# A piece of wire 60 cm in length is cut into two pieces. The first piece forms a rectangle 5 times as wide as it is long. The second piece forms a square. Where should the wire be cut to? 1)minimize the total area 2)maximize the total area

Dec 5, 2017

Max Area: ${\text{225 cm}}^{3}$
Min Area: ${\text{80.36 cm}}^{3}$

#### Explanation:

Steps:

1) Draw out a rectangle and a square

2) Label all your variables based on the information from the question

• Since the length of the rectangle is five times as the width. Therefore, we know that $5 x$ is the length and $x$ is the width.
• Since it is a square and we do not know the values of the sides, we can label the sides as $y$.

3) Find the helper equation for the equation which you are trying to optimize

• The wire $\text{60 cm}$ is the circumference for both the rectangle and the square
• $12 x + 4 y = 60$

4) Find the equation for $y$ of the helper equation in terms of $x$

• $y = 15 - 3 x$

5) Find the equation that you are trying to optimize

• $\text{Area} = A R + A S = 5 {x}^{2} + {y}^{2}$

6) Plug the equation of $y$ into the equation you are trying to optimize

• $5 {x}^{2} + {y}^{2}$
• $5 {x}^{2} + {\left(15 - 3 x\right)}^{2} = 14 {x}^{2} - 90 x + 225$

7) Take the derivative of the equation and find the critical point(s) while determine whether they are max or min

$A ' = 28 x - 90$

8) Check the endpoints!

Endpoints indicate the maximum and minimum value you can have for the rectangle

• $12 x = 60 , x = 5$ <-- indicating all material used to create the rectangle
• $x = 0$ <-- indicating all material used to create the square

9) Find the answers by plugging in endpoints and critical number into the original area equation

• $A \left(3.2143\right) = {\text{80.36 cm}}^{3}$
• $A \left(0\right) = {\text{225 cm}}^{3}$
• $A \left(5\right) = {\text{125 cm}}^{3}$

Therefore, the maximum of the total area we have is ${\text{225 cm}}^{3}$ and the minimum value we have is ${\text{80.36 cm}}^{3}$.

Dec 5, 2017

$38.4 \mathmr{and} 21.6 c m$ for a minimum area of $80.4 c {m}^{2}$
The limit as $x \to 0$ is $225 c {m}^{2}$

#### Explanation:

The total length is 60. The first rectangle has dimensions of $x \times 5 x$, and the perimeter is thus $2 \left(x + 5 x\right) = 12 x$. The remaining length for the square is then 60 – 12x. This is equal to the square perimeter, $4 y$, so we can express the square side ‘y’ in terms of ‘x’ as 4y = 60 – 12x ; y = 15 – 3x. The square area in terms of ‘x’ is thus (15 – 3x)^2 = 9x^2 – 90x +225.

Combined with the area of the rectangle we obtain the total area as:
A_t = 5x^2 + 9x^2 – 90x +225 = 14x^2 – 90x + 225 This quadratic indicates that it will only have one extrema – the minimum for a given length of wire. That point is where the first derivative of the area quadratic equation is zero. f’(x) = 28x – 90 Minimum at $x = 3.2$

The derived dimensions follow:
Rectangle short side: $x = 3.2 c m$
Rectangle long side: $5 x = 16 c m$
Rectangle area: $51.2 c {m}^{2}$
Square side: y = 15 – 3x = 5.4
Square area: $29.2 c {m}^{2}$
Total Area: $80.4 c {m}^{2}$
Total lengths: $2 \left(3.2 + 16\right) + 4 \left(5.4\right) = 38.4 + 21.6 = 60$ Correct. Cut the wire at a length of 60 – 12x = 60 – 38.4 = 21.6cm or 38.4cm