Question

You have a wire that is 38 cm long. you wish to cut it into two pieces. one piece will be bent into the shape of a square. the other piece will be bent into the shape of a circle. let a represent the total area of the square and the circle. what is the circumference of the circle when a is a minimum?

Answer 1

circumference + perimeter =38
x = circumference of circle
find r in terms of x
2$\pi$r=x
r=$\frac{x}{2 \pi }$
Area=$\pi$r^2
so..
A=$\pi$($\frac{x}{2 \pi }$)^2
=$\pi$($\frac{x^{2} }{4 \pi ^{2} }$)
=$\frac{x^{2} }{4 \pi }$

(38-x) is perimeter
then 
$\frac{38-x}{4}$
($\frac{38-x}{4}$)^2
$\frac{x^{2} }{4 \pi }$ + ($\frac{38-x}{4}$)^2
then you graph it and it equals
16.716 cm the circumference of the circle

Answer 2

The minimum point of a function gives the values of the input that gives the minimum output

• The circumference of the circle when the area is minimum is approximately 16.716 cm

Reason:

Given length of the wire = 38 cm

The number of pieces the wire will be cut = 2 pieces

Shape in which one piece will be bent = A square

Shape in which the other shape will be cut = A circle

The total area of the of the square and the circle = a

The circumference of the circle when a is a minimum = Required

Solution:

Let x, and y, represent the two piece of the wire, we have;

Where;

x = The piece for the circle

y = The piece for the square

x + y = 38 cm

• The area of the square is $\dfrac{(38 - x)^2}{16}$

Circumference of a circle = 2·π·r

The circumference of the circle = x

∴ x = 2·π·r

$r = \dfrac{x}{2 \cdot \pi}$

Area of a circle = π·r²

The area of the circle, A = $\pi \times \left (\dfrac{x}{2 \cdot \pi}\right)^2$

The sum of the areas is therefore; a =  $\dfrac{(38 - x)^2}{16} + \left (\dfrac{x^2}{4 \cdot \pi}\right)$

$a = \dfrac{(38 - x)^2}{16} + \left (\dfrac{x^2}{4 \cdot \pi}\right) = \dfrac{(x^2-76\cdot x + 1444) \cdot \pi + 4 \cdot x^2}{16 \cdot \pi}$

We get;

• a = 0.142077471535·x²-4.75·x+90.25

The value of coefficient of x² is positive, therefore, the function of the area has a minimum point

The x-coordinates of the value of the minimum point of a quadratic function is $x = \dfrac{b}{2 \cdot a}$

The x-coordinates of the value of the minimum point is therefore;

b = -4.75

a = 0.14207747151535

$x = \dfrac{b}{2 \cdot a}$

• $x = -\dfrac{-4.75}{2 \times 0.14207747151535} \approx 16.716$

• The length of the wire, x, which is the circumference of the circle, when the area is minimum, is approximately 16.716 cm

Therefore;

The length of the wire, y ≈ 38 - 16.716 = 21.284

The length of the wire, y ≈ 21.284 cm

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