Question

A piece of wire 23 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (Give your answers correct to two decimal places.)(a) How much wire should be used for the square in order to maximize the total area?m(b) How much wire should be used for the square in order to minimize the total area?m

Answer 1

Answer:

A) All the wire of 23 m should be used in order to maximize the total area.

B) The minimum area is obtained when 10m of wire is used for the square.

Step-by-step explanation:

Let's say the perimeter one side of the square is x metres.

Then, the length of the side of the square will be; = x/4 metres and consequently, its area (A1) = x/4 • x/4 = x²/16 m²

Now, since x is used for the square, the remaining will be 23 - x and thus, if one side of the equilateral triangle is given as "a", hence;

a = (23 − x)/3 m

It's area will be; A2 = (√3)a²/4

=((√3)/36) • (23 − x)² m²

Hence the total area is;

A = A1 + A2

So, A = x²/16 + ((√3)/36) • (23 − x)²)

Now let's find the first and second derivatives of this total area;

dA/dx = x/8 − ((√3)/18)(23 - x)

and;

d²A/dx² = 1/8 + ((√3)/18) = 0.22

Thus, d²A/dx² > 0

Therefore, the extremum is at

dA/dx = 0

So,

x/8 − ((√3)/18)(23 - x) = 0

x/8 = ((√3)/18)(23 - x)

x = ((8√3)/18)(23 - x)

x = 0.7698(23-x)

x = 17.7054 - 0.7698x

x + 0.7698x = 17.7054

1.7698x = 17.7054

x = 17.7054/1.7698 ≈ 10m

Since d²A/dx² > 0, this is a minimum.

Thus, the minimum area is obtained when 10m of wire is used for the square.

B) For a given perimeter, a square normally has more area than a triangle. Therefore, to get the largest or maximum area we need to use all of the wire to construct the square. Which is 23m

Answer 2

Answer:

For maximum area, all of the wire should be used to construct the square.

The minimum total area is obtained when length of the wire is 10m

Step-by-step explanation:

For maximum,  we use the whole length

For minimum,

supposed the x length was used for the square,

the length of the side of the square = x/4m

Area = $\frac{x^{2} }{16}$

For the equilateral triangle, the length of the side =  $\frac{23 - x}{3}$

Area = $\frac{\sqrt{3} }{4} a^{2} = \frac{\sqrt{3} }{4} (\frac{23 - x}{3} )^{2}$

Total Area = $\frac{x^{2} }{16}$  + $\frac{\sqrt{3} }{36} (23-x)^{2}$

$\frac{dA}{dx} = \frac{x}{8} - \frac{\sqrt{3} }{18} (23 - x)$

$\frac{d^{2}A }{dx^{2} } = \frac{1}{8} + \frac{\sqrt{3} }{18} > 0$, therefore it is minimum

$\frac{dA}{dx} = 0$

$\frac{x}{8} - \frac{\sqrt{3} }{18} (23 - x) = 0$

x = 10.00m