Question

A piece of wire 8 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.  8

b. 7

Step-by-step explanation:

Let $x$ be the length of wire for the square and $y = 8 -x$ (eq. 1) be the length of the wire for the equilateral triangle.

We know the $x$ is the total length of the wire used, hence the length of a side of the square will be $\frac{x}{4}$ . The area of the square will be: $(\frac{x}{4}) ^ 2$

Similarly for the equilateral triangle, perimeter is $y$. Hence the area will be $\frac {\sqrt {3}\ y^2}{4}$.

The total area of both shapes will be:  $A = (\frac{x}{4})^2 + \frac{\sqrt {3} \ y^2}{4}$

We will substitute the value of y from eq. 1:

$A = (\frac{x}{4})^2 + \frac{\sqrt {3} \ (8-x)^2}{4}$

We find the derivative of the above function to find maximum and minimum: $f'(x) = 0$ ⇒ $A'(x) = 0$

$A' = \frac{x}{8} - \frac{\sqrt {3} \ (8-x)}{2}$

$A' = \frac{x}{8} - \frac{\sqrt {3} \ (8-x)}{2} = 0\\\\\frac{x}{8} - \frac{8\sqrt{3} -\sqrt{3} x}{2} =0\\\\x - 4(8\sqrt{3} -\sqrt{3}x)=0\\\\x-32\sqrt3 - 4\sqrt3 x = 0\\\\x - 4\sqrt3 = 32 \sqrt3 \\\\(1 - 4\sqrt3)x = 32 \sqrt3 \\\\x = \frac{32 \sqrt3 }{(1 - 4\sqrt3)}\\\\x =6.99 \approx 7.00$

We find $A''(7)$ to check whether $x = 7$ is the minimum or maximum of the function.

$A'' = \frac {1}{8} +\frac {\sqrt {3}}{2} \\\\A'' = 0.99 \approx 1.00$

Hence, $x = 7$ is the minimum and $x = 8$ will be the maximum