A piece of wire 16cm long is cut into 2 lengths, one of which is bent into a circle, the other into a rectangle with one side three times the other.
a) Determine the ratio of the longer length of the rectangle to the radius of the circle if the sum of the areas of the circle and the square is minimum and find this minimum area.
b) If you add 6pi/(3pi+16) to the value that makes the area minimum, how much area will increase?
width of rectangle ---- x
length of rectangle --- 3x
radius of circle ----- r
so 2x + 6x + 2πr = 16
4x + πr = 8 ----> x = 2 - πr/4
let S be the sum of the areas of the rectangle , (you had "square" , I assume that was an error)
S = πr^2 + 3x^2
= πr^2 + 3(1 - πr/4)^2
= πr^2 + 3 - 3πr/2 + π^2 r^2 / 16
dS/dr = 2πr - 3π/2 + (1/8)π^2 r
= 0 for a max/min
2πr - 3π/2 + (1/8)π^2 r
16πr - 12π + π^2 r = 0
divide by π
16r - 12 + πr = 0
r(16+π) = 12
r = 12/(16+π)
then x = 2 - πr/4 = .....
ratio of longer side : radius
= 3x / r
= 3x(16+π)/12
= x(16+π) : 4
check my arithmetic
b)
"add ... to the value that makes the area minimum" ???
do we add it to the x or to the r ??
You decide, find the new dimensions, then calculate the current and the new area.
That would be just arithmetic.
Take a look at the many question of the same type below in Related Questions.
a) Let's represent the lengths of the sides of the rectangle as x and 3x (since one side is three times the other). The length of wire used to form the rectangle is given by the perimeter of the rectangle, which is 2x + 2(3x) = 16 cm.
Simplifying the equation: 2x + 6x = 16 cm. Combining like terms: 8x = 16 cm. Dividing both sides by 8, we get x = 2 cm.
So, the lengths of the sides of the rectangle are 2 cm and 6 cm. The perimeter of the rectangle is 2(2) + 2(6) = 16 cm, which matches the given wire length.
The area of the rectangle is given by A = length × width, so the area is 2 cm × 6 cm = 12 cm^2.
The length of wire used to form the circle is equal to the circumference of the circle, which is given by 2πr, where r is the radius of the circle.
Let the radius of the circle be r cm. Therefore, the remaining wire length (after forming the rectangle) is 16 cm - 2(2) - 2(6) = 16 cm - 4 cm - 12 cm = 0 cm.
Since there is no remaining wire length, the circumference of the circle must be equal to 16 cm. Therefore, the circumference formula can be written as 2πr = 16. Solving for r, we have r = 8/π cm.
The area of the circle is given by A = πr^2. Substituting the value of r, we get A = π(8/π)^2 = 64/π cm^2.
The sum of the areas of the circle and the rectangle is: 12 cm^2 + 64/π cm^2.
To find the minimum sum of these areas, we need to differentiate this expression with respect to x and set it equal to 0.
d(Area sum)/dx = d(12 + 64/π)/dx = 0.
Since the expression does not depend on x, the derivative is 0.
Therefore, this function reaches its minimum value when x = 2 cm.
Substituting x = 2 cm into the expression for the sum of the areas, we get the minimum area as follows:
12 cm^2 + 64/π cm^2 = 12 cm^2 + 64/π cm^2 = 12 cm^2 + (64/3.14) cm^2 ≈ 12 cm^2 + 20.382 cm^2 ≈ 32.382 cm^2.
So, the minimum area of the sum of the circle and the rectangle is approximately 32.382 cm^2.
b) The value that makes the area minimum is x = 2 cm. Adding 6π/(3π + 16) to this value, we get:
x + 6π/(3π + 16) = 2 cm + 6π/(3π + 16) cm.
The area of the rectangle is given by A = length × width, so the original area is:
A = (2 cm)(6 cm) = 12 cm^2.
After adding 6π/(3π + 16) to x, the new value of x is 2 cm + 6π/(3π + 16) cm. The new area is:
A_new = (2 cm + 6π/(3π + 16) cm)(6 cm) = (12 cm^2) + 36π/(3π + 16) cm^2.
To find the increase in area, we subtract the original area from the new area:
Increase in area = A_new - A = [(12 cm^2) + 36π/(3π + 16) cm^2] - 12 cm^2.
Simplifying, we get:
Increase in area = (36π/(3π + 16)) cm^2.
Therefore, the increase in area is 36π/(3π + 16) cm^2.
To solve this problem, let's break it down into smaller steps:
Step 1: Express the lengths of the wire in terms of variables.
Let's assume the longer length of the rectangle is x and the shorter length is y. Since one side of the rectangle is three times the other, we have x = 3y. Also, the wire length is given as 16 cm, so we have the equation: x + y + 2πr = 16, where r is the radius of the circle.
Step 2: Express the area equations for the circle and rectangle.
The area of a circle is given by A = πr^2, and the area of the rectangle is given by A = xy.
Step 3: Express the sum of the areas as a function of a single variable.
Since we need to find the minimum sum of areas, we can express it as a function of just one variable. We can use the equation from Step 1 to substitute for either x or y. Let's use x = 3y, so the sum of the areas becomes: A = 3y * y + πr^2.
Step 4: Solve for the variable in terms of the other variable.
To minimize the sum of the areas, we need to find the critical points of the function A. Since A depends on both y and r, we need to find y in terms of r, or r in terms of y. We can solve the equation x + y + 2πr = 16 for r, by substituting x = 3y from Step 1. This gives us: 3y + y + 2πr = 16, or 4y + 2πr = 16. Rearranging this equation, we get: r = (16 - 4y) / (2π).
Step 5: Substitute the variable expression into the equation for the sum of the areas.
Substituting the expression for r into the equation A = 3y * y + πr^2, we get: A = 3y^2 + π * [(16 - 4y) / (2π)]^2. Simplifying this equation gives us the final equation for the sum of the areas.
Step 6: Find the minimum value of the sum of areas.
To find the minimum value of the sum of the areas, we can take the derivative of the equation from Step 5 with respect to y, set it equal to zero, and solve for y. Once we find the value of y that minimizes the sum of areas, we can substitute it back into the equation to find the corresponding value of r.
Step 7: Calculate the ratio of the longer length of the rectangle to the radius of the circle and find the minimum area.
Once we have the values of y and r that minimize the sum of areas, we can calculate the ratio x/r and the value of the sum of areas A.
For part b), we can calculate the new value of the sum of areas by adding 6π/(3π+16) to the value that makes the area minimum. Then we can calculate the difference between the new and old values of the sum of areas.