + Area of triangle
Let's differentiate this expression with respect to x and set the derivative equal to zero to find the critical point:
(d / dx)((π(x^2) / (4π^2)) + ((sqrt(3) / 4) * ((100 - x)^2) / 9) = 0
This simplifies to:
(πx) / (2π^2) - (sqrt(3) / 2) * (100 - x) / 9 = 0
Simplifying further:
x / (2π) - (sqrt(3) / 2) * (100 - x) / 9 = 0
Solving for x, we find:
x = 100 / (4π + 9sqrt(3))
Now, we need to check if this critical point yields a maximum.
By taking the second derivative and evaluating it at x, we can determine the concavity of the function:
(d^2 / dx^2)((π(x^2) / (4π^2)) + ((sqrt(3) / 4) * ((100 - x)^2) / 9)
This simplifies to:
1 / (2π) + (sqrt(3) / 2) / 9
Since the second derivative is positive for all x, this indicates that the critical point is a minimum, not a maximum.
Therefore, to maximize the sum of the areas of the circle and equilateral triangle, we would need to allocate the entire 100-inch wire to the circle.
Alternatively, if we are only allowed to cut the wire once, we can calculate the value of x at which the sum of the areas is maximized.
Using the expression for x derived earlier, we find:
x = 100 / (4π + 9sqrt(3))
x ≈ 16.97 inches
So, the wire should be cut into two pieces of approximately 16.97 inches and 100 - 16.97 = 83.03 inches.
To maximize the sum of the areas of the circle and equilateral triangle, we need to figure out the optimal lengths for each piece of wire.
Let's denote the length of the wire used for the circle as "x" inches. Then the length of the wire used for the equilateral triangle would be (100 - x) inches.
The formula for the area of a circle is given by A_c = πr^2, where "r" is the radius of the circle. Since the circumference of a circle is given by C = 2πr, we can find the radius in terms of "x" by dividing x by 2π: r = x / (2π).
The formula for the area of an equilateral triangle is A_t = (√3/4) * s^2, where "s" is the side length of the triangle. Since the perimeter of an equilateral triangle is given by P = 3s, we can find the side length in terms of (100 - x) by dividing (100 - x) by 3: s = (100 - x) / 3.
Now, substitute the expressions for "r" and "s" into the area formulas:
A_c = π(x / (2π))^2 = x^2 / (4π)
A_t = (√3/4) * ((100 - x) / 3)^2 = (√3/4) * ((100 - x)^2 / 9)
The goal is to maximize the sum of these two areas, so we need to find the value of "x" that maximizes A_c + A_t.
Taking the derivative of (A_c + A_t) with respect to "x" and setting it equal to zero will give us the critical points where the function reaches a maximum. After solving for "x", we can substitute it back into the area formulas to find the maximum areas.
Differentiating (A_c + A_t) with respect to "x":
d/dx (A_c + A_t) = d/dx (x^2 / (4π)) + d/dx ((√3/4) * ((100 - x)^2 / 9))
To simplify the equation, we can calculate the derivatives separately:
d/dx (x^2 / (4π)) = (2x) / (4π) = x / (2π)
d/dx ((√3/4) * ((100 - x)^2 / 9)) = (√3/4) * (2(100 - x)(-1) / 9) = (√3/4) * (200 - 2x) / 9
Setting the derivative equal to zero:
x / (2π) + (√3/4) * (200 - 2x) / 9 = 0
To solve this equation, we can simplify and isolate "x":
9x / (2π) + (√3/2) * (200 - 2x) = 0
9x + (π√3)(200 - 2x) = 0
9x + (π√3)(200) - (π√3)(2x) = 0
7x = (π√3)(200)
x = (π√3)(200) / 7
Now that we have the value of "x", we can substitute it back into the area formulas to find the maximum areas:
A_c = ((π√3)(200) / 7)^2 / (4π)
A_t = (√3/4) * ((100 - (π√3)(200) / 7) / 3)^2
Calculating these values will give you the maximum areas for the circle and equilateral triangle.