A wire 100 centimeters in length is cut into two separate pieces (not necessarily equal). ON epiece is bent to form a square and the other to form a circle. Let x equal the length of the wire used to form the square.
a. Write the function that represents the are of the two figures combined.
b. Determine the domain of the function.
c. Find the value(s) of x that yield a max and minimum area.
d. explain your reasoning.
let me say right away, that defining x to be the length of the wire used for the square is a poor choice. But ...
let the piece used for the square be x
then the piece used for the circle = 100-x
side of square = x/4
area of square = x^2 /16
perimeter of circle = 100-x
perimiter of circle = 2πr
2πr = 100-x
r = (100-x)/(2π)
area of circle = π(100-x)^2/(4π^2) = (100-x)^2/(4π)
total area = (1/16)x^2 + (1/(4π))(100-x)^2
domain: clearly 0 < x < 100
Since you labeled your problem as "Pre-Calc" I can't take the derivative and set it equal to zero, like I want to.
So, expanding and simplifying the area equation I got
Area = (1/16 + 1/(4π))x^2 - (50/π)x + 2500/π
which is a parabola opening upwards, the vertex would represent a minimum area
the x of the vertex is -b/(2a)
= (50/π) / (2(1/16 + 1/(4π))) = appr 56 cm
so a min total will be obtained when x = 56 cm
clearly the maximum will be obtained if x = 100, that is all is used for a circle.
Here is how I would do this:
let the radius be x, and the side of the square be y
4y + 2πx = 100
4y = 100 - 2πr
y = 25 - πx/2
area = πx^2 + (25-πx/2)^2
=πx^2 + 625 - 25πx + π^2 x^2/4
= (π + π^2/4)x^2 - 25πx + 625
the x coordinate of this parabola is 25π/(2π + π^2/2)
= 25/(2 + π/2) = appr 7.00
so the length for the circle = 2πx = 44
and the circle uses 100-44 or 56 cm, same as above.
a. To find the function that represents the combined area of the square and circle, we need to consider the formulas for the area of each shape.
1. Square:
The length of each side of the square will be x/4 since we are using x units of the wire for it. Therefore, the formula for the area of the square is A_square = (x/4)^2.
2. Circle:
The remaining wire, after using x units for the square, will be (100 - x) units. This wire will be used to form the circumference of the circle. The formula for the circumference of a circle is C_circle = 2πr, where r is the radius. Since the circumference is equal to the remaining wire, we have 2πr = (100 - x). Solving for r, we get r = (100 - x) / (2π). The formula for the area of the circle is A_circle = πr^2.
Therefore, the combined area function is given by f(x) = A_square + A_circle = (x/4)^2 + π((100 - x) / (2π))^2.
b. The domain of the function f(x) is determined by the constraints of the problem. The wire cannot have a negative length or a length greater than 100 cm. Mathematically, this means 0 ≤ x ≤ 100.
c. To find the x values that yield a maximum and minimum area, we can apply calculus. We need to find the critical points of the function f(x), which are the points where its derivative is either zero or undefined.
Taking the derivative of f(x) with respect to x, we get:
f'(x) = 2(x/4)*(1/4) - 2π(100 - x) / (2π)^2
To find the critical points, we set f'(x) = 0 and solve for x:
2(x/4)*(1/4) - 2π(100 - x) / (2π)^2 = 0
x/8 - (100 - x) / (2π^2) = 0
Simplifying this equation, we have:
x/8 - (100 - x) / (2π^2) = 0
x/8 - (200π^2 - 2π^2x) / (2π^2) = 0
x - (200π^2 - 2π^2x) / π^2 = 0
π^2x + 8x - 200π^2 = 0
(π^2 + 8)x - 200π^2 = 0
x = 200π^2 / (π^2 + 8)
We need to check if this critical point falls within the domain 0 ≤ x ≤ 100. If it does, then it represents a potential minimum or maximum area.
d. In this case, the critical point (x = 200π^2 / (π^2 + 8)) falls within the domain 0 ≤ x ≤ 100, so it represents a potential minimum or maximum area. To determine whether it corresponds to a minimum or maximum, we need to analyze the behavior of the function f(x) around that point. We can do this by considering the second derivative of f(x).
Taking the second derivative of f(x), we have:
f''(x) = (2/16) - (2π^2 / (2π^2))^2
f''(x) = (1/8) - 1
f''(x) = -7/8
Since the second derivative is negative, it signifies that the point x = 200π^2 / (π^2 + 8) corresponds to a maximum area.
Therefore, x = 200π^2 / (π^2 + 8) represents the value that yields the maximum area.